analytical geometry: 2d and 3d

ANALYTICALGEOMETRY

2Dand3D

P.R.VittalVisitingProfessor

DepartmentofStatisticsUniversityofMadras

Chennai

Chennai•Delhi

BriefContents

AbouttheAuthor

Preface

1CoordinateGeometry

2TheStraightLine

3PairofStraightLines

4Circle

5SystemofCircles

6Parabola

7Ellipse

8Hyperbola

9PolarCoordinates

10TracingofCurves

11ThreeDimension

12Plane

13StraightLine

14Sphere

15Cone

16Cylinder

Contents

AbouttheAuthor

Preface

1CoordinateGeometry

1.1Introduction

1.2SectionFormula

IllustrativeExamples

Exercises

2TheStraightLine

2.1Introduction

2.2SlopeofaStraightLine

2.3Slope-interceptFormofaStraightLine

2.4InterceptForm

2.5Slope-pointForm

2.6TwoPointsForm

2.7NormalForm

2.8ParametricFormandDistanceForm

2.9PerpendicularDistanceonaStraightLine

2.10IntersectionofTwoStraightLines

2.11ConcurrentStraightLines

2.12AnglebetweenTwoStraightLines

2.13EquationsofBisectorsoftheAnglebetweenTwoLines


Exercises

3PairofStraightLines

3.1Introduction

3.2HomogeneousEquationofSecondDegreeinxandy

3.3AnglebetweentheLinesRepresentedbyax2+2hxy+by2=0

3.4EquationfortheBisectoroftheAnglesbetweentheLinesGivenbyax2+2hxy+by2=0

3.5ConditionforGeneralEquationofaSecondDegreeEquationtoRepresentaPairofStraightLines


Exercises

4Circle

4.1Introduction

4.2EquationofaCirclewhoseCentreis(h,k)andRadiusr

4.3CentreandRadiusofaCircleRepresentedbytheEquationx2+y2+2gx+2fy+c=0

4.4LengthofTangentfromPointP(x1,y1)totheCirclex2+y2+2gx+2fy+c=0

4.5EquationofTangentat(x1,y1)totheCirclex2+y2+2gx+2fy+c=0

4.6EquationofCirclewiththeLineJoiningPointsA(x1,y1)andB(x2,y2)astheendsofDiameter

4.7ConditionfortheStraightLiney=mx+ctobeaTangenttotheCirclex2+y2=a2

4.8EquationoftheChordofContactofTangentsfrom(x1,y1)totheCirclex2+y2+2gx+2fy+c

=0

4.9TwoTangentscanAlwaysbeDrawnfromaGivenPointtoaCircleandtheLocusofthePointofIntersectionofPerpendicularTangentsisaCircle

4.10PoleandPolar

4.11ConjugateLines

4.12EquationofaChordofCirclex2+y2+2gx+2fy+c=0inTermsofitsMiddlePoint

4.13CombinedEquationofaPairofTangentsfrom(x1,y1)totheCirclex2+y2+2gx+2fy+c=

0

4.14ParametricFormofaCircle


Exercises

5SystemofCircles

5.1RadicalAxisofTwoCircles

5.2OrthogonalCircles

5.3CoaxalSystem

5.4LimitingPoints

5.5Examples(RadicalAxis)

5.6Examples(LimitingPoints)

Exercises

6Parabola

6.1Introduction

6.2GeneralEquationofaConic

6.3EquationofaParabola

6.4LengthofLatusRectum

6.5DifferentFormsofParabola

IllustrativeExamplesBasedonFocusDirectrixProperty

6.6ConditionforTangency

6.7NumberofTangents

6.8PerpendicularTangents

6.9EquationofTangent

6.10EquationofNormal

6.11EquationofChordofContact

6.12PolarofaPoint

6.13ConjugateLines

6.14PairofTangents

6.15ChordIntermsofMid-point

6.16ParametricRepresentation

6.17ChordJoiningTwoPoints

6.18EquationsofTangentandNormal

6.19PointofIntersectionofTangents

6.20PointofIntersectionofNormals

6.21NumberofNormalsfromaPoint

6.22IntersectionofaParabolaandaCircle

IllustrativeExamplesBasedonTangentsandNormals

IllustrativeExamplesBasedonParameters

Exercises

7Ellipse

7.1StandardEquation

7.2StandardEquationofanEllipse

7.3FocalDistance

7.4PositionofaPoint

7.5AuxiliaryCircle

IllustrativeExamplesBasedonFocus-directrixProperty

7.6ConditionforTangency

7.7DirectorCircleofanEllipse

7.8EquationoftheTangent

7.9EquationofTangentandNormal

7.10EquationtotheChordofContact

7.11EquationofthePolar

7.12ConditionforConjugateLines

IllustrativeExamplesBasedonTangents,Normals,Pole-polarandChord

7.13EccentricAngle

7.14EquationoftheChordJoiningthePoints

7.15EquationofTangentat‘θ’ontheEllipse

7.16ConormalPoints

7.17ConcyclicPoints

7.18EquationofaChordinTermsofitsMiddlePoint

7.19CombinedEquationofPairofTangents

7.20ConjugateDiameters

7.21Equi-conjugateDiameters

IllustrativeExamplesBasedonConjugateDiameters

Exercises

8Hyperbola

8.1Definition

8.2StandardEquation

8.3ImportantPropertyofHyperbola

8.4EquationofHyperbolainParametricForm

8.5RectangularHyperbola

8.6ConjugateHyperbola

8.7Asymptotes

8.8ConjugateDiameters

8.9RectangularHyperbola

Exercises

9PolarCoordinates

9.1Introduction

9.2DefinitionofPolarCoordinates

9.3RelationbetweenCartesianCoordinatesandPolarCoordinates

9.4PolarEquationofaStraightLine

9.5PolarEquationofaStraightLineinNormalForm

9.6Circle

9.7PolarEquationofaConic

Exercises

10TracingofCurves

10.1GeneralEquationoftheSecondDegreeandTracingofaConic

10.2ShiftofOriginwithoutChangingtheDirectionofAxes

10.3RotationofAxeswithoutChangingtheOrigin

10.4RemovalofXY-term

10.5Invariants

10.6ConditionsfortheGeneralEquationoftheSecondDegreetoRepresentaConic

10.7CentreoftheConicGivenbytheGeneralEquationoftheSecondDegree

10.8EquationoftheConicReferredtotheCentreasOrigin

10.9LengthandPositionoftheAxesoftheCentralConicwhoseEquationisax2+2hxy+by2=1

10.10AxisandVertexoftheParabolawhoseEquationisax2+2hxy+by2+2gx+2fy+c=0

Exercises

11ThreeDimension

11.1RectangularCoordinateAxes

11.2FormulaforDistancebetweenTwoPoints

11.3CentroidofTriangle

11.4CentroidofTetrahedron

11.5DirectionCosines


Exercises

12Plane

12.1Introduction

12.2GeneralEquationofaPlane

12.3GeneralEquationofaPlanePassingThroughaGivenPoint

12.4EquationofaPlaneinInterceptForm

12.5EquationofaPlaneinNormalForm

12.6AnglebetweenTwoPlanes

12.7PerpendicularDistancefromaPointonaPlane

12.8PlanePassingThroughThreeGivenPoints

12.9ToFindtheRatioinwhichthePlaneJoiningthePoints(x1,y1,z1)and(x2,y2,z2)isDividedbythePlaneax+by+cz+d=0.

12.10PlanePassingThroughtheIntersectionofTwoGivenPlanes

12.11EquationofthePlaneswhichBisecttheAnglebetweenTwoGivenPlanes

12.12ConditionfortheHomogenousEquationoftheSecondDegreetoRepresentaPairofPlanes


Exercises

13StraightLine

13.1Introduction

13.2EquationofaStraightLineinSymmetricalForm

13.3EquationsofaStraightLinePassingThroughtheTwoGivenPoints

13.4EquationsofaStraightLineDeterminedbyaPairofPlanesinSymmetricalForm

13.5AnglebetweenaPlaneandaLine

13.6ConditionforaLinetobeParalleltoaPlane

13.7ConditionsforaLinetoLieonaPlane

13.8ToFindtheLengthofthePerpendicularfromaGivenPointonaLine

13.9CoplanarLines

13.10SkewLines

13.11EquationsofTwoNon-intersectingLines

13.12IntersectionofThreePlanes

13.13ConditionsforThreeGivenPlanestoFormaTriangularPrism


IllustrativeExamples(CoplanarLinesandShortestDistance)

Exercises

14Sphere

14.1DefinitionofSphere

14.2Theequationofaspherewithcentreat(a,b,c)andradiusr

14.3EquationoftheSphereontheLineJoiningthePoints(x1,y1,z1)and(x2,y2,z2)asDiameter

14.4LengthoftheTangentfromP(x1,y1,z1)totheSpherex2+y2+z2+2ux+2vy+2wz+d=0

14.5EquationoftheTangentPlaneat(x1,y1,z1)totheSpherex2+y2+z2+2ux+2vy+2wz+d

=0

14.6SectionofaSpherebyaPlane

14.7EquationofaCircle

14.8IntersectionofTwoSpheres

14.9EquationofaSpherePassingThroughaGivenCircle

14.10ConditionforOrthogonalityofTwoSpheres

14.11RadicalPlane

14.12CoaxalSystem


Exercises

15Cone

15.1DefinitionofCone

15.2EquationofaConewithaGivenVertexandaGivenGuidingCurve

15.3EquationofaConewithitsVertexattheOrigin

15.4ConditionfortheGeneralEquationoftheSecondDegreetoRepresentaCone

15.5RightCircularCone

15.6TangentPlane

15.7ReciprocalCone

Exercises

16Cylinder

16.1Definition

16.2EquationofaCylinderwithaGivenGeneratorandaGivenGuidingCurve

16.3EnvelopingCylinder

16.4RightCircularCylinder


Exercises

AbouttheAuthor

P.R.VittalwasapostgraduateprofessorofMathematicsatRamakrishnaMissionVivekanandaCollege,Chennai,fromwhereheretiredasPrincipalin1996.HewasavisitingprofessoratWesternCarolinaUniversity,USA,andhasvisitedanumberofuniversitiesintheUSAandCanadainconnectionwithhisresearchwork.Heis,atpresent,avisitingprofessorattheDepartmentofStatistics,UniversityofMadras;InstituteofCharteredAccountantsofIndia,Chennai;TheInstituteofTechnologyandManagement,Chennai;andNationalManagementSchool,Chennai,besidesbeingaresearchguideinManagementScienceatBITS,Ranchi.ProfessorVittalhaspublished30researchpapersinjournalsofnationaland

internationalreputeandguidedanumberofstudentstotheirM.Phil.andPh.D.degrees.AfellowofTamilNaduAcademyofSciences,hisresearchtopicsareprobability,stochasticprocesses,operationsresearch,differentialequationsandsupplychainmanagement.Hehasauthoredabout30booksinmathematics,statisticsandoperationsresearch.

TomygrandchildrenAaravandAdvay

Preface

Asuccessfulcourseinanalyticalgeometrymustprovideafoundationforfutureworkinmathematics.Ourteachingresponsibilitiesaretoinstilcertaintechnicalcompetenceinourstudentsinthisdisciplineofmathematics.Agoodtextbook,aswithagoodteacher,shouldaccomplishtheseaims.Inthisbook,youwillfindacrisp,mathematicallyprecisepresentationthatwillallowyoutoeasilyunderstandandgraspthecontents.Thisbookcontainsbothtwo-dimensionalandthree-dimensionalanalytical

geometry.Insomeofthefundamentalresults,vectortreatmentisalsogivenandtherefromthescalarformoftheresultshasbeendeduced.Thefirst10chaptersdealwithtwo-dimensionalanalyticalgeometry.In

Chapter1,allbasicresultsareintroduced.Theconceptoflocusiswellexplained.Usingthisidea,inChapter2,differentformsfortheequationofastraightlineareobtained;allthecharacteristicsofastraightlinearealsodiscussed.Chapter3dealswiththeequationofapairofstraightlinesanditsproperties.InChapters4and5,circleandsystemofcircles,includingcoaxialsystemandlimitingpointsofacoaxialsystem,areanalysed.Chapters6,7and8dealwiththeconicsections—parabola,ellipseand

hyperbola.Apartfromtheirpropertiessuchasfocusanddirectrix,theirparametricequationsarealsoexplained.Specialpropertiessuchasconormalpointsofallconicsaredescribedindetails.Conjugatediametersinellipseandhyperbolaandasymptotesofahyperbolaandrectangularhyperbolaarealsoanalysedwithanumberofexamples.Ageneraltreatmentofconicsandtracingofconicsisalsoprovided.InChapter9,wedescribepolarcoordinates,whichareusedtomeasure

distancesforsomespecialpurposes.Chapter10examinestheconditionsforthegeneralequationoftheseconddegreetorepresentthedifferenttypesofconics.

InChapters11to16,westudythethree-dimensionalanalyticalgeometry.Thebasicconcepts,suchasdirectionalcosines,areintroducedinChapter11.InChapter12,allformsofplaneareanalysedwiththehelpofexamples.Chapter13introducesastraightlineasanintersectionofapairofplanes.Differentformsofastraightlinearestudied;especially,coplanarlinesandtheshortestdistancebetweentwoskewlines.Chapter14dealswithspheresandsystemofspheres.InChapters15and16,twospecialtypesofconicoids—coneandcylinder—arediscussed.Anumberofillustrativeexamplesandexercisesforpracticearegiveninall

these16chapters,tohelpthestudentsunderstandtheconceptsinabettermanner.Ihopethatthisbookwillbeveryusefulforundergraduatestudentsand

engineeringstudentswhoneedtostudyanalyticalgeometryaspartoftheircurriculum.

Chapter1

CoordinateGeometry

1.1INTRODUCTION

LetXOX′andYOY′betwofixedperpendicularlinesintheplaneofthepaper.ThelineOXiscalledtheaxisofXandOYtheaxisofY.OXandOYtogetherarecalledthecoordinateaxes.ThepointOiscalledtheoriginofthecoordinateaxes.LetPbeapointinthisplane.DrawPMperpendiculartoXOX′.ThedistanceOMiscalledthex-coordinateorabscissaandthedistanceMPiscalledthey-coordinateorordinateofthepointP.

IfOM=xandMP=ythen(x,y)arecalledthecoordinatesofthepointP.ThecoordinatesoftheoriginOare(0,0).ThelinesXOX′andYOY′dividetheplaneintofourquadrants.TheyareXOY,YOX′,X′OY′andY′OX′.ThelengthsmeasuredinthedirectionsOXandOYareconsideredpositiveandthelengthsmeasuredinthedirectionsOX′andOY′areconsiderednegative.Thenatureofthecoordinatesinthedifferentquadrantsisasfollows:

Quadrant x-coordinate y-coordinate

First+ +

Second− +

− +

Third− −

Fourth+ −

ThemethodofrepresentingapointbymeansofcoordinateswasfirstintroducedbyRenaDescartesandhencethisbranchofmathematicsiscalledtherectangularCartesiancoordinatesystem.Usingthiscoordinatesystem,onecaneasilyfindthedistancebetweentwo

pointsinaplane,thecoordinatesofthepointthatdividesalinesegmentinagivenratio,thecentroidofatriangle,theareaofatriangleandthelocusofapointthatmovesaccordingtoagivengeometricallaw.

1.1.1DistancebetweenTwoGivenPoints

LetPandQbetwopointswithcoordinates(x1,y1)and(x2,y2).

DrawPLandQMperpendicularstothex-axis,anddrawQNperpendiculartoPL.Then,

Note1.1.1:ThedistanceofPfromtheoriginOis

Example1.1.1

IfPisthepoint(4,7)andQis(2,3),then

Example1.1.2

ThedistancebetweenthepointsP(2,−5)andQ(−4,7)is

1.2SECTIONFORMULA

1.2.1CoordinatesofthePointthatDividestheLineJoiningTwoGivenPointsinaGivenRatio

LetthetwogivenpointsbeP(x1,y1)andQ(x2,y2).

LetthepointRdividePQinternallyintheratiol:m.DrawPL,QMandRNperpendicularstothex-axis.DrawPSperpendiculartoRNandRTperpendiculartoMQ.LetthecoordinatesofRbe(x,y).RdividesPQinternallyintheratiol:m.Then,

TrianglesPSRandRTQaresimilar.

Also

Hence,thecoordinatesofRare

1.2.2ExternalPointofDivision

IfthepointR′dividesPQexternallyintheratiol:m,then

Choosingmnegative,wegetthecoordinatesofR′.Therefore,thecoordinatesof

R′are

Note1.2.2.1:Ifwetakel=m=1intheinternalpointofdivision,wegetthecoordinatesofthemidpoint.Therefore,thecoordinatesofthemidpointofPQ

are

1.2.3CentroidofaTriangleGivenitsVertices

LetABCbeatrianglewithverticesA(x1,y1),B(x2,y2)andC(x3,y3).

LetAA′,BB′andCC′bethemediansofthetriangle.ThenA′,B′,C′arethemidpointsofthesidesBC,CAandAB,respectively.ThecoordinatesofA′are

Weknowthatthemediansofatriangleareconcurrentatthe

pointGcalledthecentroidandGdivideseachmedianintheratio2:1.ConsideringthemedianAA′,thecoordinatesofGare

1.2.4AreaofTriangleABCwithVerticesA(x1,y1),B(x2,y2)andC(x3,y3)

LettheverticesoftriangleABCbeA(x1,y1),B(x2,y2)andC(x3,y3).

DrawAL,BMandCNperpendicularstoOX.Then,areaΔoftriangleABCiscalculatedas

Note1.2.4.1:Theareaispositiveornegativedependingupontheorderinwhichwetakethepoints.Sincescalarareaisalwaystakentobeapositivequantity,wetake

Note1.2.4.2:Iftheverticesofthetriangleare(0,0),(x1,y1)and(x2,y2),then

Note1.2.4.3:Iftheareaofthetriangleiszero,i.e.Δ=0,thenwenotethatthepointsarecollinear.Hence,theconditionforthepoints(x1,y1),(x2,y2)and(x3,y3)tobecollinearis

x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=0

1.2.5AreaoftheQuadrilateralGivenitsVertices

LetABCDbethequadrilateralwithverticesA(x1,y1),B(x2,y2),C(x3,y3)andD(x4,y4).DrawAP,BQ,CRandDSperpendicularstothex-axis.Then,

Note1.2.5.1:Thisresultcanbeextendedtoapolygonofnsideswithvertices(x1,y1),(x2,y2)……(xn,yn)as

Locus

Whenapointmovessoastosatisfysomegeometricalconditionorconditions,thepathtracedoutbythepointiscalledthelocusofthepoint.

thepathtracedoutbythepointiscalledthelocusofthepoint.Forexample,ifapointmoveskeepingaconstantdistancefromafixedpoint,

thelocusofthemovingpointiscalledcircleandthefixeddistanceiscalledtheradiusofthecircle.Moreover,ifapointmovessuchthatitsdistancefromtwofixedpointsareequal,thenthelocusofthepointistheperpendicularbisectorofthelinejoiningthetwofixedpoints.IfAandBaretwofixedpointsandpointP

movessuchthat thenthelocusofPisacirclewithABasthe

diameter.Itispossibletorepresentthelocusofapointbymeansofanequation.SupposeapointP(x,y)movessuchthatitsdistancefromtwofixedpoints

A(2,3)andB(5,−3)areequal.ThenthegeometricallawisPA=PB⇒PA2=PB2

Here,thelocusofPisastraightline.

ILLUSTRATIVEEXAMPLES

Example1.1

Findthedistancebetweenthepoints(4,7)and(−2,5).

Solution

LetPandQbethepoints(4,7)and(−2,5),respectively.

Example1.2

Provethatthepoints(4,3),(7,−1)and(9,3)aretheverticesofanisoscelestriangle.

Solution

LetA(4,3),B(7,−1),C(9,3)bethethreegivenpoints.Then

Sincethesumoftwosidesisgreaterthanthethird,thepointsformatriangle.Moreover,AB=AC=5.Therefore,thetriangleisanisoscelestriangle.

Example1.3

Showthatthepoints(6,6),(2,3)and(4,7)aretheverticesofarightangledtriangle.

Solution

LetA,B,Cbethepoints(6,6),(2,3)and(4,7),respectively.

Hence,thepointsaretheverticesofarightangledtriangle.

Example1.4

Showthatthepoints(7,9),(3,−7)and(−3,3)aretheverticesofarightangledisoscelestriangle.

Solution

LetA,B,Cbethepoints(7,9),(3,−7),(−3,3),respectively.

Hence,thepointsareverticesofarightangledtriangle.Also,BC=AC.Therefore,itisarightangledisoscelestriangle.

Example1.5

Showthatthepoints(4,−4),(−4,4)and aretheverticesofan

equilateraltriangle.

Solution

LetA,B,Cbethepoints(4,−4),(−4,4)and ,respectively.

Hence,thepointsA,BandCaretheverticesofanequilateraltriangle.

Example1.6

Showthatthesetofpoints(−2,−1),(1,0),(4,3)and(1,2)aretheverticesofaparallelogram.

Solution

LetA,B,C,Dbethepoints(−2,−1),(1,0),(4,3)and(1,2),respectively.Aquadrilateralisaparallelogramiftheoppositesidesareequal.

SincetheoppositesidesofthequadrilateralABCDareequal,thefourpointsformaparallelogram.

Example1.7

Showthatthepoints(2,−2),(8,4),(5,7)and(−1,1)aretheverticesofarectangletakeninorder.

Solution

Aquadrilateralinwhichtheoppositesidesareequalandthediagonalsareequalisarectangle.LetA(2,−2),B(8,4),C(5,7)andD(−1,1)bethefourgivenpoints.

Thus,theoppositesidesareequalandthediagonalsarealsoequal.Hence,thefourpointsformarectangle.

Example1.8

Provethatthepoints(3,2),(5,4),(3,6)and(1,4)takeninorderformasquare.

Solution

Aquadrilateralinwhichallsidesareequalanddiagonalsareequalisasquare.LetA,B,C,Dbethepoints(3,2),(5,4),(3,6),(1,4),respectively.

Thus,allsidesareequalandalsothediagonalsareequal.Hence,thefourpointsformasquare.

Example1.9

FindthecoordinatesofthecircumcentreofatrianglewhoseverticesareA(3,−2),B(4,3)andC(−6,5).Also,findthecircumradius.

Solution

LetA(3,−2),B(4,3),andC(−6,5)bethegivenpoints.LetS(x,y)bethecircumcentreofΔABC.ThenSA=SB=SC=circumradius.Now

Hence,thecircumcentreis

Now

Therefore,circumradius units.

Example1.10

Showthatthepoints(3,7),(6,5)and(15,−1)lieonastraightline.

Solution

LetA(3,7),B(6,5)andC(15,−1)bethethreepoints.Then

Hence,thethreegivenpointslieonastraightline.

Example1.11

Showthat(4,3)isthecentreofthecirclethatpassesthroughthepoints(9,3),(7,−1)and(1,−1).Finditsradius.

Solution

LetA(9,3),B(7,−1),C(1,−1)andP(4,3)bethegivenpoints.Then

Hence,PisthecentreofthecirclepassingthroughthepointsA,B,C;itsradiusis5.

Example1.12

IfOistheoriginandthecoordinatesofAandBare(x1,y1)and(x2,y2),

respectively,provethatOA·OBcosθ=x1x2+y1y2where

Solution

Bycosineformula

Hence,OA·OBcosθ=x1x2+y1y2

Example1.13

Iftanα,tanβandtanγbetherootsoftheequationx3−3ax2+3bx−c=0andtheverticesofthetriangleABCare(tanα,cotα),(tanβ,cotβ)and(tanγ,cotγ)showthatthecentroidofthetriangleis(a,b).

Solution

Giventaα,tanβandtanγaretherootsoftheequation

Thendividing(1.5)by(1.6),

ThecentroidofΔABCis

(i.e.)(a,b)from(1.4)and(1.7).

Example1.14

Iftheverticesofatrianglehaveintegralcoordinates,provethatitcannotbeanequilateraltriangle.

Solution

Theareaofthetrianglewithvertices(x1,y1),(x2,y2)and(x3,y3)is

Also,theareaofΔABCis

whereaisthesideoftheequilateraltriangle.Iftheverticesofthetrianglehaveintegralcoordinates,thenΔisarationalnumber.However,from(1.9)weinfer

thattheareais timesarationalnumber.Hence,iftheverticesofatriangle

haveintegralcoordinates,itcannotbeequilateral.

Example1.15

Ift1,t2andt3aredistinct,thenshowthatthepoints and

a≠0cannotbecollinear.

Solution

sincet1,t2andt3aredistinct.Hence,thethreegivenpointscannotbecollinear.

Example1.16

TheverticesofatriangleABCare(2,3),(4,7),(−5,2).FindthelengthofthealtitudethroughA.

Solution

TheareaofΔABCisgivenby

Weknowthat

Example1.17

TheverticesofatriangleABCareA(x1,x1tanα),B(x2,x2tanβ)andC(x3,x3

tanγ).If istheorthocentreandS(0,0)isthecircumcentre,thenshowthat

Solution

IfristhecircumradiusofΔABC,SA=SB=SC=r,SA2=r2

Then,thecoordinatesofA,BandCare(rcosα,rsinα),(rcosβ,rsinβ)and(rcosγ,rsinγ).Thecentroidofthetriangleis

Theorthocentreis andthecircumcentreisS(0,0).Geometricallywe

knowthatH,GandSarecollinear.Therefore,theslopeofSGandGHareequal.

Example1.18

AlinejoiningthetwopointsA(2,0)andB(3,1)isrotatedaboutAintheanticlockwisedirectionthroughanangleof15°.IfBgoestoCinthenewposition,findthecoordinateofC.

Solution

GiventhatCisthenewpositionofB.DrawCLperpendiculartoOXandlet(x1,y1)bethecoordinatesofC.Now

ABmakes45°withx-axisand Then

Hence,thecoordinatesofCare

Example1.19

ThecoordinatesofA,BandCare(6,3),(−3,5)and(4,−2),respectively,andP

isanypoint(x,y).ShowthattheratiooftheareaofΔPBCandΔABCis

Solution

ThepointsA,B,CandPare(6,3),(−3,5),(4,−2)and(x,y),respectively.

Example1.20

Findthecoordinatesofthepointthatdividesthelinejoiningthepoints(2,3)and(−4,7)(i)internally(ii)externallyintheratio3:2.

Solution

LetRandR′respectivelydividePQinternallyandexternallyintheratio3:2.

i. ThecoordinatesofRare

ii. ThecoordinatesofR′are

(i.e)(−16,15)

Example1.21

Findtheratioinwhichthelinejoiningthepoints(4,7)and(−3,2)isdividedbythey-axis.

Solution

Letthey-axismeetthelinejoiningthejointsP(4,7)andQ(−3,2)atR.LetthecoordinatesofRbe(0,y).LetRdividePQintheratiok:1.

ThecoordinatesofRisgivenby

Hence,theratioinwhichRdividesPQis4:3.

Example1.22

Showthatthepoints(−2,−1),(1,0),(4,3)and(1,2)formtheverticesofaparallelogram.

Solution

Aquadrilateralinwhichthediagonalsbisecteachotherisaparallelogram.

ThemidpointofACis (i.e.)(1,1).

ThemidpointofBDis (i.e.)(1,1).

Sincethediagonalsbisecteachother,ABCDisaparallelogram.

Example1.23

Find(x,y)if(3,2),(6,3),(x,y)and(6,5)aretheverticesofaparallelogramtakeninorder.

Solution

LetthefourpointsbeA,B,CandD,respectively.SinceABCDisaparallelogram,themidpointofACisthesameasthemidpointofBD.

ThemidpointofACis .ThemidpointofBDis (i.e)(6,4).

Hence(x,y)is(9,6).

Example1.24

Themidpointsofthesidesofatriangleare(6,−1),(−1,−2)and(1,4).Findthecoordinatesofthevertices.

Solution

LetD,EandFbethemidpointsofthesidesBC,CAandAB,respectively.Then,(6,−1),(−1,−2),(1,4)arethepointsD,EandF,respectively.LetA(x1,y1),B(x2,y2)andC(x3,y3)betheverticesofthetriangle.ThenBDEFisaparallelogram.

ThemidpointofDFis

ThemidpointofBEis

SinceFisthemidpointofAB,

SinceEisthemidpointofAC,

∴x3=4,y3=−7.∴Cisthepoint(4,−7)

Hence,theverticesofthetriangleare(−6,3),(8,5)and(4,−7).

Example1.25

Showthattheaxesofcoordinatestrisectthestraightlinejoiningthepoints(2,−2)and(−1,4).

Solution

Letthelinejoiningthepoints(2,−2)and(−1,4)meetx-axisandy-axisatAandB,respectively.LetthecoordinatesofAandBbe(x,0)and(0,y),respectively.LetAdividethelineintheratiok:1.Thenthex-coordinateofAisgivenby

∴−k+2=0⇒k=2Hence,Adividesthelineintheratio2:1.

LetBdividethelineintheratiol:1.Then,

Hence,

Bdividesthelineintheratio1:2.Hence,AandBtrisectthelinejoiningthepoints(2,−4)and(−1,4).

Example1.26

TheverticesofatriangleareA(3,5),B(−7,9)andC(1,−3).Findthelengthofthethreemediansofthetriangle.

Solution

LetD,EandFbethemidpointsofthesidesofBC,CAandAB,respectively.

ThecoordinatesofDare (i.e.)(−3,3).ThecoordinatesofEare

(i.e.)(2,1).ThecoordinatesofFare (i.e.)(−2,7).

Hence,thelengthsofthemediansare

units.

Example1.27

Twooftheverticesofatriangleare(4,7)and(−1,2)andthecentroidisattheorigin.Findthethirdvertex.

Solution

Letthethirdvertexofthetrianglebe(x,y).Then

Hence,thethirdvertexis(−3,−9).

Example1.28

Showthatthemidpointofthehypotenuseoftherightangledtrianglewhoseverticesare(8,−10),(7,−3)and(0,−4)isequidistantfromthevertices.

Solution

LetthethreegivenpointsbeA(8,−10),B(7,−3)andC(0,−4).

AB2+BC2=AC2

Hence,ABCisarightangledtrianglewithACashypotenuse.

ThemidpointofACis (i.e)(4,−7).

Hence,themidpointofthehypotenuseisequidistantfromthevertices.

Example1.29

Findtheratioinwhichthelinejoiningthepoints(1,−1)and(4,5)isdividedbythepoint(2,1).

Solution

LetthepointR(2,1)dividethelinejoiningthepointsP(1,−1)andQ(4,5)intheratiok:1.Then

Therefore,R(2,1)dividesPQintheratio1:2.

Example1.30

Findthelocusofthepointthatisequidistantfromtwogivenpoints(2,3)and(−4,1).

Solution

LetP(x,y)beapointsuchthatPA=PBwhereAandBarethepoints(2,3)and(−4,1),respectively.

Example1.31

Findthelocusofthepointthatmovesfromthepoint(4,3)keepingaconstantdistanceof5unitsfromit.

Solution

LetC(4,3)bethegivenpointandP(x,y)beanypointsuchthatCP=5.Then

Example1.32

Theendsofarodoflengthlmoveontwomutuallyperpendicularlines.Showthatthelocusofthepointontherodthatdividesitintheratio1:2is9x2+36y2=l2.

Solution

LetABbearodoflengthlwhoseendsAandBareonthecoordinateaxes.LetthecoordinatesofAandBbeA(a,0)andB(0,b).LetthepointP(x1,y1)divideABintheratio1:2.

ThenthecoordinatesofPare

Hence,thelocusof(x1,y1)is9x2+36y2=l2.

Example1.33

Apointmovessuchthatthesumofitsdistancesfromtwofixedpoints(al,0)

and(−al,0)isalways2a.Provethattheequationofthelocusis

Solution

LetthetwofixedpointsbeA(al,0)andB(0,−al).LetP(x1,y1)beamovingpointsuchthatPA+PB=2a.Giventhat

Then

Adding(1.10)and(1.11),

Squaringonbothsides,weget

Dividingbya2(1−l2),weget

Therefore,thelocusof(x1,y1)is

Example1.34

ArightangledtrianglehavingtherightangleatCwithCA=aandCB=bmovessuchthattheangularpointsAandBslidealongthex-axisandy-axis,respectively.FindthelocusofC.

Solution

LetthepointsAandBbeonthex-axisandy-axis,respectively.LetAandBhavecoordinates(α,0)and(0,β).LetCbethepointwithcoordinates(x1,y1).

Then

ThenAB2=a2+b2.AlsoAB2=α2+β2

Hence,α2+β2=a2+b2

Hence,thelocusofc(x1,y1)isa2x2−b2y2=0.

Example1.35

TwopointsPandQaregiven.RisavariablepointononesideofthelinePQ

suchthat isapositiveconstant2α.FindthelocusofthepointP.

Solution

LetPQbethex-axisandtheperpendicularthroughthemidpointofPQbethey-axis.LetPandQbethepoints(a,0)and(−a,0),respectively.LetRbethepoint

(x1,y1).Let Then

(i.e.)θ−ϕ=2α.Thentan(θ−ϕ)=tan2α

Hence,thelocusof(x1,y1)isx2−y2−2xycot2α=a2.

Exercises

1. Showthattheareaofthetrianglewithvertices(a,b),(x1,y1)and(x2,y2)wherea,x1andx2areingeometricprogressionwithcommonratiorandb,y1andy2areingeometricprogressionwith

commonratiosis

2. IfP(1,0),Q(−1,0)andR(2,0)arethreegivenpoints,thenshowthatthelocusofthepointS

satisfyingtherelationSQ2+SR2−2SP2isastraightlineparalleltothey-axis.

3. Showthatthepoints(p+1,1),(2p+1,3)and(2p+2,2)arecollinearifp=2or

4. Showthatthemidpointoftheverticesofaquadrilateralcoincideswiththemidpointoftheline

joiningthemidpointofthediagonals.

5. Showthatift1andt2aredistinctandnonzero,then and(0,0)arecollinear.

6. Ifthepoints arecollinearforthreedistinctvaluesa,

bandc,thenshowthatabc−(bc+ca+ab)+3(a+b+c)=0.7. PerpendicularstraightlinesaredrawnthroughthefixedpointC(a,a)tomeettheaxesofxandyatAandB.AnequilateraltriangleisdescribedwithABasthebaseofthetriangle.Provethatthe

equationofthelocusofCisthecurvey2=3(x2+a2).8. TheendsAandBofastraightlinesegmentofconstantlengthcslidesuponthefixedrectangular

axesOXandOY,respectively.IftherectangleOAPBiscompleted,thenshowthatthelocusofthe

footoftheperpendiculardrawnfromPtoABis .

9. ThepointAdividesthelinejoiningP(1,−5)andQ(3,5)intheratiok:1.FindthetwovaluesofkforwhichtheareaofthetriangleABCisequalto2unitsinmagnitudewhenthecoordinatesofBandCare(1,5)and(7,−2),respectively.

10. ThelinesegmentjoiningA(3,0)andB(0,2)isrotatedaboutapointAintheanticlockwisedirectionthroughanangleof45°andthusBmovestoC.IfpointDbethereflectionofCinthey-axis,findthecoordinatesofD.

Ans.:

11. If(a,b),(h,k)and(p,q)bethecoordinatesofthecircumcentre,thecentroidandtheorthocentreofatriangle,provethat3h=p+2α.

12. Provethatinarightangledtriangle,themidpointofthehypotenuseisequidistantfromitsvertices.

13. IfGisthecentroidofatriangleABC,thenprovethat3(GA2+GB2+GC2)=AB2+BC2+CA2.14. Showthatthelinejoiningthemidpointofanytwosidesofatriangleishalfofthethirdside.15. Provethatthelinejoiningthemidpointsoftheoppositesidesofaquadrilateralandthelinejoining

themidpointsofthediagonalsareconcurrent.

16. IfΔ1andΔ2denotetheareaofthetriangleswhoseverticesare(a,b),(b,c),(c,a)and(bc−a2,

ca−b2),(ca−b2,ab−c2)and(ab−c2,bc−a2),respectively,thenshowthatΔ2=(a+b+

c)2Δ1.17. Provethatiftwomediansofatriangleareequal,thetriangleisisosceles.

18. Ifa,bandcbethepth,qthandrthtermsofaHP,thenprovethatthepointshavingcoordinates(ab,r),(bc,p)and(ca,q)arecollinear.

19. Provethatapointcanbefoundthatisatthesamedistancefromeachofthefourpoints

20. If(x1,y1)(x2,y2)(x3,y3)and(x4,y4)betheverticesofaparallelogramandx1x3+y1y3=x2x1+y2y1thenprovethattheparallelogramisarectangle.

21. InanyΔABC,provethatAB2+AC2=2(AD2+DC2)whereDisthemidpointofBC.22. IfGisthecentroidofatriangleABCandObeanyotherpoint,thenprovethat

i. AB2+BC2+CA2=3(GA2+GB2+GC2)

ii. OA2+OB2+OC2=GA2+GB2+GC2+3GO2

23. Findtheincentreofthetrianglewhoseverticesare(20,7),(−36,7)and(0,−8).

Ans.:

24. IfA,BandCarethepoints(−1,5),(3,1)and(5,7),respectively,andD,EandFarethemidpointsofBC,CAandAB,respectively,provethatareaofΔABCisfourtimesthatofΔDEF.

25. IfD,EandFdividethesidesBC,CAandABofΔABCinthesameratio,provethatthecentroidofΔABCandΔDEFcoincide.

26. AandBarethefixedpoints(a,0)and(−a,0).FindthelocusofthepointPthatmovesinaplanesuchthat

i. PA2+PB2=2k2

ii. PA2−PB2=2PC2whereCisthepoint(c,0)

Ans.:(i)2ax+k2=0

(ii)2cx=c2−a2

27. If(xi,yi),i=1,2,3aretheverticesoftheΔABCanda,bandcarethelengthsofthesidesBC,CA

andAB,respectively,showthattheincentreofthetriangleABCis

28. Showthatthepoints(−a,−b),(0,0),(a,b)and(a2,b2)areeithercollinear,theverticesofaparallelogramortheverticesofarectangle.

29. ThecoordinatesofthreepointsO,AandBare(0,0),(0,4)and(6,0),respectively.ApointPmovessothattheareaofΔPOAisalwaystwicetheareaofΔPOB.FindtheequationofthelocusofP.

Ans.:x2−9y2=0

30. ThefourpointsA(x1,0),B(x2,0),C(x3,0)andD(x4,0)aresuchthatx1,x2aretherootsofthe

equationax2+2hx+b=0andx3,x4aretherootsoftheequationa1x2+2h1x+b1=0.Show

thatthesumoftheratiosinwhichCandDdivideABiszero,providedab1+a1b=2hh1.

31. If thenshowthatthetrianglewithvertices(xi,yi),i=1,2,3and(ai,bi),i=

1,2,3arecongruent.32. Thepoint(4,1)undergoesthefollowingthreetransformationssuccessively:

i. Reflectionabouttheliney=xii. Transformationthroughadistanceof2unitsalongthepositivedirectionofx-axis

iii. Rotationthroughanangleof abouttheoriginintheanticlockwisedirection.

Findthefinalpositionofthepoint.33. ShowthatthepointsP(2,−4),Q(4,−2)andR(1,1)lieonastraightline.Find(i)theratioPQ:QR

and(ii)thecoordinatesoftheharmonicconjugationofQwithrespecttoPandR.34. Ifapointmovessuchthattheareaofthetriangleformedbythatpointandthepoints(2,3)and

(−3,4)is8.5squareunits,showthatthelocusofthepointisx+5y−34=0.35. Showthattheareaofthetrianglewithvertices(p+5,p−4),(p−2,p+3)and(p,p)is

independentofp.

Chapter2

TheStraightLine

2.1INTRODUCTION

Inthepreviouschapter,wedefinedthatthelocusofapointisthepathtracedoutbyamovingpointaccordingtosomegeometricallaw.Weknowthatthelocusofapointwhichmovesinsuchawaythatitsdistancefromafixedpointisalwaysconstant.

2.1.1DeterminationoftheGeneralEquationofaStraightLine

SupposethepointP(x,y)movessuchthatP(x,y),A(4,−1),andB(2,3)forma

straightline.Then,

⇒x(−4)−y(2)+14=0

(i.e.)4x+2y−14=0or2x+y−7=0,whichisafirstdegreeequationinxandythatrepresentsastraightline.Thegeneralequationofastraightlineisax+by+c=0.Supposeax+by+c

=0isthelocusofapointP(x,y).IfthislocusisastraightlineandifP(x1,y1)andQ(x2,y2)beanytwopointsonthelocusthenthepointRwhichdividesPQwithratioλ:1isalsoapointontheline.SinceP(x1,y1)andQ(x2,y2)lieonthelocusax+by+c=0,

Onmultiplyingequation(2.2)byλandaddingwithequation(2.1),weget

λ(ax2+by2+c)+(ax1+by1+c)=0.(i.e.)a(λx2+x1)+b(λy2+y1)+c(λ+1)=0.

Ondividingbyλ+1,weget

Equation(2.3)showsthatthepoint liesonthelocusax+by

+c=0.ThisshowsthatthepointwhichdividesPQintheratioλ:1alsoliesonthelocuswhichisthedefinitionforastraightline.∴ax+by+c=0alwaysrepresentsastraightline.

Note2.1.1.1:Theaboveequationcanbewrittenintheform

whichisoftheformAx+By+1=0.Hence,therearetwoindependentconstantsinequationofastraightline.Now,welookintovariousspecialformsoftheequationofastraightline.

2.1.2EquationofaStraightLineParalleltoy-axisandataDistanceofhunitsfromx-axis

LetPQbethestraightlineparalleltoy-axisandataconstantdistancehunitsfromy-axis.TheneverypointonthelinePQhasthex-coordinateh.HencetheequationofthelinePQisx=h.

Note2.1.2.1:

1. Similarly,theequationofthelineparalleltox-axisandatadistancekfromitisy=k.2. Theequationofx-axisisy=0.3. Theequationofy-axisisx=0.

2.2SLOPEOFASTRAIGHTLINE

Ifastraightlinemakesanangleθwiththepositivedirectionofx-axisthentanθiscalledtheslopeofthestraightlineandisdenotedbym.

∴m=tanθ.

Wecannowdeterminetheslopeofastraightlineintermsofcoordinatesoftwopointsontheline.LetP(x1,y1)andQ(x2,y2)bethetwogivenpointsonaline.DrawPLandQMperpendicularstox-axis.LetPQmakeanangleθwithOX.

DrawQRperpendiculartoLP.Then .

2.3SLOPE-INTERCEPTFORMOFASTRAIGHTLINE

Findtheequationofthestraightline,whichmakesanangleθwithOXandcutsoffaninterceptconthey-axis.

LetP(x,y)beanypointonthestraightlinewhichmakesanangleθwithx-axis.

,OB=c=y-intercept.DrawPLperpendiculartox-axisandBN

perpendiculartoLP.Then, .BN=OL=x.

∴NP=LP−LN=LP−OB=y−c.

InΔNBP,

ThisequationistrueforallpositionsofPonthestraightline.Hence,thisistheequationoftherequiredline.

2.4INTERCEPTFORM

Findtheequationofthestraightline,whichcutsoffinterceptsaandb,respectivelyonxandyaxes.

LetP(x,y)beanypointonthestraightlinewhichmeetsxandyaxesatAandB,respectively.LetOA=a,OB=b,ON=x,andNP=y;NA=OA−ON=a−x.TrianglesPNAandBOAaresimilar.Therefore,

.ThisresultistrueforallpositionsofP

onthestraightlineandhencethisistheequationoftherequiredline.

2.5SLOPE-POINTFORM

Findtheequationofthestraightlinewithslopemandpassingthroughthegivenpoint(x1,y1).Theequationofthestraightlinewithagivenslopemis

Here,cisunknown.Thisstraightlinepassesthroughthepoint(x1,y1).Thepointhastosatisfytheequationy=mx+c.∴y1=mx1+c.Substitutingthevalueofcinequation(2.4),wegetthe

equationofthelineas

y=mx+y1−mx1⇒y−y1=m(x−x1).

2.6TWOPOINTSFORM

Findtheequationofthestraightlinepassingthroughtwogivenpoints(x1,y1)and(x2,y2).

where,misunknown.Theslopeofthestraightlinepassingthroughthepoints

Bysubstitutingequation(2.6)inequation(2.5),wegettherequiredstraightline

2.7NORMALFORM

Findtheequationofastraightlineintermsoftheperpendicularpfromtheorigintothelineandtheanglethattheperpendicularlinemakeswithaxis.

DrawOL⊥AB.LetOL=p.

Let

Therefore,theequationofthestraightlineABis

(i.e.)xcosα+ysinα=p

2.8PARAMETRICFORMANDDISTANCEFORM

Letastraightlinemakeanangleθwithx-axisandA(x1,y1)beapointontheline.DrawAL,PMperpendiculartox-axisandAQperpendiculartoPM.Then,

InΔPAQ,x−x1=rcosθ;y−y1=rsinθ.

Thesearetheparametricequationsofthegivenline.

Note2.8.1:Anypointonthelineisx=x1+rcosθ,y=y1+rsinθ.

Note2.8.2:risthedistanceofanypointonthelinefromthegivenpointA(x1,y1).

2.9PERPENDICULARDISTANCEONASTRAIGHTLINE

Findtheperpendiculardistancefromagivenpointtothelineax+by+c=0.

LetR(x1,y1)beagivenpointandax+by+c=0bethegivenline.ThroughRdrawthelinePQparalleltoAB.DrawOSperpendiculartoABmeetingPQatT.

LetOS=pandPT=p1.Let .ThentheequationofABis

whichisthesameas

Equations(2.7)and(2.8)representthesamelineand,therefore,identifying

weget

TheequationofthelinePQisxcosα+ysinα=p.SincethepointR(x1,y1)liesonthelinex1cosα+ysinα−p1=0.

∴p1=x1cosα+y1sinα.Then,thelengthoftheperpendicularlinefromRtoAB

Note2.9.1:Theperpendiculardistancefromtheoriginonthelineax+by+

2.10INTERSECTIONOFTWOSTRAIGHTLINES

Letthetwointersectingstraightlinesbea1x+b1y+c1=0anda2x+b2y+c2=0.Letthestraightlinesintersectatthepoint(x1,y1).Then(x1,y1)liesonboththelinesandhencesatisfytheseequations.Then

Solvingtheequations,weget

Therefore,thepointofintersectionis

Findtheratioatwhichthelineax+by+c=0dividesthelinejoiningthepoints(x1,y1)and(x2,y2).

Letthelineax+by=c=0dividethelinejoiningthepointsP(x1,y1)andQ(x2,y2)intheratioλ:1.Then,thecoordinatesofthepointofdivisionRare

Thispointliesonthelineax+by+c=0

Note2.10.1:

1. Ifλispositivethenthepoints(x1,y1)and(x2,y2)lieontheoppositesidesofthelineax+by+c=0.

2. Ifλisnegativethenthepoints(x1,y1)and(x2,y2)lieonthesamesideofthelineax+by+c=0.3. Inotherwords,iftheexpressionsax1+by1+candax2+by2+c2areofoppositesignsthenthe

point(x1,y1)and(x2,y2)lieontheoppositesidesofthelineax+by+c=0.Iftheyareofthesamesignthenthepoints(x1,y1)and(x2,y2)lieonthesamesideofthelineax+by+c=0.

Findtheequationofastraightlinepassingthroughintersectionofthelinesa1x+b1y+c=0anda2x+b2y+c=0.Considertheequation

Considertheequation

Thisisalinearequationinxandyandhencethisequationrepresentsastraightline.Let(x1,y1)bethepointofintersectionofthelinesa1x+b1y+c1=0anda2x+b2y+c2=0.Then(x1,y1)hastosatisfythetwoequations:

Onmultiplyingequation(2.11)byλandaddingwithequation(2.10)weget,

Thisequationshowsthatthepointx=x1andy=y1satisfiesequation(2.9).Hencethepoint(x1,y1)liesonthestraightlinegivenbytheequation(2.9),whichisalinepassingthroughtheintersectionofthelinesa1x+b1y+c1=0anda2x+b2y+c2=0.

2.11CONCURRENTSTRAIGHTLINES

Considerthreestraightlinesgivenbyequations:

Thepointofintersectionoflinesgivenbyequations(2.12)and(2.13)is

Ifthethreegivenlinesareconcurrent,theabovepointshouldlieonthestraightlinegivenbyequation(2.14).

Thisistherequiredconditionforthethreegivenlinestobeconcurrent.The

aboveconditioncanbeexpressedindeterminantform

Ifl,m,andnareconstantssuchthatl(a1x+b1y+c1)+m(a2x+b2y+c2)+n(a3x+b3y+c3)vanishesidenticallythenprovethatthelinesa1x+b1y+c1=0,a2x+b2y+c2=0,anda3x+b3y+c3=0areconcurrent.Letthelinesa1x+b1y+c1=0anda2x+b2y+c2=0meetatthepoint(x1,

y1).

Forallvaluesofxandygiventhat,

Thenitwillbetrueforx=x1andy=y1.∴l(a1x1+b1y1+c1)+m(a2x1+b2y1+c2)+n(a3x1+b3y1+c3)=0.

Usingequations(2.15)and(2.16),wegeta3x1+b3y1+c3=0.Thatis,thepoint(x1,y1)liesonthelinea3x+b3y+c3=0.Therefore,thelinesa1x+b1y+c1=0,a2x+b2y+c2=0,a3x+b3y+c3=0

areconcurrentat(x1,y1).

2.12ANGLEBETWEENTWOSTRAIGHTLINES

Letθbetheanglebetweentwostraightlines,whoseslopesarem1andm2.Letthetwolineswithslopesm1andm2makeanglesθ1andθ2withx-axis.Then,m1=tanθ1,m2=tanθ2.Also,θ=θ1−θ2

IftheRHSispositive,thenθistheacuteanglebetweenthelines.IfRHSisnegative,thenθistheobtuseanglebetweenthelines.

Note2.12.1:Ifthelinesareparallelthenθ=0andtanθ=tan0=0.

Note2.12.2:Ifthelinesareperpendicularthen,

Therefore,

1. Iftwolinesareparallelthentheirslopesareequal.2. Ifthetwolinesareperpendicularthentheproductoftheirslopesis−1.

2.13EQUATIONSOFBISECTORSOFTHEANGLEBETWEENTWOLINES

LetABandCDbethetwointersectingstraightlinesintersectingatP.Lettheselinesberepresentedbytheequationsa1x+b1y+c1=0anda2x+b2y+c2=0.

LetPRandPR′bethebisectorsofangles and ,respectively.Thenthe

perpendiculardistancesfromR(orR′)AB,andCDareequal.

Ifc1andc2arepositive,thentheequationsofthebisectorcontainingtheoriginisgivenby

Theequationofthebisectornotcontainingtheoriginis

Ifc1andc2arenotpositivethentheequationsoflinesshouldbewritteninsuchawaythatc1andc2arepositive.

Note2.13.1:Wecaneasilyobservethatthetwobisectorsareatrightangles.


Example2.1

Findtheequationofthestraightlinewhichisatadistanceof10unitsfromx-axis.

Solution

Theequationoftherequiredstraightlineisx=10orx−10=0.

Example2.2

Findtheequationofthestraightlinewhichisatadistanceof−15unitsfromy-axis.

Solution

Theequationoftherequiredlineisy=−15ory+15=0.

Example2.3

Findtheslopeofthelinejoiningthepoints(2,3)and(4,−5).

Solution

Theslopeofthelinejoiningthetwogivenpoints(x1,y1)and(x2,y2)is

Therefore,theslopeofthelinejoiningthetwogivenpointsis

Example2.4

Findtheslopeoftheline2x−3y+7=0.

Solution

Theequationofthelineis2x−3y+7=0(i.e.)3y=2x+7.

Therefore,slopeoftheline= .

Example2.5

Findtheequationofthestraightlinemakinganangle135°withthepositivedirectionofx-axisandcuttingofanintercept5onthey-axis.

Solution

Theslopeofthestraightlineis

yintercept=c=5.Therefore,theequationofthestraightlineis

Example2.6

Findtheequationofthestraightlinecuttingofftheintercepts2and−5ontheaxes.

Solution

Theequationofthestraightlineis .Here,a=2andb=−5.

Therefore,theequationofthestraightlineis or5x−2y=10.

Example2.7

Findtheequationofthestraightlinepassingthroughthepoints(7,−3)andcuttingoffequalinterceptsontheaxes.

Solution

Lettheequationofthestraightlinebe

(i.e.)x+y=a.

Thisstraightlinepassesthroughthepoint(7,−3).Therefore,7−3=a(i.e.)a=4.∴Theequationofthestraightlineisx+y=4.

Example2.8

Findtheequationofthestraightline,theportionofwhichbetweentheaxesisbisectedatthepoint(2,−5).

Solution


LetthelinemeetthexandyaxesatAandB,respectively.Thenthecoordinates

ofAandBare(a,0)and(0,b).ThemidpointofABis .However,the

midpointisgivenas(2,−5).

Therefore,

∴a=4andb=−10.

Hence,theequationofthestraightlineis

(i.e.)5x−2y=20.

Example2.9

Findtheequationofthestraightlineoftheportionofwhichbetweentheaxesisdividedbythepoint(4,3)intheratio2:3.

Solution


LetthislinemeetthexandyaxesatAandB,respectively.ThecoordinatesofAandBare(a,0)and(0,b),respectively.ThecoordinatesofthepointthatdividesABintheratio2:3are

Thispointisgivenas(4,3).

Therefore,

∴Theequationofthestraightlineis (i.e.)9x+8y=60.

Example2.10

Findtheequationstothestraightlineseachofwhichpassesthroughthepoint(3,2)andintersectthexandyaxesatAandBsuchthatOA−OB=2.

Solution

Lettheequationofthestraightlinebe .Thisstraightlinepassesthrough

thepoint(3,2).

Also,giventhatOA−OB=2

Therefore,b=a−2.Substitutingthisinequation(2.20)weget3(a−2)+2a=a(a−2).

∴Thetwostraightlinesare and

(i.e.)x−y=1and2x+3y=12.

Example2.11

ShowthatthepointsA(l,1),B(5,−9),andC(−l,6)arecollinear.

Solution

TheslopeofABis

SincetheslopesofABandBCareequalandBisthecommonpoint,thepointsarecollinear.

Example2.12

Provethatthetrianglewhoseverticesare(−2,5),(3,−4),and(7,10)isarightangledisoscelestriangle.Findtheequationofthehypotenuse.

SolutionLetthepointsbeA(−2,5),B(3,−4),andC(7,10).AB2=(−2−3)2+(5+4)2=25+81=106,BC2=(3−7)2+(−4−10)2=16+196=212.AC2=(−2−7)2+(5−10)2=81+25=106.Therefore,AB2+AC2=BC2

andAB=AC.Hence,the∆ABCisarightangledisoscelestriangle.TheequationofthehypotenuseBCis

Example2.13

Findtheequationofthestraightlinewhichcutsoffinterceptsontheaxesequalinmagnitudebutoppositeinsignandpassingthroughthepoint(4,7).

Solution

Lettheequationofthestraightlinecuttingoffinterceptsequalinmagnitudebut

oppositeinsignbe (i.e.)x−y=a.

Thispassesthroughthepoint(4,7).Therefore,4−7=a(i.e.)a=−3.Hence,theequationofthestraightlineisx−y+3=0.

Example2.14

Findtheratioinwhichtheline3x−2y+5=0dividesthelinejoiningthepoints(6,−7)and(−2,3).

Solution

Lettheline3x−2y+5=0dividethelinejoiningthepointsA(6,−7)andB(−2,

3)intheratiok:1.Thenthecoordinatesofthepointofdivisionare

.Sincethispointliesonthestraightline3x−2y+5=0,weget

∴Therequiredratiois37:7.

Example2.15

Provethatthelines3x−4y+5=0,7x-8y+5=0,and4x+5y=45areconcurrent.

Solution

Given

Solvingequations(2.22)and(2.23),wegetthepointofintersectionofthetwolines.

∴Fromequation(2.22),15−4y=−5.∴y=5.Hence,thepointofintersectionofthelinesis(5,5).Substitutingx=5andy=

5,inequation(2.24),weget20+25=45whichistrue.∴Thethirdlinealsopassesthroughthepoints(5,5).Henceitisprovedthat

thethreelinesareconcurrent.

Example2.16

Findthevalueofasothatthelinesx−6y+a=0,2x+3y+4=0,andx+4y+1=0areconcurrent.

Solution

Given

Solvingtheequations(2.26)and(2.27)weget,

Onsubtracting,weget5y=2

∴Fromequation(2.27)

Hence,thepointofintersectionofthelinesis .Sincethelinesare

concurrentthispointshouldlieonx−6y+a=0.

Example2.17

Provethatforallvaluesofλthestraightlinex(2+3λ)+y(3−λ)−5−2λ=0passesthroughafixedpoint.Findthecoordinatesofthefixedpoint.

Solution

x(2+3λ)+y(3−λ)−5−2λ=0.Thisequationcanbewrittenintheform

Thisequationrepresentsastraightlinepassingthroughtheintersectionoflines

forallvaluesofλ.

Onadding,weget11x=11⇒x=1andhencefromequation(2.29)wegety=1.Therefore,thepointofintersectionofstraightlines(2.29)and(2.30)is(1,1).

Thestraightline(2.28)passesthroughthepoint(1,1)forallvaluesofλ.Hence(2.28)passesthroughthefixedpoint(1,1).

Example2.18

Findtheequationofthestraightlinepassingthroughtheintersectionofthelines3x−y=5and2x+3y=7andmakinganangleof45°withthepositivedirectionofx-axis.

SolutionSolvingtheequations,

Weget,

Onadding,weget11x=22.

∴x=2.Fromequation(2.31),6−y=5.

∴y=1.Hence(2,1)isthepointofintersectionofthelines(2.31)and(2.32).

Theslopeoftherequiredlineism=tanθ,m=tan45°=1.Therefore,theequationoftherequiredlineisy−y=m(x−x1)(i.e.)y−1=1(x−2)⇒x−y=1.

Example2.19

Findtheequationofthestraightlinepassingthroughtheintersectionofthelines7x+3y=7and2x+y=2andcuttingoffequalinterceptsontheaxes.

Solution

Thepointofintersectionofthelinesisobtainedbysolvingthefollowingtwoequations:

Onsubtracting,wegetx=1andhencey=0.Therefore,thepointofintersection

is(1,0).Theequationofthestraightlinecuttingoffequalinterceptsis

(i.e)x+y=a.Thisstraightlinepassesthrough(1,0).Therefore,1+0=a(i.e.)a=1.

Hence,theequationoftherequiredstraightlineisx+y=1.

Example2.20

Findtheequationofthestraightlineconcurrentwiththelines2x+3y=3andx+2y=2andalsoconcurrentwiththelines3x−y=1andx+5y=11.

Solution

Thepointofintersectionofthelines2x+3y=3andx+2y=2isobtainedbysolvingthefollowingtwoequations:

Onsubtracting,wegety=1andhencex=0.Therefore,thepointofintersectionis(0,1).

Onadding,weget16x=16whichgivesx=1andhencey=2.Thepointofintersectionofthesecondpairoflinesis(1,2).Theequationofthelinejoiningthetwopoints(0,1)and(1,2)is

Example2.21

Findtheanglebetweenthelines

Solution

Theslopeoftheline .Therefore, (i.e.)θ1=

60°.Theslopetheline Therefore,

.Theanglebetweenthelinesisθ1−θ2=30°.

Example2.22

Findtheequationoftheperpendicularbisectorofthelinejoiningthepoints(−2,6)and(4,−6).

Solution

Theslopeofthelinejoiningthepoints(−2,6)and(4,−6)is .

Therefore,theslopeoftheperpendicularlineis .Themidpointoftheline

joiningthepoints(−2,6)and(4,−6)is (i.e.)(1,0).

Therefore,theequationoftheperpendicularbisectorisy−y1=m(x−x1)

(i.e.)y−0= (x−1)⇒2y=x−1orx−2y−1=0.

Example2.23

A(4,1),B(7,4),andC(5,−2)aretheverticesofatriangle.FindtheequationoftheperpendicularlinefromAtoBC.

Solution

TheslopeofthelineBCis

Therefore,theslopeoftheperpendicularADtoBCis− .Hence,theequationof

theperpendicularfromA(4,1)onBCisy−y1=m(x−x1)

Example2.24

Thefootoftheperpendicularfromthepoint(1,2)onalineis(3,–4).Findtheequationoftheline.

Solution

LetABbethelineandD(3,−4)bethefootoftheperpendicularfromC(1,2)

TheslopeofthelineCDis

Therefore,theslopeofthelineABis .

TheequationofthelineABisy−y1=m(x−x1)

Example2.25

Findtheequationoftherightbisectorofthelinejoiningthepoints(2,3)and(4,5).

Solution

Therightbisectoristheperpendicularbisectorofthelinejoiningthepoints(2,

3)and(4,5).Themidpointofthelineis

Therefore,theslopeofthegivenlineis

∴Theslopeoftherightbisectoris−1.Theequationoftherightbisectorisy−y1=m(x−x1)⇒y−4=−1(x−3)ory−4=−x+3orx+y=7.

Example2.26

Findthepointontheline3y−4x+11=0whichisequidistantfromthepoints(3,2)and(−2,3).

Solution

LetP(x1,y1)bethepointontheline3y−4x+11=0whichisequidistantfromthepointsA(3,2)andB(−2,3).

Since,

Sincethe(x1,y1)liesontheline,

Substitutingy1=5x1in(2.40),weget15x1−4x1+11=0.∴x1=−1andhencey1=−5.Therefore,therequiredpointis(−1,−5).

Example2.27

Findtheequationofthelinepassingthroughthepoint(2,3)andparallelto3x−4y+5=0.

Solution

Theslopeoftheline3x−4y+5=0is .

Therefore,theslopeoftheparallellineisalso .

Hencetheequationoftheparallellinethrough(2,3)isy−y1=m(x−x1)

Example2.28

Findtheequationofthelinepassingthroughthepoint(4,−5)andisperpendiculartotheline7x+2y=15.

Solution

Theslopeoftheline

Therefore,theslopeoftheperpendicularlineis .Theequationofthe

perpendicularlinethrough(4,−5)isy−y1=m(x−x1)

Example2.29

Findtheequationofthelinethroughtheintersectionof2x+y=8and3x+7=2yandparallelto4x+y=11.

Solution

Thepointofintersectionofthelines2x+y=8and3x+7=2yisobtainedbysolvingthefollowingtwoequations:

Onadding,weget

Therefore,thepointofintersectionis

Theslopeoftheline4x+y=11is−4.Theslopeoftheparallellineisalso−4.Theequationoftheparallellineis

(i.e.)28x+7y=74.

Example2.30

Findtheimageoftheoriginontheline3x−2y=13.

Solution

LetO′(x1,y1)betheimageofOonthelineAB.ThenCisthemidpointofOO′.

TheslopeofthelineOO′is

TheequationofthelineOO′is

Solvingtheequations

TogetthecoordinatesofC:

Therefore,Cis(3,−2).CbeingthemidpointofOO′.

Therefore,theimageis(6,−4).

Example2.31

Findtheequationofthestraightlinepassingthroughtheintersectionofthelines3x+4y=17and4x−2y=8andperpendicularto7x+5y=12.

Solution

(2.45)×1+(2.46)×2gives

From(2.45),9+4y=17.Therefore,y=2.Hence(3,2)isthepointof

intersectionofthelines(2.45)and(2.46).Theslopeoftheline

Therefore,theslopeoftheperpendicularlineis

7y−14=5x−15or5x−7y=1.

Example2.32

Findtheorthocentreofthetrianglewhoseverticesare(5,−2),(−1,2),and(1,4).

Solution

Slopeof Therefore,slopeoftheperpendicularADis−1.

TheequationofthelineADisy+2=−1(x−5).

Slopeof Therefore,slopeofBEis

TheequationofBEis

Solvingtheequations(2.47)and(2.48),wegetthecoordinatesoftheorthocentre:

Onadding5x=1orx= .

From(2.47),

∴Theorthocentreis

Example2.33

Thepoints(1,3)and(5,1)aretwooppositeverticesofarectangle.Theothertwoverticeslieontheliney=2x+c.Findcandtheremainingtwovertices.

Solution

LetABCDbetherectanglewithAandCasthepointswithcoordinates(1,3)and(5,1),respectively.Inarectanglethediagonalsbisecteachother.

ThemidpointACis

AsthispointliesonBDwhoseequationisy=2x+c.Weget2=6+corc=−4.Therefore,theequationofthelineBDisy=2x−4.Therefore,thecoordinatesofanypointonthislineis(x,2x−4).IfthisisthepointBthenAB2+BC2=AC2.

Asy=2x−4,thecorrespondingvaluesofy=0,4.Therefore,thecoordinatesofBandDare(2,0)and(4,4).

Example2.34

Ifa,b,andcaredistinctnumbersdifferentfrom1thenshowthatthepoints

arecollinearifab+bc+ca−abc=

3(a+b+c).

Solution

LetA,B,andClieonthestraightlinepx+qy+r=0.

Thentheequationofthelinesatisfiesthecondition

wheret=a,b,andc(i.e.)pt3+qt2+rt−3q−r=0.Here,a,b,andcaretherootsofthisequation.

Example2.35

Avertexofanequilateraltriangleisat(2,3)andtheequationoftheoppositesideisx+y=2.Findtheequationsoftheothersides.

Solution

TheslopeofBCis−1.LetmbeslopeofABorAc.Then

Therefore,theequationofothertwosidesare and

.

Example2.36

Onediagonalofasquareistheportionoftheline interceptedbetween

theaxes.Findtheequationoftheotherdiagonal.

Solution

TheslopeofABis LetmbeslopeofAC.

TheequationofthesideOCis

TheequationofthesideBDis

Theequationoftheotherdiagonalis

Example2.37

IftheverticesofΔABCare(xi,yi)i=1,2,3.Showthattheequationofthe

medianthroughAisgivenby

Solution

ThecoordinatesofthemidpointofBCare

TheequationofthemedianADisgivenby Sincetheareaof

alineiszero.

Example2.38

If(x,y)isanarbitrarypointontheinternalbisectorofverticalangleAofΔABC,where(xi,yi),i=1,2,3aretheverticesofA,B,andC,respectively,anda,b,andcarethelengthofthesidesBC,CA,andAB,respectively,provethat

Solution

InΔABC,ADistheinternalbisectorof .Weknowthat

ThecoordinatesofDare

TheequationofADisgivenby

Example2.39

Findtheorthocentreofthetrianglewhoseverticesare(a,0),(0,b),and(0,0).

Solution

Theorthocentreisthepointofconcurrenceofaltitudes.SinceOAandOBareperpendiculartoeachother,OAandOBarethealtitudesthroughAandBofΔABC.Therefore,0istheorthocentre.Hence,thecoordinatesoftheorthocentreis(0,0).

Example2.40

Provethattheorthocentreofthetriangleformedbythethreelines

liesonthelinex+a=0.

Solution

Theequationofthelinepassingthroughtheintersectionofthelines

TheslopeofthelineADis TheslopeofthelineBCist1.SinceADis

perpendiculartoBC,

TheequationofthelineADis

SimilarlytheequationofthelineBEis

Subtractingequations(2.52)from(2.53),t3(t1−t2)x+at3(t1−t2)=0.Sincet1≠t2,x+a=0theorthocentreliesonthelinex+a=0.

Example2.41

Showthatthereflectionofthelinepx+qy+r=0,onthelinelx+my+n=0is(px+qy+r)(l2+m2)−2(lp+mq)(lx+my+n)=0.

Solution

LetADbethereflectionofthelinepx+qy+r=0inthelinelx+my+n=0.ThentheequationoflineADispx+qy+r+k(lx+my+n)=0.ThentheperpendicularfromanypointonACtoABandADareequal.

SincethepointPliesonAC,lx+my+n=0,lx1+my1+n=0

HencetheequationofthelineADis(l2+m2)(px+qy+r)−2(pl+qm)(lx+my+n)=0.

Example2.42

Thediagonalsoftheparallelogramaregivenbythesidesu=p,u=q,v=r,v=swhereu=ax+by+candv=a1x+b1y+c1.Showthattheequationofthediagonalwhichpassesthroughthepointsofintersectionofu=p,v=randu=

qandr=sisgivenby

Solution

Consider

Thisisalinearequationinxandyand,therefore,itrepresentsastraightline.ThecoordinatesofBaregivenbytheintersectionofthelinesu=pandv=s.However,u=pandv=rsatisfiestheequation(2.54).Inaddition,u=q,andv=ssatisfytheequation(2.54)andhencetheequation(2.54)isthelinepassingthroughBandDandrepresentstheequationofthediagonalBD.

Example2.43

AlinethroughthepointA(−5,−4)meetsthelinesx+3y+2=0,2x+y+4=0,

andx−y−5=0atthepointsB,C,andD,respectively.If

findtheequation.

Solution

Theequationofthelinepassingthroughthepoint(−5,−4)is

Anypointonthelineis(rcosθ−5,rsinθ−4).Thepointmeetsthelinex+3y+2=0atBthenAB=r·(rcosθ−5)+3(rsinθ−4)+2=0.

Iftheline(2.55)meetstheline2x+y+4=0atCthen2(rcosθ−5)+(rsinθ−4)+4=0.Now,r=ACand

Iftheline(2.55)meetsthelinex−y−5=0then(rcosθ−5)−(rsinθ−4)−5=0andhereAD=r.

Giventhat

Hence,theequationofthelineis

Example2.44

AvariablestraightlineisdrawnthroughOtocuttwofixedlinesL1andL2atA1

andA2.ApointAistakenonthevariablelinesuchthat Showthat

thelocusofPisastraightlinepassingthroughthepointofintersectionofL1andL2.

Solution

LettheequationofthelineOXbe whereOA=r.Anypointonthis

lineis(rcosθ,rsinθ).LetOA1=r1andOA2=r2.A1is(r1cosθ,r1sinθ)andA2is(r2cosθ,r2sinθ).LetthetwofixedstraightlinesbeL1:l1x+m1y−1=0andL2:l2x+m2y−1=0.SincethepointsA1andA2lieonthetwolines,respectively,

(i.e.)p(l1x+m1y−1)+q(l2x+m2y−1)=0whichisastraightlinepassingthroughthepointofintersectionofthetwofixedstraightlinesL1=0andL2=0.

Example2.45

Iftheimageofthepoint(x1,y1)withrespecttothelinemy+lx+n=0isthe

point(x2,y2)showthat

Solution

Q(x2,y2)isthereflectionofP(x1,y1)onthelinelx+my+n=0.Themidpoint

ofPQliesonthelinelx+my+n=0.TheslopeofPQis Theslopeofthe

lineislx+my+n=0is Sincethesetwolinesareperpendicular,

Example2.46

ProvethattheareaofthetrianglewhoserootsareLr=arx+bry+cr(r=1,2,3)

is whereCiisthecofactorofci(i=1,2,3)inAgivenby

Solution

LetAr,Br,andCrbethecofactorsofar,br,andcrinD.Thepointofintersection

ofthelinesa1x+b1y+c1=0,a2x+b2y+c2=0is

∴Theverticesofthetriangleare Thentheareaof

thetriangleisgivenby,

whereDisthedeterminantformedbythecofactors.

Example2.47

AstraightlineLintersectsthesidesBC,CA,andABofatriangleABCinD,E,

andF,respectively.Showthat

Solution

LetDEFbethestraightlinemeetingBC,CA,andABatD,E,andF,respectively.LettheequationsofthelineDEFbelx+my+n=0.

LetDdivideBCintheratioλ:1.ThenthecoordinatesofDare

Asthispointliesonthelinelx+my+n=0.

Similarly

Multiplyingthesethreeweget

Example2.48

Astraightlineissuchthatthealgebraicsumofperpendicularsdrawnuponitfromanynumberoffixedpointsiszero.Showthatthestraightlinepassesthroughafixedpoint.

Solution

Let(x1,y1),(x2,y2),…,(xn,yn),benfixedpointsandax+by+c=0beagivenline.Thealgebraicsumoftheperpendicularsfrom(xi,yi),i=1,2,...,ntothislineiszero.

Thisequationshowsthatthepoint liesonthelineax

+by+c=0.Therefore,thelinepassesthroughafixedpoint.

Example2.49

Determineallthevaluesofαforwhichthepoint(α,α2)liesinsidethetriangleformedbythelines2x+3y−1=0,x+2y−3=0,and5x−6y−1=0.

Solution

∴L1(α,α2)=2a+3α2−1>0ifpointsAand(a,α2)liesonthesamesideoftheline.3α2+2a−1>0⇒(3α−1)(α+1)>0.

FromtheconditionsI,II,andIII,wehave

Example2.50

Findthedirectioninwhichastraightlinemustbedrawnthroughthepoint(1,4)sothatitspointofintersectionwiththelinex+y+5=0maybeatadistance

units.

Solution

Lettheequationofthelinethroughthepoint(1,4)be

Anypointonthislineis(rcosθ+1,rsinθ+4).Ifthispointliesonthelinex+y+5=0thenrcosθ+1+rsinθ+4+5=0.

∴Therequiredstraightlinemakesanangleof withthepositivedirectionsof

x-axisandpassesthroughthepoint(1,4).

Exercises

1. Findtheareaoftriangleformedbytheaxes,thestraightlineLpassingthroughthepoints(1,1)

and(2,0)andthelineperpendiculartotheLandpassingthrough

Ans.:

2. Theline3x+2y=24meetsy-axisatAandx-axisatB.TheperpendicularbisectorofABmeetsthelinethrough(0,−1)paralleltox-axisatC.FindtheareaΔABC.

Ans.:91sq.units

3. If(x,y)beanarbitrarypointonthealtitudethroughAofΔABCwithvertices(xi,yi),i=1,2,3

thentheequationofthealtitudethroughAis

4. Arayoflightissentalongthelinex−2y−3=0.Uponreachingtheline3x−2y−5=0therayisreflectedfromit.Findtheequationofthelinecontainingthereflectedray.

Ans.:29x−2y−31=0

5. Theextremitiesofthediagonalsofasquareare(1,1)and(−2,−1).Obtaintheequationoftheotherdiagonal.

Ans.:6x+4y+3=0

6. Thestraightline3x+4y=5and4x−3y=15intersectatthepointA.Onthisline,thepointsBandCarechosensothatAO=AC.FindthepossibleequationsofthelineBCpassingthroughthepoint(1,2).

Ans.:x−7y+13=0and7x+y−9=0

7. Theconsecutivesidesofaparallelogramare4x+5y=0and7x+2y=0.Iftheequationofonediagonalis11x+7y=9,findtheequationoftheotherdiagonal.

Ans.:x−y=0

8. Showthatthelinesax±by±c=0enclosearhombusofarea

9. IftheverticesofaΔOBCareO(0,0),B(−3,−1),andC(−1,−3),findtheequationoftheline

paralleltoBCandintersectingsidesOBandOCwhoseperpendiculardistancefrom(0,0)is .

Ans.:2x+2y+ =0

10. Findthelocusofthefootoftheperpendicularfromtheoriginuponthelinejoiningthepoints(acosθ,bsinθ)and(−asinθ,bcosθ)whereaisavariable.

Ans.:a2x2+b2y2=2(x2+y2)2

11. Showthatthelocusgivenbyx+y=0,(a-b)x+(a+b)y=2aband(a+b)x+(a−b)y=2ab

formanisoscelestrianglewhoseverticalangleis Determinethecentroidofatriangle.

Ans.:

12. Thesidesofaquadrilateralhavetheequations,x+2y=3,x=1,x−3y=4,and5x+y+12=0.Showthatthediagonalsofthequadrilateralareatrightangles.

13. GivennstraightlinesandafixedpointO.ThroughOastraightlineisdrawnmeetingtheselines

inthepointA1,A2,…,AnandapointAsuchthat Provethatthelocus

ofthepointAisastraightline.14. Findtheequationofthelinejoiningthepoint(3,5)tothepointofintersectionofthelines4x+y−

1=0and7x−3y−35=0andprovethatthelineisequidistantfromtheoriginandthepointsA,B,C,andD.

15. Findtheequationofthelinepassingthroughthepoint(2,3)andmakinginterceptsoflength2unitsandbetweenthelines.

Ans.:3x+4y−8=0andx−2=0

16. Ifxcosα+ysinα=pwhere beastraightline,provethattheperpendicularsp1,p2,and

p3onthelinefromthepoint(m2,2m),(mm′,m+m′),and(m′2,2m′),respectively,areinG.P.

17. Provethatthepoints(a,b),(c,d),and(a−c,b−d)arecollinearif(ad=bc).Also,showthatthestraightlinepassingthroughthesepointspassesthroughtheorigin.

18. Onediagonalofasquareisalongtheline8x−15y=0andoneofitsverticesis(1,2).Findtheequationsofthesidesofthesquarethroughthisvertex.

Ans.:2x+y=4andx−2y+3=0

19. Findtheorthocentreofatriangleformedbylineswhoseequationsarex+y=1,2x+8y=6,and4x−y+4=0.

Ans.:

20. Thesidesofatriangleareur=xcosαr+ysinαr−pr=0,r=1,2,3.Showthatitsorthocentreisgivenbyu1cos(α2−α3)=u2cos(α3−α1)=u3cos(α1−α2).

21. Findtheequationofstraightlinespassingthroughthepoint(2,3)andhavinganinterceptoflength2unitsbetweenthestraightlines2x+3y=3and2x+y=5.

Ans.:x=2,3x+4y=18

22. LetalineLhasinterceptsaandbonthecoordinateaxes.Whentheaxesarerotatedthroughanangle,keepingtheoriginfixed,thesamelineLhasinterceptspandq.Obtaintherelationbetweena,b,p,andq.

Ans.:

23. AlinethroughthevariablepointA(k+1,2k)meetstheline7x+y−16=0,5x−y+8=0,x−5y+8=0atB,C,andD,respectively.ProvethatAC,AB,andADareinG.P.

24. Findtheequationofthestraightlinespassingthrough(−2,−7)andhavinganinterceptoflength3betweenthestraightlines4x+3y=12and4x+3y=3.

Ans.:x+2=0,7x−24y+182=0

25. Alineissuchthatitssegmentbetweenthestraightlines5x−y−4=0and3x+4y−4=0isbisectedatthepoint(1,5).Obtainitsequation.

Ans.:83x−35y+92=0.

26. Provethatthe(a−b)x+(b−c)y+(c−a)=0,(a−c)x+(c−a)y+(a−b)=0,and(c−a)x+(a−b)y+(b−c)=0areconcurrent.

27. Twoverticesofatriangleare(5,−1)and(−2,3).Iftheorthocentreofthetriangleisattheorigin,findthecoordinatesofthethirdvertex.

Ans.:(−4,−7)

28. Alineintersectsx-axisatA(7,0)andy-axisatB(0,−5).AvariablelinePQwhichisperpendiculartoABintersectsx-axisatPandy-axisatQ.IfAQandBPintersectatR,thenfindthelocusofR.

Ans.:x2+y2−7x+5y=0

29. ArectanglePQRShasitssidePQparalleltotheliney=mxandverticesP,Q,andSonthelinesy=a,x=b,andx=−b,respectively.FindthelocusofthevertexR.

Ans.:(m2−1)x−my+b(m2+1)+am=0

30. Determinetheconditiontobeimposedonβsothat(O,β)shouldbeonorinsidethetrianglehavingsidesy+3x+2=0,3y−2x−5=0,and4y+x−14=0.

Ans.:

31. Showthatthestraightlines7x−2y+10=0,7x+2y−10=0,andy=2formanisoscelestriangleandfinditsarea.

Ans.:14sq.units

32. TheequationsofthesidesBC,CA,andABofatriangleABCareKr=arx+bry+cr=0,r=1,2,3.ProvethattheequationofalinedrawnthroughAparalleltoBCisK3(a2b1−a1b2)=K2(a3b1−a1b3).

33. ThesidesofatriangleABCaredeterminedbytheequationur=arx+bry+cr=0,r=1,2,3.ShowthatthecoordinatesoftheorthocentreofthetriangleABCsatisfytheequationλ1u1=λ2u2+λ3u3whereλ1=a2a3+b2b3,λ2=a3a1+b3b1,andλ3=a1a2+b1b2.

34. ProvethatthetwolinescanbedrawnthroughthepointP(P,Q)sothattheirperpendiculardistancesfromthepointQ(2a,2a)willbeequaltoaandfindtheirequations.

Ans.:y=a,4x−3y+3a=0.

35. Findthelocusofapointwhichmovessuchthatthesquareofitsdistancefromthebaseofanisoscelestriangleisequaltotherectangleunderitsdistancesfromtheothersides.

Ans.:

36. Provethatthelinesgivenby(b+c)x−bcy=a(b2+bc+c2),(c+a)x−cay=b(c2+ca+a2),

and(a+b)x−aby=c(a2+ab+b2)areconcurrent.37. Showthattheareaofthetriangleformedbythelinesy=m1x+c1,y=m2x+c2,andy=m3x+

c3is

38. Findthebisectoroftheacuteanglebetweenthelines3x+4y=1whichisthebisectorcontainingtheorigin.

Ans.:11x+3y−17=0(originliesintheobtuseanglebetweenthelines.)

39. Ifa1a2+b1b2>0provethattheoriginliesattheobtuseanglebetweenthelinesa1x+b1y+c1=0anda2x+b2y+c2=0,wherec1andc2bothbeingofthesamesign.

40. Findtheequationtothediagonalsoftheparallelogramformedbythelinesax+by+c=0,ax+by

+d=0,a′x+b′y+c′=0,a′x+b′y−d′=0.Showthattheparallelogramwillbearhombusif(a2

+b2)(c′−d′)2=(a′2+b′2)(c−d)2.41. AvariablelineisataconstantdistancepfromtheoriginandmeetscoordinateaxesinAandB.

ShowthatthelocusofthecentroidoftheΔOABisx−2+y−2=p−2.42. Amovinglineislx+my+n=0wherel,m,andnareconnectedbytherelational+bm+cn=0,

anda,b,andcareconstants.Showthatthelinepassesthroughafixedpoint.43. Findtheequationofbisectorofacuteanglebetweenthelines3x−4y+7=0and12x+5y−2=0.44. Qisanypointonthelinex−a=0andOistheorigin.IfAisthepoint(a,0)andQR,thebisector

meetsx-axisonR.ShowthatthelocusofthefootoftheperpendicularfromRtoOQisthe

(x−2a)(x2+y2+a2x)=0.45. Thelinesax+by+c=0,bx+cy+a=0,andcx+ay+b=0areconcurrentwherea,b,andcare

thesidesoftheΔABCinusualnotationandprovethatsin3A+sin3B+sin3C=3sinAsinBsinC.

46. AvariablestraightlineOPQpassesthroughthefixedpointO,meetingthetwofixedlinesinpointsPandQ.InthestraightlineOPQ,apointRistakensuchthatOP,OR,andOQareinharmonicprogression.ShowthatthelocusofpointQisastraightline.

47. Arayoflightissetalongthelinex−2y−3=0.Onreachingtheline3x−2y−5=0,therayisreflectedfromit.Findtheequationofthelinecontainingthereflectedray2qx−2y−31=0.

Ans.:2qx−2y−31=0.

48. LetΔABCbeatrianglewithAB=AC.IfDisthemidpointofBC,andEisthefootoftheperpendiculardrawnfromDtoACandFisthemidpointofBE.ProvethatAFisperpendiculartoBE.

Ans.:14x+23y−40=0.

49. TheperpendicularbisectorsofthesidesABandACofatriangleABCarex−y+5=0andx+2y=0,respectively.IfthepointAis(1,−2),findtheequationoftheline14x+23y−40=0.

50. Atriangleisformedbythelinesax+by+c=0,lx+my+n=0,andpx+qy+r=0.Showthat

thestraightline passesthroughtheorthocentreofthetriangle.

51. Provethatthediagonalsoftheparallelogramformedbythelinesax+by+c=0,ax+by+c′=

0,a′x+b′y+c=0,anda′x+b′y+c′=0willbeatrightanglesifa2+b2=a′2+b′2.

52. Onediagonalofasquareistheportionoftheline interceptedbetweentheaxes.Show

thattheextremitiesoftheotherdiagonalare

53. Showthattheoriginliesinsideatrianglewhoseverticesaregivenbytheequations7x−5y−11=0,8x+3y+31=0,andx+3y−19=0.

54. ArayoflighttravellingalongthelineOA,Obeingtheorigin,isreflectedbythelinemirrorx−y+1=0,thepointofincidenceAis(1,2).Thereflectedrayisagainreflectedbythemirrorx−y=1,thepointofincidencebeingB.IfthereflectedraymovesalongBC,findtheequationofBC.

Ans.:2x−y−6=0

55. Ifthelinesp1x+q1y=1,p2x+q2y=1,andp3x+q3y=1areconcurrent,provethatthepoints(p1,q1),(p2,q2),and(p3,q3)arecollinear.

56. Ifp,q,andrbethelengthoftheperpendicularsfromtheverticesA,B,andCofatriangleonany

straightline,provethata2(p–q)(p–r)+b2(q–r)(q–p)+c2(r–p)(r–q)=4Δ2.57. Provethattheareaoftheparallelogramformedbythestraightlinea1x+b1y+c1=0,a1x+b1y

+d1=0,a2x+b2y+c2=0,anda2x+b2y+d2=0is

58. Arayoflightissentalongtheline2x−3y=5.Afterrefractingacrossthelinex+y=1,itenterstheoppositesidesafterturningby15°awayfromthelinex+y=1.Findtheequationofthelinealongwhichtherefractedraytravels

Ans.:(15 −20)x−(30−10 )y+(11−18 )=0.

59. Twosidesofanisoscelestrianglearegivenbytheequations7x−y+3=0andx+y−7=0anditsthirdsidepassesthroughthepoint(1,−10).Determinetheequationofthethirdside.

Ans.:x−3y−31=0,3x+y+7=0.

60. Findallthosepointsonthelinex+y=4whichareatcunitdistancefromtheline4x+3y=10.61. Arethepoints(3,4)and(2,−6)onthesameoroppositesidesoftheline3x−4y=8?

Ans.:oppositesides

62. Howmanycirclescanbedrawneachtouchingallthethreelinesx+y=1,y=x,and7x−y=6?Findthecentreandradiusofoneofthecircles.

Ans.:Focus:(0,7)Incentre

63. Showthat beanypointonalinethentherangeofvaluesoftforwhichthe

pointpliesbetweentheparallellinesx+2y=1and2x+

64. Showthata,b,andcareanythreetermsofAPthenthelineax+by+c=0alwayspassesthroughafixedpoint.

65. Showthatifa,b,andcareinG.P.,thenthelineax+by+c=0formsatrianglewiththeaxes,whoseareaisaconstant.

Chapter3

PairofStraightLines

3.1INTRODUCTION

Weknowthateverylinearequationinxandyrepresentsastraightline.ThatisAx+By+C=0,whereA,BandCareconstants,representsastraightline.Considertwostraightlinesrepresentedbythefollowingequations:

Alsoconsidertheequation

If(x1,y1)isapointonthestraightlinegivenby(3.1)then

l1x1+m1y+n1=0Thisshowsthat(x1,y1)isalsoapointonthelocusof(3.3).Therefore,every

pointonthelinegivenby(3.1)isalsoapointonthelocusof(3.3).Similarly,everypointonthelinegivenby(3.2)isalsoapointonthelocusof(3.3).Therefore,(3.3)satisfiesallpointsonthestraightlinesgivenby(3.1)and(3.2).Hence,wesay(3.3)representsthecombinedequationofthestraightlinesgivenby(3.1)and(3.2).Itispossibletorewrite(3.3)as

Thepairofstraightlinesgivenby(3.1)and(3.2)isingeneralrepresentedintheform(3.4).However,wecannotsaythateveryequationofthisformwillrepresentapairofstraightlines.Wewillfindtheconditionthatisnecessaryandsufficientfortheequationoftheform(3.4)torepresentapairofstraightlines.Beforethatwewillseethateveryseconddegreehomogeneousequationinxandywillrepresentapairofstraightlines.

3.2HOMOGENEOUSEQUATIONOFSECONDDEGREEINxANDy

Everyhomogeneousequationofseconddegreeinxandyrepresentsapairofstraightlinespassingthroughtheorigin.Considertheequationax2+2hxy+by2=0,a≠0.

Dividingbyx2,weget Thisisaquadraticequationin

andhencetherearetwovaluesfor saym1andm2.Then

(i.e.)b(y−m1x)(y−m2x)=0.

Buty−m1x=0andy−m2x=0arestraightlinespassingthroughtheorigin.Therefore,ax2+2hxy+by2=0representsapairofstraightlinespassing

throughtheorigin.

Note3.2.1:ax2+2hxy+by2=b(y−m1x)(y−m2x)Equatingthecoefficientsofx2andxy,weget

3.3ANGLEBETWEENTHELINESREPRESENTEDBYax2+2hxy+by2=0

Lety–m1x=0andy−m2x=0bethetwolinesrepresentedbyax2+2hxy+by2

=0.

Letθbetheanglebetweenthelinesgivenbyax2+2hxy+by2=0.Thentheanglebetweenthelinesisgivenby

Thepositivesigngivestheacuteanglebetweenthelinesandthenegativesigngivestheobtuseanglebetweenthem.

Note3.3.1:Ifthelinesareparallelorcoincident,thenθ=0.Thentanθ=0.Therefore,from(3.7),wegeth2=ab.

Note3.1.3:Ifthelinesareperpendicularthen andsowegetfrom(3.7)

Thismeansa+b=0.Hence,theconditionforthelinestobe

parallelorcoincidentish2=abandtheconditionforthelinestobeperpendicularisa+b=0(i.e.)Coefficientofx2+Coefficientofy2=0.

3.4EQUATIONFORTHEBISECTOROFTHEANGLESBETWEENTHELINESGIVENBYax2+

2hxy+by2=0

Wewillnowderivetheequationforthebisectoroftheanglesbetweenthelinesgivenbyax2+2hxy+by2=0.Thecombinedequationofthebisectorsofthe

anglesbetweenthelinesgivenbyax2+2hxy+by2=0is

LetOAandOBbethetwolinesy−m1x=0andy−m2x=0representedbyax2

+2hxy+by2=0.LetthelinesOAandOBmakeanglesθ1andθ2withthex-axis.Then,weknowthat

LetθbetheanglemadebytheinternalbisectorOPwithOX.

Then istheanglemadebytheexternalbisectorOQwithOX.The

combinedequationofthebisectorsis

From(3.8)and(3.9),weget

Hence,thecombinedequationofthepairofbisectorsis

Aliter:Let(x1,y1)beapointonthebisectorOP.Then

Also,2θ=θ1+θ2.Accordingto(3.9)

From(3.9)and(3.10),

Thelocusof(x1,y1)is

Thisisthecombinedequationofthebisectors.

3.5CONDITIONFORGENERALEQUATIONOFASECONDDEGREEEQUATIONTOREPRESENTAPAIROFSTRAIGHTLINES

Wewillnowderivetheconditionforthegeneralequationofaseconddegreeequationtorepresentapairofstraightlines.Theconditionforthegeneralequationoftheseconddegreeax2+2hxy+by2+2gx+2fy+c=0torepresentapairofstraightlinesisabc+2fgh−af2−bg2−ch2=0.

Method1:Considerthegeneralequationoftheseconddegree

Letlx+my+n=0andl1x+m1y+n1=0betheequationsoftwolinesrepresentedby(3.11).Then

Comparingthecoefficients,weget

Weknowthat

Bymultiplyingthetwodeterminants,weget

Bymultiplyingthetwodeterminants,weget

Substitutingthevaluesfrom(3.12)in(3.13),weget

Expandingthedeterminant,weget

Thisistherequiredcondition.

Method2:

ax2+2hxy+by2+2gx+2fy+c=0Writingthisequationintheformby2+2hxy+2fy+(ax2+2gx+c)=0andsolvingforyweget

Thisequationwillrepresenttwostraightlinesifthequadraticexpressionundertheradicalsignisaperfectsquare.Theconditionforthisis4(hf−bg)2−4(h2−ab)(f2−bc)=0

Sinceb≠0,

abc+2fgh−af2−bg2−ch2=0Thisistherequiredcondition.

Method3:Lettheequationax2+2hxy+by2+2gx+2fy+c=0representapairofstraightlinesandlet(x1,y1)betheirpointofintersection.Shiftingtheorigintothepoint(x1,y1),weget

wherethenewaxesOXandOYareparallelto(Ox,Oy).

As(3.16)representsapairofstraightlinespassingthroughtheneworigin,ithastobeahomogeneousequationinXandY.Hence,

Substituting(3.17)and(3.18)in(3.20),weget

gx1+fy1+c=0(3.21)Eliminatingx1andy1from(3.17),(3.18)and(3.21),weget

Expanding,wegetabc+2fgh−af2−bg2−ch2=0.

Note3.5.1:Solving(3.17)and(3.18),weget

Hence,thepointofintersectionofthelinesrepresentedby(3.11)is

Note3.5.2:Iflx+my+n=0andl1x+m1y+n1=0arethetwostraightlinesrepresentedby(3.11),thenlx+my=0andl1x+m1y=0willrepresenttwostraightlinesparalleltothelinesrepresentedby(3.11)andpassingthroughtheorigin.Theircombinedequationis

Therefore,ifax2+2hxy+by2+2gx+2fy+c=0representsapairofstraightlines,thentheequationax2+2hxy+by2=0willrepresentapairoflinesparalleltothelinesgivenby(3.11).Weknowthateveryhomogeneousequationofseconddegreeinxandy

representsapairofstraightlinespassingthroughtheorigin.Wenowusethisideatogetthecombinedequationofthepairoflinesjoiningtheorigintothepointofintersectionofthecurveax2+2hxy+by2+2gx+2fy+c=0andthelinelx+my=1.

Theequationofthecurve

andtheline

lx+my=1.(3.23)willmeetattwopointssayPandQ.Let(x1,y1)beoneofthepointsofintersection,sayP.Then

andlx1+my1=1Letushomogenise(3.22)withthehelpof(3.23).Then,wewrite

Ifwesubstitutex=x1andy=y1in(3.25),weget

becauseof(3.23)and(3.24).ThereforeP(x1,y1)liesonthelocusof(3.25).SimilarlywecanshowthatthepointQ(x2,y2)alsoliesonthelocusof(3.25).However,aseconddegreehomogeneousequationrepresentsapairofstraightlinespassingthroughorigin.Hence,(3.25)isthecombinedequationofthepairoflinesOPandOQ.Hence,homogensingtheseconddegreeequation(3.22)withthehelpof

(3.23),wegetapairofstraightlinespassingthroughtheorigin.


Example3.1

Thegradientofoneofthelinesax2+2hxy+by2=0istwicethatoftheother.Showthat8h2=9ab.

Solution

Theequationax2+2hxy+by2=0representsapairofstraightlinespassingthroughtheorigin.Letthelinesbey−m1x=0andy−m2x=0.Then

ax2+2hxy+by2=b(y–m1x)(y=m2x)Equatingthecoefficientsofxyandx2onbothsides,weget

Here,ithasbeengiventhatm2=2m1.


Example3.2

Provethatoneofthelinesax2+2hxy+by2=0willbisectananglebetweenthecoordinateaxesif(a+b)2=4h2.

Solution

Lety–m1x=0andy–m2x=0bethetwolinesrepresentedbya2+2hxy+by2

=0.

Then

Sinceoneofthelinesbisectstheanglebetweentheaxes,wetakem1=±1.Then

Example3.3

Findthecentroidofthetriangleformedbythelinesgivenbytheequations12x2

–20xy+7y2=0and2x–3y+4=0.

Solution

Therefore,thesidesofthetrianglearerepresentedby

Thepointofintersectionofthelines(3.28)and(3.29)is(0,0).Letussolve(3.29)and(3.30).

Thus,thepointofintersectionofthesetwolinesis(7,6).Now,letussolve(3.28)and(3.30).

Thus,thepointofintersectionofthesetwolinesis(1,2).Then,thecentroidof

thetrianglewithvertices(0,0),(7,6)and(1,2)is (i.e)

Example3.4

Findtheproductofperpendicularsdrawnfromthepoint(x1,y1)onthelinesax2

+2hxy+by2=0.

Solution

Letthelinesbey–m1x=0andy−m2x=0.Then

Letp1andp2betheperpendiculardistancesfrom(x1,y1)onthetwolinesy–m1x=0andy–m2x=0,respectively.

Example3.5

Ifthelinesax2+2hxy+by2=0bethetwosidesofaparallelogramandthelinelx+my=1beoneofthediagonals,showthattheequationoftheotherdiagonalisy(bl–hm)y=(am–h)lx.Showthattheparallelogramisarhombusifh(a2–b2)=(a–h)lm.

Solution

ThediagonalACnotpassingthroughtheoriginislx+my=1.

TheequationofthelinesOAandOCbey−m1x=0andy−m2x=0.ThenthecorrespondingcoordinatesofAaregotbysolvingy–m1x=0andlx

+my=1.

Sincediagonalsbisecteachotherinaparallelogram,theequationofthediagonal

OBis

IfOABCisarhombus,thenthediagonalsareatrightanglesHence,theproductoftheirslopesis–1.

Example3.6

Provethattheareaofthetriangleformedbythelinesy=x+candthestraight

lines

Solution

Letthetwolinesrepresentedbyax2+2hxy+by2=0bey–m1x=0andy–m2x=0.Solvingtheequationsy–m1x=0andy=x+c,wegetthecoordinatesofAtobe

Example3.7

LandMarethefeetoftheperpendicularsfrom(c,0)onthelinesax2+2hxy+by2=0.ShowthattheequationofthelineLMis(a–b)x+2hy+bc=0.

Solution

LettheequationofLMbelx+my=1.SinceLandMarethefeetoftheperpendicularsfromA(c,0)onthetwolines

y−m1x=0andy–m2x=0,thepointsO,A,LandMareconcyclic.TheequationofthecirclewithOAasdiameterisx(x–c)+y2=0orx2+y2–cx=0.ThecombinedequationofthelinesOLandOMisgotbyhomogenisingthe

equationofthecirclewiththehelpoflinelx+my=1.Hence,thecombinedequationofthelinesOLandLMis

ButthecombinedequationofthelinesOLandOMis

ax2+2hxy+by2=0Boththeseequationsrepresentthesamelines.Thereforeidentifyingtheseequations,weget

Therefore,thelinelx+my–1=0is

(i.e.)(a–b)x+2hy+bc=0

Example3.8

Showthatfordifferentvaluesofpthecentroidofthetriangleformedbythestraightlinesax2+2hxy+by2=0arexcosα+ysinα=pliesonthelinex(atanα–h)+y(htanα–b)=0.

Solution

LetOAandOBbethelinesrepresentedbyax2+2hxy+by2=0andtheirequationsbey–m1x=0andy–m2x=0.TheequationofthelineABisxcosα+ysinα=p.ThecoordinatesofAare

ThecoordinatesofBare

Themidpoint(x1,y1)ofABis

Example3.9

Findtheconditionthatoneofthelinesgivenbyax2+2hxy+by2=0maybeperpendiculartooneofthelinesgivenbya1x2+2h1xy+by2=0.

Solution

Lety=mxbealineofax2+2hxy+by2=0.Then

Then

Hence,


Hence,therequiredconditionis(aa1–bb1)2+4(ha1+h1b)(bh1+a1h)=0.

Example3.10

Twosidesofatriangleliealongy2–m2x2=0anditsorthocentreis(c,d).Showthattheequationofitsthirdsideis(1–m2)(cx+dy)=c2–m2d2.

Solution

LetOA,OBandABbethelines

EquationofODisbx–ay=0.ThispassesthroughH(c,d).∴bc=ad.(1)EquationofAHis

ThecoordinatesofAare

Thatpointliesx–my=c–md

From(3.33),

Hence,theequationofthelineAB(ax+by=1)becomes(1–m2)(cx+dy)=c2

–m2d2.

Example3.11

Showthattheequationm(x3−3xy2)+y3–3x2y=0representsthreestraightlinesequallyinclinedtooneanother.

Solution

y3–3x2y=m(3xy2–x3)Dividingbyx3,weget

Thesevaluesofθshowthatthelinesareequallyinclinedtooneanother.

Example3.12

Showthatthestraightlines(A2–3B2)x2+8ABx+(B2–3A2)=0formwiththe

lineAx+By+C=0anequilateraltriangleofarea

Solution

Thesidesofthetrianglearegivenby

Theanglebetweenthelines(3.34)and(3.36)is

Similarlytheanglebetweenthelines(3.35)and(3.36)is

Sincethethreesidesformatriangle, istheonlypossibility.

Hence,thetriangleisequilateral.

Example3.13

Showthattwoofthestraightlinesax3+bx2y+cxy2+dy3=0willbeperpendiculartoeachotherifa2+d2+bd+ac=0.

Solution

ax3+bx2y+cxy2+dy3=0Thisbeingathirddegreehomogeneousequation,itrepresentsthreestraightlinespassingthroughorigin.Letthethreelinesbey–m1x=0,y–m2x=0andy–m3x=0.Ifmistheslopeofanylinethen


Sincem3isarootof(3.37)

Exercises

1. Showthattheequationofpairoflinesthroughtheoriginandperpendiculartothepairoflinesax2

+2hxy+by2=0isbx2–2hxy+ay2=0.2. ThroughapointAonthex-axis,astraightlineisdrawnparalleltothey-axissoastomeetthepair

ofstraightlinesax2+2hxy+by2=0inBandC.IfAB=BC,provethat8h2=9ab.3. FromapointA(1,1),straightlinesALandAMaredrawnatrightanglestothepairofstraightlines

3x2+7xy–2y2=0.FindtheequationofthepairoflinesALandAM.AlsofindtheareaofthequadrilateralALOMwhereOistheoriginofthecoordinate.

4. Showthattheareaofthetriangleformedbythelinesax2+2hxy+by2=0andlx+my=1is

5. Showthattheorthocentreofthetriangleformedbythelinesax2+2hxy+by2=0andlx+my=1

isgivenby

6. Showthatthecentroid(x1,y1)ofthetriangleformedbythelinesax2+2hxy+by2=0andlx+

my=1is

7. Atrianglehasthelinesax2+2hxy+by2=0fortwoofitssidesandthepoint(c,d)forits

orthocentre.Provethattheequationofthethirdsideis(a+b)(cx+dy)=ad2–2hbd+bc2.

8. Iftheslopeofoneofthelinesgivenbyax2+2hxy+by2=0isktimestheother,provethat4kh2

=abc(1+k)2.9. Ifthedistanceofthepoint(x1,y1)fromeachoftwostraightlinesthroughtheoriginisd,prove

thattheequationofthestraightlinesis(x1y–xy1)2=d2(x2+y2).

10. Astraightlineofconstantlength2lhasitsextremitiesoneoneachofthestraightlinesax2+2hxy

+by2=0.Showthatthelineofmidpointis(ax+by)2(hx+by)+(ab–h2)2l2a.

11. Provethatthetriangleformedbythelinesax2+2hxy+by2=0andlx+my=1isrightangledif

(a+b)/al2+2hlm+bm2)=0.

12. Showthatiftwoofthelinesax3+bx2y+cxy2+dy3=0makecomplementaryangleswithx-axisinanticlockwisedirection,thena(a–c)+d(b–d)=0.

13. Iftheslopeofthelinesgivenbyax2+2hxy+by2=0isthesquareoftheother,showthatab(a+

h)–6ahb+8h3=0.

14. Showthatthelineax+by+c=0andthetwolinesgivenby(ax+by)2=3(bx–ay)2forman

equilateraltriangleofarea

15. Ifoneofthelinegivenbyax2+2hxy+by2=0iscommonwithoneofthelinesofa1x2+2h1xy

+b1y2=0.showthat(ab1–a1b)

2+4(ah1–a1h).(bh1–b1h)=0.

16. Apointmovessothatitsdistancebetweenthefeetoftheperpendicularsfromitonthelinesax2+

2hxy+by2=0isaconstant2k.Showthatthelocusofthepointis(x2+y2)(h2–ab)=k2[(a–b)2

+4h2].17. Showthatthedistancefromtheorigintotheorthocentreofthetriangleformedbythelines

andax2+2hxy+by2=0is

18. Aparallelogramisformedbythelinesax2+2hxy+by2=0andthelinesthrough(p,q)paralleltothem.Showthattheequationofthediagonalnotpassingthroughtheoriginis(2x–p)(ap+hq)+(2y–q)(hp+bq)=0.

19. Ifthelinesgivenbylx+my=1andax2+2hxy+by2=0formanisoscelestriangle,showthat

h(l2–m2)=lm(a–b).

Example3.14

Findλsothattheequationx2+5xy+4y2+3x+2y+λ=0representsapairoflines.Findalsotheirpointofintersectionandtheanglebetweenthem.

Solution

Considertheseconddegreetermsx2+5xy+4y2.

x2+5xy+4y2=(x+y)(x+4y)Letthetwostraightlinesbex+y+l=0andx+4y+m=0.Then

Equatingthecoefficientsofx,yandconstantterms,weget

Solving(3.40)and(3.41),weget

From(3.42),

Thenthetwolinesare and3x+12y+

10=0.

Theanglebetweenthelinesisgivenby

Example3.15

Findthevalueofλsothattheequationλx2–10xy+12y2+5x–16y–3=0representsapairofstraightlines.Findalsotheirpointofintersection.

Solution

λx2–10xy+12y2+5x–16y–3=0Comparingwiththeequationax2+2hxy+by2+2gx+2fy+c=0wegeta=λ,2h=−10,b=12,2g=5,2f=–16,c=–3

Theconditionforthegivenequationtorepresentapairofstraightlinesisabc+2fgh–af2–bg2–ch2=0

–36λ+200–64λ–75+75=0⇒λ=2Then2x2–10xy+12y2+5x–16y–3=(2x–4y+l)(x−3y+m)Equatingthecoefficientsofx,yandconstantterms,

Therefore,thetwolinesarex–2y+3=0and2x−6y−1=0.Solvingthesetwo

equations,wegetthepointofintersectionas

Example3.16

Findthevalueofλsothattheequationx2−λxy+2y2+3x−5y+2=0representsapairofstraightlines.

Solution

Example3.17

Provethatthegeneralequationoftheseconddegreeax2+2hxy+by2+2gx+2fy+c=0representsparallelstraightlinesifh2=abandbg2=af2.Provethat

thedistancebetweenthetwostraightlinesis

Solution

Lettheparallellinesbelx+my+n=0andlx+my+n1=0.Thenax2+2hxy+by2+2gx+2fy+c=(lx+my+n)(lx+my+n1)

Equatingtheliketerms,weget

Also,thedistancebetweenthelineslx+my+n=0andlx+my+n1=0is

Example3.18

Ifax2+2hxy+by2+2gx+2fy+c=0representstwostraightlinesequidistantfromtheorigin,showthatf4−g4=c(bf2−ag2).

Solution

Letthetwolinesrepresentedbythegivenequationbelx+my+n=0andl1x+m1y+n1=0.Then

Perpendiculardistancesfromtheorigintothetwolinesareequal.Therefore,

Squaring

Example3.19

Iftheequationax2+2hxy+by2+2gx+by2+2gx+2fy+c=0representstwostraightlines,provethattheproductofthelengthsoftheperpendicularsfromthe

originonthestraightlinesis

Solution

Letthetwolinesbelx+my+n=0andl1x+my+n=0.Therefore

Theproductoftheperpendicularsfromtheoriginontheselines

Example3.20

Ifax2+2hxy+by2+2gx+by2+2gx+2fy+c=0representstwostraightlines,provethatthesquareofthedistanceoftheirpointofintersectionfromtheorigin

is Further,ifthetwogivenlinesareperpendicular,thenprove

thatthedistanceoftheirpointofintersectionfromtheoriginis

Solution

Letthetwostraightlinesbelx+my+n=0andl1x+m1y+n1=0.

Theirpointofintersectionis

Hence,thedistanceofthispointfromtheoriginisgivenby

Ifthelinesareperpendicularthen(a+b)=0.Then

Example3.21

Showthatthelinesgivenby12x2+7xy−12y2=0and12x2+7xy−12y2−x+7y−1=0arealongthesidesofasquare.

Solution

Theseconddegreetermsin(3.42)and(3.43)arethesame.Thisimpliesthatthetwolinesrepresentedby(3.42)areparalleltothetwolinesrepresentedby(3.43).Hence,thesefourlinesfromaparallelogram.Also,ineachoftheequationscoefficientofx2+coefficientofy2=0.Hence,eachequationformsapairofperpendicularlines.Thus,thefourlines

formarectangle.Thetwolinesrepresentedby(3.42)are3x+4y=0and4x−3y=0.Thetwolinesrepresentedby(3.43)are3x+4y−1=0and4x−3y+1=0.

Theperpendiculardistancebetween2x+4y=0and3x+4y−1=0is .

Theperpendiculardistancebetween4x−3y=0and4x−3y+1=0is .

Hence,thefourlinesformasquare.

Exercises

1. Showthattheequation6x2+17xy+12y2+22x+31y+20=0representsapairofstraightlinesandfindtheirequations.

Ans.:2x+3y+4=03x+4y+5=0

2. Provethattheequations8x2+8xy+2y2+26x+13y+15=0representstwoparallelstraightlinesandfindthedistancebetweenthem.

Ans.:

3. Provethattheequation3x2+8xy−7y2+21x−3y+18=0representstwolines.Findtheirpointofintersectionandtheanglebetweenthem.

Ans.:

4. Ifax2+2hxy+by2+2gx+2fy+c=0andax2+2hxy+by2−2gx−2fy+c=0eachrepresentsa

pairoflines,provethattheareaoftheparallelogramenclosedis

5. Showthattheequation3x2+10xy+8y2+14x−22y+15=0representstwostraightlines

intersectingatanangle

6. Theequationax2−2xy−2y2−5x+5y+c=0representstwostraightlinesperpendiculartoeachother.Findaandc.

Ans.:a=2,c=−3

7. Findthedistancebetweentheparallellinesgivenby4x2+12xy+9y2−6x−9y+1=0.

Ans.:

8. Showthatthefourlines2x2+3xy−2y2=0and2x2+3xy−2y2−3x+y+1=0formasquare.

9. Showthatthestraightlinesrepresentedbyax2+2hxy+by2=0andthoserepresentedbyax2+

2hxy+by2+2gx+2fy+c=0formarhombus,if(c−h)fg+h(f2−g2)=0.

10. Iftheequationax2+2hxy+by2+2gx+2fy+c=0representstwostraightlinesandparallellinestothesetwolinesaredrawnthroughtheoriginthenshowthattheareaoftheparallelogramso

formedis

11. Ifthestraightlinesgivenbyax2+2hxy+by2+2gx+2fy+c=0intersectsonthey-axisthen

showthat2fgh−hg2−ch2=0.

12. Aparallelogramissuchthattwoofitsadjacentsidesarealongthelinesax2+2hxy+by2=0anditscentreis(a,b).Findtheequationoftheothertwosides.

Ans.:a(x−2a)+2h(x−2a)(y−2b)+b(y−2b)2=0

Example3.22

Showthatthepairoflinesgivenby(a−b)(x2−y2)+4hxy=0andthepairoflinesgivenbyh(x2−y2)=(a−b)xyaresuchthateachpairbisectstheanglebetweentheotherpairs.

Solution

Thecombinedequationofthebisectorsofthepairoflinesgivenby(3.44)is

(i.e.)h(x2−y2)=xy(a−b)

whichis(3.45).Thecombinedequationofthebisectorsoftheanglebetweenlinesgivenby

(3.45)is

whichis(3.44).Hence,eachpairbisectstheanglebetweentheother.

Example3.23

Ifthebisectorsofthelinex2−2pxy−y2=0arex2−2qxy−y2=0showthatpq+1=0.

Solution

Thecombinedequationofthebisectorsof(3.46)is

Butequationofthebisectorisgivenby

x2−2qxy−y2=0(3.47)

Comparing(3.46)and(3.47),weget

∴pq+1=0

Example3.24

Provethatifoneofthelinesgivenbytheequationax2+2hxy+by2=0bisectstheanglebetweenthecoordinateaxesthen(a+b)2=4h2.

Solution

Thebisectorsofthecoordinateaxesaregivenbyy=xandy=−x.Ify=xisoneofthelinesofax2+2hxy+by2=0thenax2+2hx2+bx2=0.

(i.e.)a+b=–2hIfy=–xisoneofthelinesofax2+2hxy+by2=0,thena+b=2h.Fromthesetwoequations,weget(a+b)2=4h2.

Example3.25

Showthattheliney=mxbisectstheanglebetweenthelinesax2+2hxy+by2=0ifh(1−m2)+m(a−b)=0.

Solution

Thecombinedequationofthebisectorsoftheanglesbetweenthelinesax2−2hxy+by2=0is

Ify=mxisoneofthebisectors,thenithastosatisfytheaboveequation.

Example3.26

Showthatthepairofthelinesgivenbya2x2+2h(a+b)xy+b2y2=0isequallyinclinedtothepairgivenbyax2+2hxy+by2=0.

Solution

Inordertoshowthatthepairoflinesgivenbya2x2+2h(a+b)xy+b2y2=0isequallyinclinedtothepairoflinesgivenbyax2+2hxy+by2=0,wehavetoshowthatboththepairshavethesamebisectors.Thecombinedequationsofthe

bisectorsofthefirstpairoflinesis whichisthe

combinedequationofthesecondpairoflines.

Exercises

1. Ifthepairoflinesx2−2axy−y2=0bisectstheanglesbetweenthelinesx2−2pxy−y2=0thenshowthatthelatterpairalsobisectstheanglebetweentheformerpair.

2. Ifoneofthebisectorsofax2+2hxy+by2=0passesthroughthepointofintersectionofthelines

ax2+2hxy+by2+2gx+2fy+c=0thenshowthath(f2−g2)+(a−b)fg=0.

3. Ifthepairofstraightlinesax2+2hxy+by2=0andbx2+2gxy+by2=0besuchthateachbisectstheanglebetweentheotherthenprovethathg−b=0.

4. Provethattheequations6x2+xy−12y2−14x+47y−40=0and14x2+xy−4y2−30x+15y=0representtwopairsoflinessuchthatthelinesofthefirstpairareequallyinclinedtothoseofthesecondpair.

5. Provethattwoofthelinesrepresentedbytheequationax4+bx2y+cx2y2+dxy3+ay4=0willbisecttheanglebetweentheothertwoifc+ba=0andb+d=0.

Example3.27

Ifthestraightlinesjoiningtheorigintothepointofintersectionof3x2−xy+3y2

+2x−3y+4=0and2x+3y=kareatrightangles,provethat6k2−5k+52=0.

Solution

Let

Thecombinedequationofthelinesjoiningtheorigintothepointofintersectionofthelinesgiven(3.48)and(3.49)isgotbyhomogenising(3.48)withthehelpof(3.49).Hence,thecombinedequationofthelinesjoiningtheorigintothepointsofintersectionof(3.48)and(3.49)is

Sincethetwostraightlinesareatrightangles,coefficientofx2+coefficientofy2=0

Example3.28

Showthatthepairofstraightlinesjoiningtheorigintothepointofintersectionofthestraightlinesy=mx+candthecirclex2+y2=a2areatrightangles2c2

=a2(1+m2).

Solution

Itisgiventhatx2+y2=a2andy=mx+c.

ThecombinedequationofthelinesOPandOQisgivenby

SinceOPandOQareatrightangles,coefficientofx2+coefficientofy2=0

c2−m2a2+c2−a2=0⇒2c2=a2(1+m2)

Example3.29

Showthatthejoinoforigintotheintersectionofthelines2x2−7xy+3y2+5x+10y−25=0andthepointsatwhichtheselinesarecutbythelinex+2y−5=0aretheverticesofaparallelogram.

Solution

Letequation(3.50)representsthelinesCAandCBand(3.51)representsthelineAB.ThecombinedequationofthelinesOAandOBisgotbyhomogeniousing

(3.50)withthehelpof(3.51).

Sincetheseconddegreetermsin(3.50)and(3.52)arethesamethetwolinesrepresentedby(3.50)areparalleltothetwolinesrepresentedby(3.52).Therefore,thefourlinesformaparallelogram.

Example3.30

Ifthechordofthecirclex2+y2=a2whoseequationislx+my=1subtendsanangleof45°attheoriginthenshowthat4[a2(l2+m2)−1]=[a2(l2+m2)−2]2.

Solution

Itisgiventhat,

ThecombinedequationofthelinesOPandOQis

Then

Example3.31

Findtheequationtothestraightlinesjoiningtheorigintothepointof

intersectionofthestraightline andthecircle5(x2+y2+ax+by)=9ab

andfindtheconditionsthatthestraightlinesmaybeatrightangles.

Solution

Itisgiventhat,

Thecombinedequationofthelinesjoiningtheorigintothepointsofintersectionof(3.53)and(3.54)is

Sincethelinesareatrightangles,coefficientofx2+coefficientofy2=0

Example3.32

Thelinelx+my=1meetsthecirclex2+y2=a2inPandQ.IfOistheorigin

thenshowthat .

Solution

Theperpendicularfromtheorigintotheline OP=a

Example3.33

Thestraightliney−k=m(x+2a)intersectsthecurvey2=4a(x+a)inAandC.

Showthatthebisectorsofangle ,‘O’beingtheorigin,arethesameforall

valuesofm.

Solution

Let

ThecombinedequationofthelinesOAandOBis

Thecombinedequationofthebisectorsis

Example3.34

Provethatifallchordsofax2+2hxy+by2+2gx+2fy+c=0subtendarightangleattheorigin,thentheequationmustrepresenttwostraightlinesatrightanglesthroughtheorigin.

Solution

Lettheequationofthechordbe

lx+my=1(3.56)Letthelines(3.55)and(3.56)intersectatPandQ.ThecombinedequationofOPandOQisax2+2hxy+by2+(2gx+2fy)(lx+my)+c(lx+my)2=0.

Since ,coefficientofx2+coefficientofy2=0.

(a+2gl+cl2)+(b+2fm+cm2)=0

Sincelandmarearbitrary,coefficientsofl1,l2,m1,m2andtheconstanttermvanishseparately.Sinceg=0,f=0,c=0anda+b=0.Hence,equation(3.55)becomesax2+2hxy+by2=0whichisapairof

perpendicularlinesthroughtheorigin.

Exercises

1. Showthatthelinejoiningtheorigintothepointscommonto3x2+5xy+3y2+2x+3y=0and3x−2y=1areatrightangles.

2. Ifthestraightlinesjoiningtheorigintothepointofintersection3x2−xy+3y2+2x−3y+4=0

and2x+3y=kareatrightanglesthenshowthat6k2−5k+52=0.

3. Showthatallthechordsofthecurve3x2−y2−2x+y=0whichsubtendarightangleattheoriginpassthroughafixedpoint.

4. Ifthecurvex2+y2+2gx+2fy+c=0interceptsonthelinelx+my=1,whichsubtendsaright

angleattheoriginthenshowthata(l2+m2)+2(gl+fm+1)=0.5. Ifthestraightlinesjoiningtheorigintothepointofintersectionofthelinekx+hy=2hkwiththe

curve(x−h)2+(y−k)2=a2areatrightanglesattheoriginshowthath2+k2=a2.

6. Provethatthetriangleformedbythelinesax2+2hxy+by2=0andlx+my=1isisoscelesif(l2

−m2)h=(a−b)lm.

7. Provethatthepairoflinesjoiningtheorigintotheintersectionofthecurves bythe

linelx+my+n=0arecoincidentifa2l2+b2m2=n2.

8. Showthatthestraightlinesjoiningtheorigintothepointofintersectionofthecurvesax2+2hxy

+by2+2gx=0anda1x2+2h1xy+b1y

2+2g1x=0willbeatrightanglesifg1(a1+b1)=g(h1+b1).

9. Showthattheanglebetweenthelinesdrawnfromtheorigintothepointofintersectionofx2+2xy

+y2+2x+2y−5=0and3x−y+1=0is

Chapter4

Circle

4.1INTRODUCTION

Definition4.1.1:Acircleisthelocusofapointinaplanesuchthatitsdistancefromafixedpointintheplaneisaconstant.Thefixedpointiscalledthecentreofthecircleandtheconstantdistanceiscalledtheradiusofthecircle.

4.2EQUATIONOFACIRCLEWHOSECENTREIS(h,k)ANDRADIUSr

LetC(h,k)bethecentreofthecircleandP(x,y)beanypointonthecircle.CP=ristheradiusofthecircle.CP2=r2(i.e.)(x−h)2+(y−k)2=r2.Thisistheequationoftherequiredcircle.

Note4.2.1:Ifthecentreofthecircleisattheorigin,thentheequationofthecircleisx2+y2=r2.

4.3CENTREANDRADIUSOFACIRCLEREPRESENTEDBYTHEEQUATIONx2+y2+2gx+2fy+c=0

Addingg2+f2tobothsides,weget

Thisequationisoftheform(x−h)2+(y−k)2=r2,whichisacirclewithcentre(h,k)andradiusr.Thus,equation(4.1)representsacirclewhosecentreis(−g,

−f)andradius

Note4.3.1:Aseconddegreeequationinxandywillrepresentacircleifthecoefficientsofx2andy2areequalandthexytermisabsent.

Note4.3.2:

1. Ifg2+f2−cispositive,thentheequationrepresentsarealcircle.

2. Ifg2+f2−ciszero,thentheequationrepresentsapoint.

3. Ifg2+f2−cisnegative,thentheequationrepresentsanimaginarycircle.

4.4LENGTHOFTANGENTFROMPOINTP(x1,y1)TOTHECIRCLEx2+y2+2gx+2fy+c=0

ThecentreofthecircleisC(−g,−f)andradius .LetPTbethe

tangentfromPtothecircle.

Note4.4.1:

1. IfPT2>0thenpointP(x1,y1)liesoutsidethecircle.

2. IfPT2=0thenthepointP(x1,y1)liesonthecircle.

3. IfPT2<0thenpointP(x1,y1)liesinsidethecircle.

4.5EQUATIONOFTANGENTAT(x1,y1)TOTHECIRCLEx2+y2+2gx+2fy+c=0

Thecentreofthecircleis(−g,−f).Theslopeoftheradius

Hence,theequationoftangentat(x1,y1)is(y−y1)=m(x−x1)

Addinggx1+fy1+ctobothsides,

sincethepoint(x1,y1)liesonthecircle.Hence,theequationofthetangentat(x1,y1)isxx1+yy1+g(x+x1)+f(y+y1)

+c=0.

4.6EQUATIONOFCIRCLEWITHTHELINEJOININGPOINTSA(x1,y1)ANDB(x2,y2)ASTHEENDSOFDIAMETER

A(x1,y1)andB(x2,y2)aretheendsofadiameter.LetP(x,y)beanypointonthe

circumferenceofthecircle.Then (i.e.)AP⊥PB.

TheslopeofAPis theslopeofBPis

SinceAPisperpendiculartoPB,m1m2=−1

Thisistherequiredequationofthecircle.

4.7CONDITIONFORTHESTRAIGHTLINEy=mx+cTOBEATANGENTTOTHECIRCLEx2+

y2=a2

Method1:Thecentreofthecircleis(0,0).Theradiusofthecircleisa.Ify=mx+cisatangenttothecircle,theperpendiculardistancefromthecentreonthestraightliney=mx+cistheradiusofthecircle.


Method2:Theequationofthecircleis

x2+y2=a2(4.2)Theequationofthelineis

y=mx+c(4.3)Thex-coordinatesofthepointofintersectionofcircle(4.2)andline(4.3)aregivenby

Ify=mx+cisatangenttothecircle,thenthetwovaluesofxgivenbyequation(4.4)areequal.Theconditionforthisisthediscriminantofquadraticequation(4.4)iszero.


Note4.7.1:Anytangenttothecirclex2+y2=a2isoftheform

4.8EQUATIONOFTHECHORDOFCONTACTOFTANGENTSFROM(x1,y1)TOTHECIRCLE

x2+y2+2gx+2fy+c=0

LetQRbethechordofcontactoftangentsfromP(x1,y1).LetQandRbethepoints(x2,y2)and(x3,y3),respectively.TheequationsoftangentsatQandRare

xx2+yy2+g(x+x2)+f(y+y2)+c=0xx3+yy3+g(x+x3)+f(y+y3)+c=0

ThesetwotangentspassthroughthepointP(x1,y1).Therefore,x1x2+y1y2+g(x1+x2)+f(y1+y2)+c=0and

x1x3+y1y3+g(x+x3)+f(y+y3)+c=0

Thesetwoequationsshowthatthepoints(x2,y2)and(x3,y3)lieonthestraightline

xx1+yy1+g(x+x1)+f(y+y1)+c=0

Hence,theequationofthechordofcontactfrom(x1,y1)is

xx1+yy1+g(x+x1)+f(y+y1)+c=0

4.9TWOTANGENTSCANALWAYSBEDRAWNFROMAGIVENPOINTTOACIRCLEAND

THELOCUSOFTHEPOINTOFINTERSECTIONOFPERPENDICULARTANGENTSISACIRCLE

Lettheequationofthecirclebe

x2+y2=a2(4.6)Let(x1,y1)beagivenpoint.Anytangenttothecirclex2+y2=a2is

Ifthistangentpointsthrough(x1,y1),then

Thisisaquadraticequationinm.Hence,therearetwovaluesform,andforeachvalueofmthereisatangent.Thus,therearetwotangentsfromagivenpointtoacircle.Let(x1,y1)bethepointofintersectionofthetwotangentsfrom(x1,y1).Ifm1andm2aretheslopesofthetwotangents,then

Ifthetwotangentsareperpendicular,thenm1m2=−1.

Thelocusof(x1,y1)isx2+y2=a2+b2,whichisacircle.

4.10POLEANDPOLAR

Definition4.10.1:Thepolarofapointwithrespecttoacircleisdefinedtobethelocusofthepointofintersectionoftangentsattheextremitiesofavariablechordthroughthatpoint.Thepointiscalledthepole.

4.10.1PolarofthePointP(x1,y1)withRespecttotheCirclex2+y2+2gx+2fy

+c=0

Lettheequationofcirclebe

LetQRbeavariablechordthroughthepointP(x1,y1).LetthetangentsatQandRtothecircleintersectatT(h,k).Then,QRisthechordofcontactofthetangentsfromT(h,k).Itsequationis

xh+yk+g(x+h)+f(y+k)+c=0

ThischordpassesthroughP(x1,y1).Therefore,

Thelocusof(h,k)is

Hence,thepolarof(x1,y1)isxx1+yy1+g(x+x1)+f(y+y1)+c=0.

Note4.10.1.1:

1. Ifthepoint(x1,y1)liesoutsidethecircle,thepolarof(x1,y1)isthesameasthechordofcontactfrom(x1,y1).Ifthepointliesonthecircle,thenthetangentat(x1,y1)isthepolarofthepointP(x1,y1).

2. Thepoint(x1,y1)iscalledthepoleofthelinexx1+yy1+g(x+x1)+f(y+y1)+c=0.Line(4.12)iscalledthepolarofthepoint(x1,y1).

3. Thepolarof(x1,y1)withrespecttothecirclex2+y2=a2isxx1+yy1=a

2.

4.10.2PoleoftheLinelx+my+n=0withRespecttotheCirclex2+y2=a2

Let(x1,y1)bethepoleoftheline

lx+my+n=0(4.13)withrespecttothecirclex2+y2=a2.Then,thepolarof(x,y)is

xx1+yy1=a2(4.14)Equations(4.13)and(4.14)representthesameline.Therefore,identifyingthesetwoequations,weget

Hence,thepoleofthelinelx+my+n=0is

4.11CONJUGATELINES

Definition4.11.1:Twolinesaresaidtobeconjugatewithrespecttothecirclex2

+y2=a2ifthepoleofeitherlineliesontheotherline.

4.11.1ConditionfortheLineslx+my+n=0andl1x+m1y+n1=0tobeConjugateLineswithRespecttotheCirclex2+y2=a2

Thepoleofthelinelx+my+n=0is Sincethetwogivenlinesare

conjugatetoeachother,thispointliesonthelinel1x+m1y+n1=0.

4.12EQUATIONOFACHORDOFCIRCLEx2+y2+2gx+2fy+c=0INTERMSOFITSMIDDLEPOINT

LetPQbeachordofthecirclex2+y2+2gx+2fy+c=0andR(x1,y1)beitsmiddlepoint.Theequationofanychordthrough(x1,y1)is

Anypointonthislineisx=x1+rcosθ,y=y1+rsinθ.WhenthechordPQmeetsthecirclethispointliesonthecircle.Therefore,

ThevaluesofrofthisequationarethedistancesRPandRQ,whichareequalinmagnitudebutoppositeinsign.Theconditionforthisisthecoefficientofr=0.

Eliminatingcosθandsinθ,from(4.15)and(4.16),weget

Addinggx1+fy1+ctobothsides,weget

ThisistherequiredequationofthechordPQintermsofitsmiddlepoint(x1,y1).ThisequationcanbeexpressedintheformT=S1whereT=xx1+yy1+g(x

+x1)+f(y+y1)+cand

Note4.12.1:Tistheexpressionwehaveintheequationsofthetangent(x1,y1)tothecircleS:x2+y2+2gx+2fy+c=0andS1istheexpressionwegetbysubstitutingx=x1andy=y1intheleft-handsideofS=0.

4.13COMBINEDEQUATIONOFAPAIROFTANGENTSFROM(x1,y1)TOTHECIRCLEx2+y2+

2gx+2fy+c=0

Lettheequationofachordthrough(x1,y1)be

Anypointonthislineis(x1+rcosθ,y1+rsinθ).Ifthispointliesonthecirclex2+y2+2gx+2fy+c=0,then

Ifchord(4.17)isatangenttocircle(4.18),thenthetwovaluesofrofthisequationareequal.Theconditionforthisis

Butfrom(4.17)

Substitutingthisin(4.19),weget

Thisequationisthecombinedequationofthepairoftangentsfrom(x1,y1).

4.14PARAMETRICFORMOFACIRCLE

x=acosθ,y=asinθsatisfytheequationx2+y2=a2.Thispointisdenotedby‘θ’,whichiscalledaparameterforthecirclex2+y2=a2.

4.14.1EquationoftheChordJoiningthePoints‘θ’and‘ϕ’ontheCircleandtheEquationoftheTangentatθ

Thetwogivenpointsare(acosθ,asinθ)and(acosϕ,asinϕ).Theequationofthechordjoiningthesetwopointsis

Thischordbecomesthetangentat‘θ’ifϕ=0.Therefore,theequationofthetangentat‘θ’isxcosθ+ysinθ=a.


Example4.1

Findtheequationofthecirclewhosecentreis(3,−2)andradius3units.

Solution

Theequationofthecircleis

Example4.2

Findtheequationofthecirclewhosecentreis(a,−a)andradius‘a’.

Solution

Thecentreofthecircleis(a,−a).Theradiusofthecircleisa.Theequationofthecircleis(x−a)2+(y+a)2=a2(i.e.)x2−2ax+a2+y2+2ay+a2=a2(i.e.)x2+y2−2ax+2ay+a2=0.

Example4.3

Findthecentreandradiusofthefollowingcircles:

i. x2+y2−14x+6y+9=0

ii. 5x2+5y2+4x−8y−16=0

Solution

i.

ii.

Example4.4

Findtheequationofthecirclewhosecentreis(2,−2)andwhichpassesthroughthecentreofthecirclex2+y2−6x−8y−5=0

Solution

Thecentreoftherequiredcircleis(2,−2).Thecentreofthecirclex2+y2−6x−8y−5=0is(3,4).Theradiusoftherequiredcircleisgivenbyr2=(2−3)2+(−2−4)2=1+36=37.

Therefore,theequationoftherequiredcircleis(x−2)2+(y+2)2=37(i.e.)x2+y2−4x+4y−29=0

Example4.5

Showthattheline4x−y=17isadiameterofthecirclex2+y2−8x+2y=0.

Solution

Thecentreofthecirclex2+y2−8x+2y=0is(4,−1).Substitutingx=4andy=−1intheequation4x−y=17,weget16+1=17,whichistrue.Therefore,theline4x−y=17passesthroughthecentreofthegivencircle.Hence,thegivenlineisadiameterofthecircle.

Example4.6

Provethatthecentresofthecirclesx2+y2+4y+3=0,x2+y2+6x+8y−17=0andx2+y2−30x−16y−42=0arecollinear.

Solution

ThecentresofthethreegivencirclesareA(0,−2),B(−3,−4)andC(15,8).

TheslopeofABis

TheslopeofBCis

SincetheslopesABandBCareequalandBisacommonpoint,thepointsA,BandCarecollinear.

Example4.7

Showthatthepoint(8,9)liesonthecirclex2+y2−10x−12y+43=0andfindtheotherendofthediameterthrough(8,9).

Solution

Substitutingx=8andy=9inx2+y2−10x−12y+43=0,weget64+81−80−108+43=0(i.e.)188−188=0,whichistrue.Therefore,thepoint(8,9)liesonthegivencircle.Thecentreofthiscircleis(5,6).Let(x,y)betheotherendofthediameter.

Hence,theotherendofthediameteris(2,3).

Example4.8

Findtheequationofthecirclepassingthroughthepoints(1,1),(2,−1)and(3,2).

2).

Solution

Lettheequationofthecirclebex2+y2+2gx+2fy+c=0.Thecirclepassesthroughthepoints(1,1),(2,−1)and(3,2).

Fromequation(4.20),

Fromequation(4.20),−5−1+c=−2⇒c=4

Therefore,theequationofthecircleisx2+y2−5x−y+4=0.

Example4.9

Showthatthepoints(3,4),(0,5)(−3,−4)and(−5,0)areconcyclicandfindtheradiusofthecircle.

Solution

Lettheequationofthecirclebex2+y2+2gx+2fy+c=0.Thispassesthroughthepoints(3,4),(0,5)and(−3,−4).Therefore,

Fromequation(4.28),f=0Fromequation(4.25),c=25Hence,theequationofthecircleisx2+y2−25=0(4.30)Substitutingx=−5andy=0inequation(4.30),weget0+25−25=0,whichistrue.Therefore,(−5,0)alsoliesonthecircle.Hence,thefourgivenpointsareconcyclic.Thecentreofthecircleis(0,0)andtheradiusis5units.

Example4.10

Findtheequationofthecirclewhosecentreliesonthelinex=2yandwhichpassesthroughthepoints(−1,2)and(3,−2).

Solution

Lettheequationofthecirclebex2+y2+2gx+2fy+c=0.Thispassesthroughthepoints(−1,2)and(3,−2).Therefore,

Subtracting,weget

Substitutingthisinequation(4.33),weget

−2f+f=1⇒f=−1∴g=−2

From(4.31),4−4+c=−5⇒c=⇒−5Hence,theequationofthecircleisx2+y2−4x−2y−2y−5=0.

Example4.11

Findtheequationofthecirclecuttingoffintercepts4and6onthecoordinateaxesandpassingthroughtheorigin.

Solution

Lettheequationofthecirclebex2+y2+2gx+2fy+c=0.Thispassesthroughthepoints(0,0),(4,0)and(0,6).

Thus,theequationofthecircleisx2+y2−4x−6y=0.

Example4.12

Findtheequationofthecircleconcentricwithx2+y2−8x−4y−10=0andpassingthroughthepoint(2,3).

Solution

Twocirclesaresaidtobeconcentriciftheyhavethesamecentre.Therefore,theequationoftheconcentriccircleisx2+y2−8x−4y+k=0.Thiscirclepassesthrough(2,3).

∴4+9−16−12+k=0∴k=15Hence,theequationoftheconcentriccircleisx2+y2−8x−4y+15=0.

Example4.13

Findtheequationofthecircleonthejoiningthepoints(4,7)and(−2,5)astheextremitiesofadiameter.

Solution

Theequationoftherequiredcircleis(x−x1)(x−x2)+(y−y1)(y−y2)=0

Example4.14

Theequationoftwodiametersofacircleare2x+y−3=0andx−3y+2=0.Ifthecirclepassesthroughthepoint(−2,5),finditsequation.

Solution

Thecentreofthecircleisthepointofintersectionofthediameter.

Thecentreofthecircleisthepointofintersectionofthediameter.

Addingthesetwoequations,weget7x=7.∴x=1From(4.38),y=1.Hence,thecentreofthecircleis(1,1)andradiusis

Therefore,theequationofthecircleis

Example4.15

Findthelengthofthetangentfromthepoint(2,3)tothecirclex2+y2+8x+4y+8=0.

Solution

ThelengthofthetangentfromP(x1,y1)tothecirclex2+y2+2gx+2fy+c=0

isgivenby Here,thelengthofthetangentfrom

P(2,3)tothegivencircleis

Example4.16

Determinewhetherthefollowingpointslieoutside,onorinsidethecirclex2+y2

−4x+4y−8=0:A(0,1),B(5,9),C(−2,3).

Solution

Theequationofthecircleisx2+y2−4x+4y−8=0.

Therefore,pointAliesinsidethecircle.PointsBandClieoutsidethecircle.

Example4.17

Findtheequationofthetangentatthepoint(2,−5)onthecirclex2+y2−5x+y−14=0.

Solution

Givenx2+y2−5x+y−14=0

Therefore,theequationofthetangentis

Example4.18

Findthelengthofthechordofthecirclex2+y2−4x−6y−3=0giventhat(1,1)isthemidpointofachordofthecircle.

Solution

Centreofthecircleis(2,3)andradius PointM(1,1)isthe

midpointofthechordAB.

Therefore,thelengthofthechord units.

Example4.19

Showthatthecirclesx2+y2−2x+6y+6=0andx2+y2−5x+6y+15=0toucheachotherinternally.

Solution

Forthecirclex2+y2−2x+6y+6=0,

centreisA(1,−3)andradius units

Forthecirclex2+y2−5x+6y+15=0,

centreis andradius

Distancebetweenthecentresis

Thus,thedistancebetweenthecentresisequaltothedifferenceinradii.Hence,thetwocirclestoucheachotherinternally.

Example4.20

TheabscissaofthetwopointsAandBaretherootsoftheequationx2+2x−a2

=0andtheordinatesaretherootsoftheequationy2+4y−b2=0.FindtheequationofthecirclewithABasitsdiameter.Alsofindthecoordinatesofthecentreandthelengthoftheradiusofthecircle.

Solution

Lettherootsoftheequationx2+2x−a2=0beαandβ.Then

Letγ,δbetherootsoftheequationx2+4y−b2=0.Then

ThecoordinatesofAandBare(α,γ)and(β,δ).TheequationofthecircleonthelinejoiningthepointsAandBastheendsofadiameteris(x−α)(x−β)+(y−γ)(y−δ)=0.

Thecentreofthecirclein(−1,−2)andtheradius

Example4.21

Findtheequationofacirclethatpassesthroughthepoint(2,0)andwhosecentreisthelimitpointoftheintersectionofthelines3x+5y=1and(2+c)x+5c2y=1asc→1.

Solution

Thecentreofthecircleisthepointofintersectionofthelines

Asc→1,thex-coordinateofthecentreis .

From(4.42),

Hence,thecentreofthecircleis

Radiusisthelengthofthelinejoiningthepoints(2,0)and

Therefore,theequationofthecirclesis

Example4.22

Findthelengthinterceptedonthey-axisbythechordofthecirclejoiningthepoints(−4,3)and(12,−1)asdiameter.

Solution


Ify1andy2arethey-coordinatesofthepointofintersectionofthecircleandy-axis,then

Example4.23

Therodswhoselengthsareaandbslidealongthecoordinateaxesinsuchawaythattheirextremitiesareconcyclic.Findthelocusofthecentreofthecircle.

Solution

LetABandCBbetheportionofx-axisandy-axis,respectively,interceptedbythecircle.LetP(x1,y1)bethecentreofthecircle.DrawPLandPMperpendiculartox-axisandy-axis,respectively.Then,bysecondproperty

ThelocusofP(x1,y1)is4(x2−y2)=a2−b2.

Example4.24

Showthatthecirclesx2+y2−2x−4y=0andx2+y2−8y−4=0toucheachother.Findthecoordinatesofthepointofcontactandtheequationofthecommontangents.

Solution

ThecentresofthesetwocirclesareC1(1,2)andC2(0,4).Theradiiofthetwocirclesare

Thedistancebetweenthecentresis

∴r1−r2=C1C2.Hence,thecirclestoucheachotherinternally.ThepointofcontactCdividesC1C2internallyintheratio1:1.

IfCisthepoint(x1,y1)then

∴C(2,0)isthepointofcontact.Theslopeof

Hence,theslopeofthecommontangentis1/2.Theequationofthecommon

tangentis

Example4.25

ShowthatthegeneralequationofthecirclethatpassesthroughthepointA(x1,

y1)andB(x2,y2)maybewrittenas

Solution

LetA(x1,y1)andB(x2,y2)bethetwopointsonthecircumferenceofthecircleandA(x1,y1)beanypointonthecircumference.

Let .TheslopeofAPandBPare and

Example4.26

Showthatifthecirclex2+y2=a2cutsoffachordoflength2bontheliney=mx+c,thenc2=(1+m)2(a2−b2).

Solution

Givenx2+y2=a2.Thecentreofthecircleis(0,0).Radius=r=a.DrawOLperpendiculartoAB.Then,ListhemidpointofAB.

Example4.27

Apointmovessuchthatthesumofthesquaresofthedistancesfromthesidesofasquareofsideunityisequaltoa.Showthatthelocusisacirclewhosecentrecoincideswiththecentreofthesquare.

Solution

Letthecentreofthesquarebetheorigin.LetP(x,y)beanypoint.Then,the

equationofthesidesare

SumoftheperpendiculardistancesfromPonthesidesisequaltoa

Hence,thelocusofPisthecirclex2+y2−1=0.Thecentreofthecircleis(0,0),whichisthecentreofthesquare.

Example4.28

Ifthelinesl1x+m1y+n1=0andl2x+m2y+n2=0cutthecoordinateaxesatconcyclicpoints,provethatl1l2=m1m2.

Solution

Givenl1x+m1y+n1=0.Theinterceptsofthelineontheaxisare

IfthelinemeetstheaxesatL1andM1,then Ifthesecond

linemeetstheaxesatL2andM2,then

Example4.29

Showthatthelocusofapointwhoseratioofdistancesfromtwogivenpointsisconstantisacircle.Hence,showthatthecirclecannotpassthroughthegivenpoints.

Solution

LetthetwopointsAandBbechoseninthex-axisandthemidpointofABbe(0,0).ThenletA(a,0)andB(−a,0).GiventhatPA=K·PB⇒PA2=K2PB2wherekisaconstant.

(x−a)2+(y−0)2=K2[(x+a)2+y2]

Inthisequation,thecoefficientsofx2andy2arethesameandthereisnoxyterm.Therefore,thelocusofPisacircle.IfA(a,0)liesonthiscircle,thenO=K2[4a2]⇒a=0ork=0,whicharenotpossible.Therefore,thepointAdoesnotlieonthecircle.Similarly,thepointB(−a,0)alsodoesnotlieonthecircle.

Example4.30

Findtheequationofthecirclewhoseradiusis5andwhichtouchesthecirclex2

+y2−2x−4y−20=0atthepoint(5,5).

Solution

Givenx2+y2−2x−4y−20=0.

Centreis(1,2)andradius=

Letthecentreoftherequiredcirclebe(x1,y1).ThepointofcontactisthemidpointofAB.

∴x=9andy=8Thus,Bis(9,8).Hencetheequationoftherequiredcircleis

Example4.31

OneofthediametersofthecirclecircumscribingtherectangleABCDis4y=x+7.IfAandBarethepoints(−3,4)and(5,4),respectively,findtheareaoftherectangleABCD.

Solution

LetP(x1,y1)bethecentreofthecircleand4y=x+7betheequationofthediameterofBD.

ThemidpointofACis(1,1).TheslopeofABis0.Therefore,theslopeofPLis∞.

Hence,theareaoftherectangleABCD=8×4=32sq.cm.

Example4.32

Findtheequationofthecircletouchingthey-axisat(0,3)andmakinganinterceptof8cmonthex-axis.

Solution

Lettheequationofthecirclebex2+y2+2gx+2fy+c=0.Centreis(−g,−f),Thus,−f=3orf=−3.Whenthecirclemeetsthex-axis,y=0.

Hence,theequationofthecircleisx2+y2±10x−6y+9=0.

Example4.33

Findtheequationofthecirclepassingthroughthepoint(−4,3)andtouchingthelinesx+y=2andx−y=2.

Solution

x+y=2andx−y=2intersectatthepoint(2,0).

Moreover,theselinesareperpendicularandtheirslopesare1and−1.So,theymake45°and135°withthex-axis.Henceoneofthebisectorsisthex-axisandcentreliesononeofthebisectors.Ifx2+y2+2gx+2fy+c=0istheequationofthecircle,thenf=0.Alsotheperpendiculardistancefrom(−g,0)tothetangentsisequaltotheradius.

Since(−4,3)liesonthecircle16+9−8g+c=0

Hence,equation(4.46)becomesg2−4g−4−16g+50=0org2−20g+46=0.

Thus,therearetwocircleswhoseequationsaregivenby

Example4.34

Aisthecentreofthecirclex2+y2−2x−4y−20=0.IfthetangentsdrawnatthepointsB(1,7)andD(4,−2)onthecirclemeetatthepointC,thenfindtheareaofthequadrilateralABCD.

Solution

x2+y2−2x−4y−20=0

Centreofthiscircleis(1,2)

Theequationsoftangentsat(1,7)and(4,−2)tothecirclearex+7y−(x+1)−2(y+7)−20=0(i.e.)5y−35=0⇒y=7and

4x−2y−(x+4)−2(y−2)−20=0(i.e.)3x−4y−20=0

Sincey=7,x=16.Hence,thepointCis(16,7).AreaofthequadrilateralABCD=2×areaofΔABC

Example4.35

FromthepointA(0,3)onthecirclex2+4x+(y−3)2=0,achordABisdrawnandextendedtoapointMsuchthatAM=AB.FindtheequationofthelocusofM.

Solution

AM=2.AB

Hence,BisthemidpointofAM.ThenthecoordinatesofBare

ThispointBliesonthecirclex2+4x+(y−3)2=0.

Therefore,thelocusof(x1,y1)isx2+y2+8x−6y+9=0.

Example4.36

ABisadiameterofacircle,CDisachordparalleltoABand2CD=AB.ThetangentatBmeetsthelineACproducedatE.ProvethatAE=2.AB.

Solution

Lettheequationofthecirclebex2+y2=a2andPQbethediameteralongthex-axis.CDisparalleltoAB.LetAB=2aandpointsAandBbe(a,0)and(−a,0),respectively.Also

Example4.37

Findtheareaofthetriangleformedbythetangentsfromthepoint(h,k)tothecirclex2+y2=a2andtheirchordofcontact.

Solution

Theequationofthecircleisx2+y2=a2.LetABbethechordofcontactoftangentsfromC(x1,y1).ThentheequationofABisxx1+yy1=a2.

WeknowthatOCisperpendiculartoAB.LetABandOCmeetatL.

TheperpendiculardistancefromConAB

Example4.38

Letacirclebegivenby2x(x−a)+y(2y−b)=0,(a,b≠0).Findtheconditiononaandbiftwochordseachintersectedbythex-axiscanbedrawntothecircle

from

Solution

Thechordisbisectedbythex-axis.Letthemidpointofthechordbe(h,0).Theequationofthechordis

Thischordpassesthrough .

Sincethechordmeetsthex-axisattworeals,Discriminant>0

Example4.39

Findtheconditionthatthechordofcontactfromapointtothecirclex2+y2=a2

subtendsarightangleatthecentreofthecircle.

Solution

Theequationtothechordofcontactfrom(x1,y1)tothecircle

ThenthecombinedequationtoOAandOBisgotbyhomogenizingequation(4.48)withthehelpofequation(4.49).

ThecombinedequationofOAandOBis


Example4.40

Ify=mxbetheequationofachordofthecirclewhoseradiusisa,theoriginbeingoneoftheextremitiesofthechordandtheaxisbeingadiameterofthecircle,provethattheequationofacircleofwhichthischordisadiameteris(1+m2)(x2+y2)−2a(x+my)=0.

Solution

Letabetheradiusofthecircle.Thus(a,0)isthecentreofthecircle.Theequationofthecircleis

(x−a)2+y2=a2⇒x2+y2−2ax=0

Wheny=mxmeetsthecirclex2+m2x2−2ax=0.

Therefore,theextremitiesofthischordare(0,0)and Then,the

equationofthecirclewiththechordasadiameteris

Example4.41

Findtheequationtothecirclethatpassesthroughtheoriginandcutsoffequalchordsoflengthafromthestraightlinesy=xandy=−x.

Solution

Letthelinesy=xandy=−xmeetthecircleatP,P′andQ,Q′,respectively.

ThenOP=OQ=a=OP′=OQ′.ThecoordinatesofPandP′are

SimilarlythecoordinatesofQandQ′are

Therearefourcirclespossiblehavingcentresat

Hence,theequationsofthefourcirclesaregivenby

Example4.42

Findthelocusofthemidpointofchordsofthecirclex2+y2=a2,whichsubtendsarightangleatthepoint(c,0).

Solution

SinceABsubtends90°atC(c,0),PA=PB=PC.LetPbethepoint(x1,y1).

SincePisthemidpointofthechordAB,CP⊥AP

Since ,PC=AP.

Thelocusof(x1,y1)is2(x2+y2)−2cx1+(c2−a2)=0.

Example4.43

Findtheequationsofthecirclesthattouchthecoordinateaxesandthelinex=a.

Solution

y=0,x=0andx=aarethetangentstothecircle.Therearetwocirclesasshowninthefigure.

Thecentresare andradius .Theequationsofthecirclesare

Example4.44

Findtheshortestdistancefromthepoint(2,−7)tothecirclex2+y2−14x−10y−151=0.

Solution

x2+y2−14x−10y−151=0

Centeris(7,5)

Centeris(7,5)

Radius=

TheshortestdistanceofthepointPfromthecircle=∣CP−r∣

Example4.45

Letα,βandγbetheparametricanglesofthreepointsP,QandR,respectively,onthecirclex2+y2=a2andAbethepoint(−a,0).Ifthelengthofthechords

AP,AQandARareinAPthenshowthat arealsoinAP.

Solution

LetP(acosα,asinα),Q(acosβ,asinβ),R(acosr,asinr)Ais(a,0)

ThelengthsofchordsAP,AQ,ARareinAP.

Example4.46

LetS=x2+y2+2gx+2fy+c=0.Findthelocusofthefootoftheperpendicularfromtheoriginonanychordofthecirclethatsubtendsarightangleattheorigin.

Solution

LettheequationofthelineABbe

lx+my=1(4.50)

Let(x1,y1)bethemidpointofAB.

LetP(x1,y1)bethefootoftheperpendicularfromtheoriginonAB.Then,sinceOPisperpendiculartoAP.

Since(x1,y1)liesonthelinelx+my=1wehave

lx1+my1=1(4.52)

ThecombinedequationoflinesOAandOBisgotbyhomogenizingtheequationofthecirclex2+y2+2gx+2fy+c=0withthelinelx+my=1.

Since ,theconditioniscoefficientofx2+coefficientofy2=0.Hence,

Thelocusof(x1,y1)is2(x2+y2)+2gx+2fy+c=0.

Example4.47

Pisthepoint(a,b)andQisthepoint(b,a).FindtheequationofthecircletouchingOPandOQatPandQwhereOistheorigin.

Solution


LetC(−g,−f)bethecentreofthecircle.

WeknowthatPQisthechordofthecontactfromOandOCisperpendiculartoPQ.∴SlopeofPQ×slopeofOC=−1

TheequationofOPis

SinceCPisperpendiculartoOP,ristheperpendiculardistancefromConOP.

Thepoint(a,b)liesonthecircle(4.53).

Hencetheequationofthecircleis

Example4.48

Acircleofcircumradius3kpassesthroughtheoriginandmeetstheaxesatAandB.ShowthatthelocusofthecentroidofΔOABisthecirclex2+y2=4K2.

Solution

LetAandBbethepoints(a,0)and(0,b),respectively.Let(x1,y1)bethe

centroidofΔOAB.Thensince ,ABisadiameterofthecircle.

LetthecentroidofΔOABbe(x1,y1).Then ora=3x1andb=

3y1.Substitutingthisin(4.55),weget .Thelocusof(x1,

y1)isx2+y2=4k2.

Example4.49

Avariablelinepassesthroughafixedpoint(a,b)andcutsthecoordinateaxesat

thepointsAandB.ShowthatthelocusofthecentreofthecircleABis

Solution

LetABbeavariablelinewhoseequationbe

Thispassesthroughthepoint(a,b).

Since ABisadiameterofthecircumcircleofΔOAB.Itscentreis

If(x1,y1)bethecircumcentre,then

∴α=2x1andβ=2y1

Hence,from(4.55),weget

Thelocusof(x1,y1)is

Example4.50

If4l2−5m2+6l+1=0thenshowthatthelinelx+my+1=0touchesafixedcircle.Findthecentreandradiusofthecircle.

Solution

Lettheline

touchthecircle

Thentheperpendiculardistancefrom(h,k)toline(4.56)isequaltotheradius.

Buttheconditionisgivenby

Identifying(4.58)and(4.59),weget

Hence,thelinetouchesthefixedcircle(x−3)2+y2=5orx2+y2−6x+4=0

whosecentreis(3,0)andradiusis

Exercises

1. Findtheequationofthefollowingcircles:

i. centre(2,−5)andradius5unitsii. centre(−2,−4)andradius10unitsiii. centre(a,b)andradius(a+b)

Ans.:(i)x2+y2−4x+10y+4=0

Ans.:(ii)x2+y2+4x+8y−80=0

Ans.:(iii)x2+y2−2ax−2by=0

2. Findthecentreandradiusofthefollowingcircles:

i. x2+y2−22x−4y+25=0

ii. 4(x2+y2)−8(x−2y)+19=0

iii. 2x2+2y2+3x+y+1=0

Ans.:(i)(11,2),10

Ans.:(ii)(1,−2),

Ans.:

3. Findtheequationofthecirclepassingthroughthepoint(2,4)andhavingitscentreonthelinesx−y=4and2x+3y=8.

Ans.:x2+y2−8x−4=0

4. Findtheequationofthecirclewhosecentreis(−2,3)andwhichpassesthroughthepoint(2,−2).

Ans.:x2+y2+4x−6y−28=0

5. Showthattheline4x−y=17isadiameterofthecirclex2+y2−8x+2y=0.

6. Theequationofthecircleisx2+y2−8x+6y−3=0.Findtheequationofitsdiameterparallelto2x−7y=0.Alsofindtheequationofthediameterperpendicularto3x−4y+1=0.

Ans.:2x−7y−29=04x+3y−7=0

7. Findtheequationofthecirclepassingthroughthefollowingpoints:i. (2,1),(1,2),(8,9)ii. (0,1),(2,3),(−2,5)iii. (5,2),(2,1),(1,4)

Ans.:x2+y2−10x−10y−25=0

Ans.:3x2+3y2+2x−20y+17=0

Ans.:x2+y2−6x−6y+13=0

8. Findtheequationofthecirclethroughthepoints(1,0)and(0,1)andhavingitscentreonthelinex+y=1.

Ans.:x2+y2−x−y=0

9. Findtheequationofthecirclepassingthroughthepoints(0,1)and(4,3)andhavingitscentreontheline4x−5y−5=0.

Ans.:x2+y2−5x−2y+1=0

10. Twodiametersofacircleare5x−y=3and2x+3y=8.Thecirclepassesthroughthepoint(−1,7).Finditsequation.

Ans.:x2+y2−2x−4y=164

11. Findtheequationofthecirclecircumscribingthetriangleformedbytheaxesandthestraightline3x+4y+12=0.

Ans.:x2+y2+4x+3y=0

12. Showthatthepoints(−1,2),(−2,4),(−1,3)and(2,0)areonacircleandfinditsequation.13. Ifthecoordinatesoftheextremitiesofthediameterofacircleare(3,5)and(−7,−5),findthe

equationofthecircle.

Ans.:x2+y2+4x−3y=0

14. Findtheequationofthecirclewhenthecoordinatesoftheextremitiesofoneofitsdiametersare(4,1)and(−2,–7).

Ans.:x2+y2−2x+6y−15=0

15. Ifoneendofthediameterofthecirclex2+y2−2x+6y−15=0is(4,1),findthecoordinatesoftheotherend.

Ans.:(−2,−7)

16. Provethatthetangentsfrom(0,5)tothecirclesx2+y2+2x−4=0andx2+y2−y+1=0areequal.

17. Findtheequationofthecirclepassingthroughtheoriginandhavingitscentreat(3,4).Alsofindtheequationofthetangenttothecircleattheorigin.

Ans.:x2+y2−6x−8y=0,3x+4y=0

18. Findtheslopeoftheradiusofthecirclex2+y2=25throughthepoint(3,−4)andhencewritedowntheequationofthetangenttothecircleatthepoint.Whataretheinterceptsmadebythistangentonthex-axisandy-axis?

Ans.:

19. Onevertexofasquareistheoriginandtwoothersare(4,0)and(0,4).Findtheequationofthecirclecircumscribingthesquare.Alsofindtheequationofthetangenttothiscircleattheorigin.

Ans.:x2+y2−4x−4=0,x+y=0

20. Acirclepassesthroughtheoriginandthepoints(6,0)and(0,8).Finditsequationandalsotheequationofthetangenttothecircleattheorigin.

Ans.:x2+y2−6x−8y=0,3x+4y=0

21. AandBaretwofixedpointsonaplaneandthepointPmovesontheplaneinsuchawaythatPA=2PBalways.ProveanalyticallythatthelocusofPisacircle.

22. Doesthepoint(2,1)lie(i)on,(ii)insideor(iii)outsidethecirclex2+y2−4x−6y+9=0?

23. Showthatthecirclesx2+y2−2x+2y+1=0andx2+y2+6x−4y−3=0toucheachotherexternally.

24. Provethatthecentresofthethreecirclesx2+y2−2x+6y+1=0,x2+y2+4x−12y−9=0and

x2+y2=25lieonthesamestraightline.Whatistheequationofthisline?Ans.:3x+4y=0

25. Provethatthetwocirclesx2+y2+2ax+c2=0andx2+y2+2by+c2=0toucheachotherif

26. Showthatthecirclesx2+y2−4x+2y+1=0andx2+y2−12x+8y+43=0toucheachotherexternally.

27. Showthatthecirclesx2+y2=400andx2+y2−10x−24y+120=0touchoneanother.Findthecoordinatesofthepointofcontact.

Ans.:

28. Findthelengthofthetangentfromtheorigintothecircle4x2+4y2+6x+7y+1=0.

29. Showthatthecirclesx2+y2−26x−19=0andx2+y2+3x−8y−43=0touchexternally.Findthepointofcontactandthecommontangent.

30. Apointmovessothatthesquareofitsdistancefromthebaseofanisoscelestriangleisequaltotherectanglecontainedbyitsdistancesfromtheequalsides.Provethatthelocusisacircle.

31. Provethatthecentresofthecirclesx2+y2=1,x2+y2+4x+8y−1=0andx2+y2−6x−12y+1=0arecollinear.

32. Provethattheconstantintheequationofthecirclex2+y2+2gx+2fy+c=0isequaltotherectangleunderthesegmentsofthechordsthroughtheorigin.

33. Findtheequationofthelocusofapointthatmovesinaplanesothatthesumofthesquaresfromtheline7x−4y−10=0and4x+7y+5=0isalwaysequalto3.

Ans.:13x2+13y2−20x+30y−14=0

34. Showthatthecirclesx2+y2−10x+4y−20=0andx2+y2+14x−6y+22=0toucheachother.Findtheequationoftheircommontangentatthepointofcontactandalsothepointofcontact.

Ans.:

35. LandMarethefeetoftheperpendicularfrom(c,0)onthelinesax2+2hxy+by2=0.ShowthattheequationofLMis(a+b)x+2hy+bc=0.

36. Acirclehasradius3unitsanditscentreliesontheliney=x−1.Findtheequationofthecircleifitpassesthrough(−1,3).

37. Findtheequationofthecircleonthelinejoiningthepoints(−4,3)and(12,−1).Findalsotheinterceptsmadebyitonthey-axis.

Ans.:

38. Showthatthepoints liesoutsidethecircle3x2+3y2−5x−6y+4=0.

39. Findtheconditionthatthelinelx+my+n=0touchesthecirclex2+y2=a2.Findalsothepointofcontact.

Ans.:

40. Findtheequationofthecirclepassingthroughthepoint(3,5)and(5,3)andhavingitscentreontheline2x+3y−1=0.

Ans.:5x2+3y2−14x−14y−50=0

41. ABCDisasquarewhosesideisa.TakinglineAOastheaxisofcoordinates,provethatthe

equationofthecircumcircleofthesquareisx2+y2−ax−ay=0.42. Findtheequationofthecirclewithitscentreontheline2x+y=0andtouchingthelines4x−3y

+10=0and4x−3y−3=0.

Ans.:x2+y2−2x+4y−11=0

43. Findtheequationofthecirclethatpassesthroughthepoint(1,1)andtouchesthecirclex2+y2+4x−6y−3=0atthepoint(2,3)onit.

Ans.:x2+y2+x−6y+3=0

44. Provethatthetangenttothecirclex2+y2=5atthepoint(1,−2)alsotouchesthecirclex2+y2−8x+6y+20=0andfinditspointofcontact.

Ans.:(3,−1)

45. AvariablecirclepassesthroughthepointA(a,b)andtouchesthex-axis.Showthatthelocusofthe

otherendofthediameterthroughAis(x−c)2=4by.46. FindtheequationofthecirclepassingthroughthepointsA(−5,0),B(1,0),andC(2,1)andshow

thattheline4x−3y−5=0isatangenttotheline.47. Findtheequationofthecirclethroughtheoriginandthroughthepointofcontactofthetangents

fromtheorigintothecircle.

Ans.:2x2+2y2−11x−13y=0

48. Thecirclex2+y2−4x−4y+4=0isinscribedinatrianglethathastwoofitssidesalongthe

coordinateaxes.Thelocusofthecircumferenceofthetriangleis Findk.

Ans.:k=1

49. Acircleofdiameter13mwithcentreOcoincidingwiththeoriginofcoordinateaxeshasdiameterABonthex-axis.IfthelengthofthechordACbe5m,findtheareaofthesmallerportionboundedbetweenthecirclesandthechordAC.

Ans.:1.9m2.

50. Findtheradiusofthesmallestcirclethattouchesthestraightline3x−y=6at(1,−3)andalsotouchestheliney=x.

Ans.:

51. If formdistinctpointsonacircleshowthatm1,m2,m3,m4=1.

52. Ifthelinexcosα+ysinα=ρcutsthecirclex2+y2=a2inMandN,thenshowthatthecircle

whosediameterisMNisx2+y2−a2−2ρ(xcosα+ysinα−ρ)=0.

53. Showthatthetangentsdrawnfromthepoint(8,1)tothecirclex2+y2−2x−4y−20=0areperpendiculartoeachother.

54. Howmanycirclescanbedrawneachtouchingallthethreelinesx+y=1,y=x+1and7x−y=6?Findthecentreandradiusofallthecircles.

Ans.:

55. Findthepointsonthelinex−y+1=0,thetangentsfromwhichtothecirclex2+y2−3x=0areoflength2units.

Ans.:

56. Onthecircle16x2+16y2+48x−3y−43=0,findthepointnearesttotheline8x−4y+73=0andcalculatethedistancebetweenthispointandtheline.

Ans.:

57. Findtheequationsofthelinestouchingthecirclex2+y2+10x−2y+6=0andparalleltotheline2x+y−7=0.

Ans.:2x+y−1=0,2x+y+19=0

58. Findtheequationofthecirclewhosediameteristhechordofintersectionofthelinex+3y=6

andthecurve4x2+9y2=36.

Ans.:5(x2+y2)−12x−16y+12=0

59. Findtheequationforthecircleconcentricwiththecirclex2+y2−8x+6y−5=0andpassesthroughthepoint(−2,7).

Ans.:x2+y2−8x+6y−27=0

60. Findtheequationofthecirclethatcutsoffintercepts−1and−3onthex-axisandtouchesthey-

axisatthepoint

Ans.:

61. Findthecoordinatesofthepointofintersectionoftheline5x−y+7=0andthecirclex2+y2+3x−4y−9=0.Alsofindthelengthofthecommonsegment.

Ans.:

62. Theline4x+3y+k=0isatangenttothecirclex2+y2=4.Findthevalueofk.Ans.:k=±10

63. Findtheequationsoftangentstothecirclex2+y2−6x+4y−17=0thatareperpendicularto3x−4y+5=0.

Ans.:4x+3y+19=0,4x+3y−31=0

64. Findtheequationoftangentstothecirclex2+y2−14x+y−5=0atthepointswhoseabscissais10.

Ans.:3x+7y−93=0,3x−7y−64=0

65. Showthatthecirclesx2+y2−4x+6y+8=0andx2+y2−10x−6y+14=0toucheachother.Findthepointofcontact.

Ans.:(3,−1)

66. Showthatthetangenttothecentrex2+y2=0atthepoint(1,−2)alsotouchesthecirclex2+y2−8x+6y+20=0.Findthepointofcontact.

Ans.:(3,−1)

67. AstraightlineABisdividedatCsothatAB=3CB.CirclesaredescribedonACandCBasdiametersandacommontangentmeetsABproducedatD.ShowthatBDisequaltotheradiusofthesmallercircle.

68. Thelines3x−4y+4=0and6x−8y−7=0aretangentstothesamecircle.Findtheradiusofthiscircle.

Ans.:

69. Fromtheorigin,chordsaredrawntothecircle(x−1)2+y2=1.Findtheequationofthelocusofthemidpointofthesechords.

Ans.:x2+y2−x=0

70. Findtheequationsofthepairoftangentstothecirclex2+y2−2x+4y=0from(0,1).

Ans.:2x2−2y2+3xy−3x+4y−2=0

71. Ifthepolarofpointsonthecirclex2+y2=a2withrespecttothecirclex2+y2=b2touchthe

circlex2+y2=c2,showthata,bandcareinGP.

72. Ifthedistancesoforigintothecentresofthreecirclesx2+y2−2λx=c2whereλisavariableandcisaconstantareinG.P,provethatthelengthofthetangentdrawntothemfromanypointonthe

circlex2+y2=c2areinG.P.

73. Atangentisdrawntoeachofthecirclesx2+y2=a2andx2+y2=b2.Ifthetwotangentsaremutuallyperpendicular,showthatthelocusoftheirpointofintersectionisacircleconcentricwiththegivencircles.

74. Ifthepoleofanylinewithrespecttothecirclex2+y2=a2liesonthecirclex2+y2=9a2,then

showthatthelinewillbeatangenttothecircle .

75. Atrianglehastwoofitssidesalongthey-axis,anditsthirdsidetouchesthecirclex2+y2−2ax−

2ay+a2=0.Provethatthelocusofthecircumcentreofthetriangleis2xy−2a(x+y)+a2=0.76. Lines5x+12y−10=0and6x−11y−40=0touchacircleC,ofdiameter6.IfthecentreofC1

liesinthefirstquadrant,findtheequationofcircleC2.whichisconcentricwithC1andcutsinterceptsoflength8ontheselines.

Ans.:

77. Findtheequationofthecirclethattouchesthey-axisatadistanceof4unitsfromtheoriginandcutsoffaninterceptof6unitsfromthex-axis.

Ans.:x2+y2+10x−8y+16=0

78. Findtheequationofthecircleinwhichthelinejoiningthepoints(0,b)and(b,−a)isachordsubtendinganangle45°atanypointonitscircumference

Ans.:x2+y2−2(a+b)x+2(a−b)y+(a2+b2)

79. Fromanypointonagivencircle,tangentsaredrawntoanothercircle.Provethatthelocusofthemiddlepointofthechordofcontactisathirdcircle;thedistancebetweenthecentresofthegivencircleisgreaterthanthesumoftheirradii.

80. Apointmovessothatthesumofthesquaresoftheperpendicularsthatfallfromitonthesidesofanequilateraltriangleisconstant.Provethatthelocusisacircle.

81. AcircleofconstantradiuspassesthroughtheoriginOandcutstheaxesinAandB.Showthatthe

locusofthefootoftheperpendicularfromABis(x2+y2)2(x2+y2)=4r2.

82. Findtheequationoftheimageofthecircle(x−3)2+(y−2)2=1bythemirrorx+y=19.

Ans.:(x−17)2+(y−16)2=1

83. Findthevalueofλforwhichthecirclex2+y2+6x+5+λ(x2+y2−8x+7)=0dwindlesintoapoint.

Ans.:

84. Avariablecirclealwaystouchestheliney=xandpassesthroughthepoint(0,0).Showthatthe

commonchordsofthiscircleandx2+y2+6x+8y−7=0willpassthroughafixedpoint

85. Theequationofthecirclethattouchestheaxesofthecoordinatesandtheline and

whosecentreliesinthefirstquadrantisx2+y2−2cx−2cy+c2=0.Findthevaluesofc.Ans.:(1,6)

86. Aregioninxy-planeisboundedbythecurve andtheliney=0.Ifthepoint(a,a+1)

liesintheinterioroftheregion,findtherangeofa.Ans.:a∈(−1,3)

87. Thepoints(4,−2)and(3,6)areconjugatewithrespecttothecirclex2+y2=24.Findthevalueofb.

Ans.:b=−6

88. Ifthetwocirclesx2+y2+2gx+2fy=0andx2+y2+2g1x+2ƒ1y=0toucheachother,show

thatƒ1g=gƒ1.89. Showthatthelocusofthepointsofchordsofcontactoftangentssubtendingarightangleatthe

centreisaconcentriccirclewhoseradiusis timestheradiusofthegivencircle.Alsoshowthat

thisisalsothelocusofthepointofintersectionofperpendiculartangents.90. Showthatthepoints(xi,yi),i=1,2,3arecollinearifandonlyiftheirpoleswithrespecttothe

circlesx2+y2=a2areconcurrent.91. ThelengthofthetangentsfromtwogivenpointsAandBtoacirclearet1andt2,respectively.If

thepointsareconjugatepoints,showthat

92. Showthattheequationtothepairoftangentsdrawnfromtheorigintothecirclex2+y2+2gx+

2ƒy+c=0is(gx+ƒy)2=(ƒ2+g2).Hencefindthelocusofthecentreofthecircleifthesetangentsareperpendicular.

Ans.:x2+y2=2c

93. ThreesidesofatrianglehavetheequationsLi=y−mrx−cr=0,r=1,2,3.ThenshowthatλL2L3+μL3L1+vL1L2=0whereλ,μ,v≠0istheequationofthecircumcircleofthetriangleifΣλ(m2+m3)=0andΣλ(m2m3−m1)=0.

94. Atriangleisformedbythelineswhosecombinedequationisc(x+y−4)(xy−2x−y+2)=0.

Showthattheequationofitscircumferenceisx2+y2−3x−5y+8=0.

95. Twodistinctchordsdrawnfromthepoint(p,q)onthecirclex2+y2=px+qy,wherepq≠0,are

bisectedbythex-axis.Showthatp2>8q2.

96. Showthatthenumberofpointswithintegralcoordinatesthatareinteriortothecirclex2+y2=16is45.

97. Findthenumberofcommontangentstothecirclesx2+y2−6x−14y+48=0andx2+y2−6x=0.

Ans.:4

98. Thetangentstothecirclex2+y2=4atthepointsAandBmeetatP(−4,0).FindtheareaofthequadrilateralPAOB.

Ans.:

99. Theequationsoffourcirclesare(x±a)2+(y±a)2=a2.Findtheradiusofacircletouchingallthefourcircles.

Ans.:

100. Acircleofradius2touchesthecoordinateaxesinthefirstquadrant.Ifthecirclemakesacompleterotationonthex-axisalongthepositivedirectionofthex-axis,thenshowthattheequationofthe

circleinthenewpositionisx2+y2−4(x+y)−8λx+(2+4π)2=0.101. Twotangentsaredrawnfromtheorigintoacirclewithcentreat(2,−1).Iftheequationofoneof

thetangentsis3x+y=0,findtheequationoftheothertangent.Ans.:x−3y=0

102. Findtheequationofthechordofthecirclex2+y2=a2passingthroughthepoint(2,3)fartherfromthecentre.

Ans.:2x+3y=17

103. Anequilateraltriangleisinscribedinthecirclex2+y2=a2,withthevertexat(a,0).Findtheequationofthesideoppositetothisvertex.

Ans.:2x+a=0

104. AlineisdrawnthroughthepointP(3,1)tocutthecirclex2+y2=9atAandB.FindthevalueofPA·PB.

Ans.:121

105. C1andC2arecirclesofunitradiuswiththeircentresat(0,0)and(1,0),respectively.C3isacircleofunitradius,passingthroughthecentresofthecirclesC1andC2andhavingitscentreabovethex-axis.FindtheequationofthecommontangenttoC1andC3thatpassesthroughC2.

Ans.:

106. Achordofthecirclex2+y2−4x−6y=0passingthroughtheoriginsubtendsanangle

atthepointwherethecirclemeetsthepositivey-axis.Findtheequationofthechord.Ans.:x−2y=0

107. AcirclewithitscentreattheoriginandradiusequaltoameetstheaxisofxatAandB.PandQarerespectivelythepoints(acosα,atanα)and(acosβ,atanβ)suchthatα−β=2γ.Showthatthe

locusofthepointofintersectionofAPandBQisx2+y2−2aytanγ=−2.

108. AcircleC1ofradiustouchesthecirclex2+y2=a2externallyandhasitscentreonthepositivex-

axis.AnothercircleC2ofradiusctouchescircleC1externallyandhasitscentreonthepositivex-axis.Ifa<b<c,showthatthethreecircleshaveacommontangentifa,b,careinGP.

109. Findtheequationsofcommontangentstothecirclesx2+y2+14x−14y+28=0andx2+y2−14x+4y−28=0

Ans.:28y+45y+371=0andy−7=0.

110. Ifacirclepassesthroughthepointsofintersectionofthecoordinateaxeswiththelinex−λy+1=

0(λ≠0)andx−2y+3=0thenλsatisfiestheequation6λ2−7λ+2=0.

111. OAandOBareequalchordsofthecirclex2+y2−2x+4y=0perpendiculartoeachotherand

passingthroughtheorigin.ShowthattheslopesofOAandOBsatisfytheequation3m2−8m−3=0.

112. Findtheequationofthecirclepassingthroughthepoints(1,0)and(0,1)andhavingthesmallestpossibleradius.


113. Findtheequationofthecirclesituatedsystematicallyoppositetothecirclex2+y2−2x=0withrespecttothelinex+y=2.

Ans.:x2+y2−4x−2y+4=0

114. OisafixedpointandRmovesalongafixedlineLnotpassingthroughO.IfSistakenonORsuch

thatOR·OS=K2,thenshowthatthelocusofSisacircle.115. Showthatthecircumferenceofthetriangleformedbythelinesax+by+c=0,bx+cy+a=0

andcx+ay+b=0passesthroughtheoriginif(b2+c2)(c2+a2)(a2+b2)=abc(b+c)(c+a)(a+b).

116. Twocirclesaredrawnthroughthepoints(a,5a)and(4a,a)totouchthey-axis.Provethatthey

intersectatanangle

117. ShowthatthelocusofapointPthatmovessothatitsdistancefromthegivenpointOisalwaysinagivenration:1·(n≠−1)toitsdistanceonthelinejoiningthepointsthatdividesthelineOAinthegivenratioasdiameter.

118. Thelines3x−4y+4=0and6x−3y−7=0aretangentstothesamecircle.Findtheradiusofthecircle.

Ans.:

119. Theliney=xtouchesacircleatPsothat whereOistheorigin.Thepoint(−10,2)is

insidethecircleandlengthofthechordontheline Findtheequationofthe

line.

Ans.:x2+y2+18x−2y+32=0

120. Findtheintervalsofvaluesofaforwhichtheliney+x=0bisectstwochordsdrawnfromapoint

tothecircle

121. Showthatallchordsofthecircle3x2−y2−2x+4y=0thatsubtendarightangleattheoriginare

concurrent.Doestheresultholdforthecurve3x2+3y2−2x+4y=0?Ifyes,whatisthepointofconcurrency,andifnot,givethereason.

122. Findtheequationsofthecommontangentstothecirclesx2+y2−14x+6y+33=0andx2+y2

+30x−20y+1=0.Ans.:4x−3y−12=0,24x+7y−22=0

123. Provethattheorthocentreofthetrianglewhoseangularpointsare(acosα,asinα),(acosβ,asinβ)and(acosγ,asinγ)isthepoint[a(cosα+cosβ+cosγ),a(sinα+sinβ+sinγ)].

Chapter5

SystemofCircles

5.1RADICALAXISOFTWOCIRCLES

Definition5.1.1:Theradicalaxisoftwocirclesisdefinedasthelocusofapointsuchthatthelengthsoftangentsfromittothetwocirclesareequal.

ObtaintheequationoftheradicalaxisofthetwocirclesS≡x2+y2+2gx+2fy+c=0andS1≡x2+y2+2g1x+2fy+c1=0.

LetP(x1,y1)beapointsuchthatthelengthsoftangentstothetwocirclesareequal.

Thelocusof(x1,y1)is2(g−g1)x+2(f−f1)y+(c−c1)=0whichisastraightline.Therefore,theradicalaxisoftwogivencircleisastraightline.

Note5.1.1:IfS=0andS1=0aretheequationsoftwocircleswithunitcoefficientsforx2andy2termsthentheequationoftheradicalaxisisS−S1=0.

Note5.1.2:Radicalaxisoftwocirclesisastraightlineperpendiculartothelineofcentres.ThecentresofthetwocirclesareA(−g,−f)andB(−g1,−f1).

Theslopeofthelineofcentresis

Theslopeoftheradicalaxisis

∴m1m2=–1

Therefore,theradicalaxisisperpendiculartothelineofcentres.

Note5.1.3:IfthetwocirclesS=0andS1=0intersectthentheradicalaxisisthecommonchordofthetwocircles.

Note5.1.4:Ifthetwocirclestoucheachother,thentheradicalaxisisthecommontangenttothecircles.

Note5.1.5:Ifacirclebisectsthecircumferenceofanothercirclethentheradicalaxispassesthroughthecentreofthesecondcircle.

Showthattheradicalaxesofthreecirclestakentwobytwoareconcurrent.LetS1=0,S2=0andS3=0betheequationsofthreecircleswithunitcoefficientsforx2andy2terms.ThentheradicalaxesofthecirclestakentwobytwoareS1−S2=0,S2−S3=0andS3−S1=0.

∴(S1−S2)+(S2−S3)+(S3−S1)≡0

Sincesumofthetermsvanishesidentically,thelinesrepresentedbyS1−S2=0,S2−S3=0andS3−S1=0areconcurrent.Thecommonpointofthelinesiscalledtheradicalcentre.

5.2ORTHOGONALCIRCLES

Definition5.2.1:Twocirclesaredefinedtobeorthogonalifthetangentsattheirpointofintersectionareatrightangles.

FindtheconditionforthecirclesS≡x2+y2+2gx+2fy+c=0,S1≡x2+y2

+2g1x+2f1y+c1=0tobeorthogonal.

LetPbeapointofintersectionofthetwocirclesS=0andS1=0.ThecentresareA(−g,−f),B(−g1,−f1).

Theradiiare

Sincethetwocirclesareorthogonal,PAisperpendiculartoPB.(i.e.)APBisarighttriangle.

Showthatifacirclecutstwogivencirclesorthogonallythenitscentreliesontheradicalaxisofthetwogivencircles.LetS1=x2+y2+2g1x+2f1y+c1=0andS2=x2+y2+2g2x+2f2y+c1=0bethetwogivencircles.LetS=x2+y2+2gx+2fy+c=0cutsS1=0andS2=0orthogonally.

SinceS=0cutsS1=0andS2=0orthogonally,

Bysubtracting,weget

Thisshowsthat(−g,−f)liesontheline,2(g1−g2)x+2(f1−f2)y+(c1−c2)=0whichistheradicalaxisofthetwocircles.Therefore,thecentreofthecircleS=0liesontheradicalaxisofthecirclesS1

=0andS2=0.

5.3COAXALSYSTEM

Definition5.3.1:Asystemofcirclesissaidtobecoaxalifeverypairofthesystemhasthesameradicalaxis.

Expresstheequationofacoaxalsystemofcirclesinthesimplestform.Inacoaxalsystemofcircles,everypairofthesystemhasthesameradicalaxis.Therefore,thereisacommonradicalaxistoacoaxalsystemofcircles.Hence,inacoaxalsystemthecentresareallcollinearandthecommonradical

axisisperpendiculartothelinesofcentres.Therefore,letuschoosethelineofcentresasx-axisandthecommonradicalaxisasy-axis.Letusconsidertwocirclesofthecoaxalsystem,

Sincethecentreslieonthex-axis,f1=0andf2=0.Therefore,theequationsofthecirclesarex2+y2+2gx+c=0andx2+y2+

2g1x+c1=0.Theradicalaxisofthesetwocirclesis2(g−g1)x+(c−c1)=0.

However,thecommonradicalaxisisthey-axiswhoseequationisx=0.

∴c−c1=0orc=c1.Hence,thegeneralequationtoacoaxalsystemofcirclesisx2+y2+2gx+c=0wheregisavariableandcisaconstant.

Sotheequationofacoaxalsystemcanbeexpressedinthesimplestform

x2+y2+2λx+c=0whereλisavariableandcisaconstant.

5.4LIMITINGPOINTS

Definition5.4.1:Limitingpointsaredefinedtobethecentresofpointcirclesbelongingtoacoaxalsystem;thatis,theyarecentresofcirclesofzeroradiibelongingtoacoaxalsystem.

Obtainthelimitingpointsofthecoaxalsystemofcirclesx2+y2+2λx+c=

0.Centresare(−λ,0)andradiiare

Forpointcirclesradiiarezero.

Therefore,limitingpointsare

Theorem5.4.1:Thepolarofonelimitingpointofacoaxalsystemofcircleswithrespecttoanycircleofthesystempassesthroughtheotherlimitingpoint.

Proof:Letx-axisbethelineofcentresandy-axisbethecommonradicalaxisofacoaxalsystemofcircles.Thenanycircleofthecoaxalsystemis

whereλisavariableandcisaconstant.

Thelimitingpointsofthiscoaxalsystemofcirclesare

Thepolarofthepoint withrespecttothecircle(5.1)is

Thislinepassesthroughtheotherlimitingpoint.Foreverycoaxalsystemofcirclesthereexistsanorthogonalsystemofcircles.Letx-axisbethelineofcentresandy-axisbethecommonradicalaxis.Thentheequationtoacoaxalsystemofcirclesis

Letusassumethatthecircle

cuteverycircleofthecoaxalsystemofcirclesgivenby(5.2)orthogonally.Thentheconditionfororthogonalityis

Letusnowconsidertwocirclesofthecoaxalsystemforthedifferentvaluesofλ,sayλ1andλ2.Thecondition(5.4)becomes2gλ1=c+k,2gλ2=c+k.

∴2(λ1−λ2)g=0.Sinceλ1−λ2≠0,g=0andsok=−c.Hence,from(5.3)theequationofthecirclewhichcutseverymemberofthe

system(5.2)isx2+y2+2fy−c=0,wherefisanarbitraryconstant.Therefore,foreverycoaxalsystemofcirclesthereexistsanorthogonalsystemofcirclesgivenbyx2+y2+2fy−c=0;wherefisavariableandcisaconstant.Forthis

systemoforthogonalcirclesy-axisisthelineofcentresandx-axisisthecommonradicalaxis.

Note5.4.1:Everycircleoftheorthogonalcoaxalsystemofcirclespasses

throughthelimitingpoints .

Theorem5.4.2:IfS=x2+y2+2gx+2fy+c=0andS1=x2+y2+2g1x+2f1y+c1=0beanytwocirclesofacoaxalsystemthenanycircleofcoaxalsystemcanbeexpressedintheformS+λS1=0.

Proof:

Consider,

whereλisavariable.Inthisequation,(1+λ)isthecoefficientofx2andy2.

Dividingby(1+λ)equation(5.7)becomes inwhichthecoefficientof

x2andy2areunity.

Nowconsidertwodifferentvaluesofλ,thatis,λ1andλ2.Then, and

Theradicalaxisofthesetwocirclesis

Sinceλ1−λ2≠0andthereforeS−S1=0whichisthecommonradicalaxis.Therefore,everymemberofthecoaxalsystemcanbeexpressedintheformS+λS1=0whereλisavariable.

Theorem5.4.3:IfS=x2+y2+2gx+2fy+c=0isacircleofacoaxalsystemandL=lx+my+n=0isthecommonradicalaxisofthesystemthenS+λL=0istheequationofacircleofthecoaxalsystemofcircles.

Proof:

Considertwomembersofthesystem(5.10)forthedifferentvaluesofλ,thatis,λ1andλ2.Then,S+λ1L=0andS+λ2L=0Theradicalaxisofthesetwocirclesis(λ1−λ2)L=0.

Sinceλ1−λ2≠0,L=0whichisthecommonradicalaxis.Therefore,S+λL=0representsanycircleofthecoaxalsysteminwhichS=0isacircleandL=0isthecommonradicalaxis.

5.5EXAMPLES(RADICALAXIS)

Example5.5.1

Findtheradicalaxisofthetwocirclesx2+y2+2x+4y−7=0andx2+y2−6x+2y−5=0andshowthatitisatrightanglestothelineofcentresofthetwocircles.

Solution

TheradicalaxisofthecirclesisS−S1=0.

Theslopeoftheradicalaxisism1=−4.Thecentresofthetwocirclesare(−1,−2)and(3,−1).

Theslopeofthelineofcentresis

Therefore,theradicalaxisisperpendiculartothelineofcentres.

Example5.5.2

Showthatthecirclex2+y2+2gx+2fy+c=0willbisectthecircumferenceofthecirclex2+y2+2g1x+2f1y+c1=0,if2g1(g−g1)+2f1(f−f1)=c−c1.

Solution

Let

Theradicalaxisofthesetwocirclesis2(g−g1)c+2(f−f1)y+c−c1=0,Circle(5.14)bisectsthecircumferenceofthecircle(515).Therefore,radicalaxispassesthroughthecentreofthesecondcircle.Theradicalaxisofthetwogivencirclesbe

Example5.5.3

Showthatthecirclesx2+y2−4x+6y+8=0andx2+y2−10x−6y+14=0toucheachotherandfindthecoordinatesofthepointofcontact.

Solution

Theradicalaxisofthesetwocirclesis6x+12y−6=0.

ThecentresofthecirclesareA(2,−3)andB(5,3).

Theradiiofthecirclesare

TheperpendiculardistancefromA(2,−3)ontheradicalaxisx+2y−1=0is

radiusofthefirstcircle.

Therefore,radicalaxistouchesthefirstcircleandhencethetwocirclestoucheachother.

Theequationofthelinesofcentresis

or

Solving(5.18)and(5.19),wegetthepointofcontact.Therefore,thepointofcontactis(3,−1).

Example5.5.4

Showthatthecirclesx2+y2+2ax+c=0andx2+y2+2by+c=0touchif

Solution

Theradicalaxisofthetwogivencirclesis2ax−2by=0.Thecentreofthefirst

circleis(−a,0).Theradiusofthefirstcircleis

Ifthetwocirclestoucheachother,thentheperpendiculardistancefromthecentre(−a,0)totheradicalaxisisequaltotheradiusofthecircle.

Ondividingbya2b2c,weget

Example5.5.5

Findtheradicalcentreofthecirclesx2+y2+4x+7=0,2x2+2y2+3x+5y+9=0andx2+y2+y=0.

Solution

Let

Theradicalaxisofcircles(5.20)and(5.22)is

Theradicalaxisofthecircles(5.20)and(5.22)is

Solving(5.23)and(5.24)wegettheradicalcentreasfollows:

Therefore,theradicalcentreis(−2,−1).

Example5.5.6

Provethatifthepointsofintersectionofthecirclesx2+y2+ax+by+c=0andx2+y2+a1x+b1y+c1=0bythelinesAx+By+C=0andA1x+B1y+C1=0areconcyclicif

Solution

Let

Ax+By+C=0meetsthecircle(5.25)atPandQandA1x+B1y+C1=0meetsthecircle(5.26)atRandS.SinceP,Q,RandSareconcyclic,theequationofthiscirclebe

Theradicalaxisofthecircles(5.25)and(5.26)is



Sincethesethreeradicalaxesareconcurrentwegetfromequations(5.30),(5.31)and(5.32),

Example5.5.7

Provethatthedifferenceofthesquareofthetangentstotwocirclesfromanypointintheirplanevariesasthedistanceofthepointfromtheirradicalaxis.

Solution

LetP(x1,y1)beanypointandthetwocirclesbe

Theequationtotheradicalaxisofthesetwocirclesbe

Theperpendiculardistanceofthepointfromtheradicalaxisis

Fromequations(5.36)and(5.37),weget

Example5.5.8

Provethatforallconstantsλandμ,thecircle(x−a)(x−a+λ)+(y−b)(y−b+μ)=r2bisectsthecircumferenceofthecircle(x−a)2+(y−b)2=r2.

Solution


Thecentreofthesecondcircleis(a,b).Substitutingx=a,y=bin(5.40),weget

λa+μb−λa−μb=0.

∴(a,b)liesontheradicalaxis.Therefore,theradicalaxisbisectsthecircumferenceofthesecondcircle.

Example5.5.9

Provethatthelengthofcommonchordofthetwocirclesx2+y2+2λx+c=0

and

Solution

Thetwogivencirclesare

CentresareA(−λ,0)andB(−μ,0),radiiare

Theradicalaxisisλx−μy+c=0.TheperpendiculardistancefromAon

Therefore,thelengthofcommonchord

Example5.5.10

Showthatthecirclex2+y2−8x−6y+21=0isorthogonaltothecirclex2+y2

−2y−15=0.Findthecommonchordandtheequationofthecirclepassingthroughthecentresandintersectingpointsofthecircles.

Solution

Theconditionfororthogonalityis2gg1+2ff1=c+c1.

(i.e.)2(−4)(0)+2(−3)(−1)=21−150+6=6whichistrue.

Therefore,thetwocirclescuteachotherorthogonally.TheequationofthecommonchordisS−S1=0.

AnycirclepassingthroughtheintersectionofthecirclesisS+λL=0.

(i.e.)x2+y2−8x−6y+21+λ(2x+y−9)=0.

Thispassesthroughthecentre(4,3)ofthefirstcircle.

Therefore,theequationoftherequiredcircleisx2+y2−8x−6y+21+2(2x+y−9)=0.

(i.e.)x2+y2−4x−4y+3=0

Example5.5.11

Findtheequationtothecirclewhichcutsorthogonallythethreecirclesx2+y2+2x+17y+4=0,x2+y2+7x+6y+11=0andx2+y2−x+22f+33=0.

Solution

Lettheequationofthecirclewhichcutsorthogonallythethreegivencirclesbex2+y2+2gx+2fy+c=0.Thentheconditionsfororthogonalityare

From(5.45),weget3g+10=1

From(5.41),weget−6−34=c+4orc=−44Therefore,theequationofthecirclewhichcutsorthogonallythethreegiven

circlesisx2+y2−6x−4y−44=0.

Aliter



Therefore,radicalcentreis(3,2).

Therefore,radicalcentreis(3,2).IfRisthelengthofthetangentfrompoints(3,2)tothefirstcirclethenR2=9

+4+6+34+4=57.Therefore,theequationoftherequiredcircleis(x−3)2+(y−2)2=57.

(i.e.)x2+y2−6x−4y−44=0

Example5.5.12

Findtheequationofthecirclewhichpassesthroughtheorigin,hasitscentreonthelinex+y=4andcutsorthogonallythecirclex2+y2−4x+2y+4=0.

Solution

Lettheequationoftherequiredcirclepassingthroughtheoriginbe

Thiscirclecutsorthogonallythecircle

Thecentreofthecircle(5.48)liesonx+y=4.

Adding,weget

Therefore,theequationoftherequiredcircleisx2+y2−4x−4y=0.

Example5.5.13

IftheequationofthecircleswithradiirandRareS=0andS1=0,respectively

thenshowthatthecircles willintersectorthogonally.

Solution

Withoutlossofgenerality,wecanassumethelineofcentresofthetwocirclesasx-axisandthedistancebetweenthecentresas2a.Thenthecentresofthetwocirclesare(a,0)and(−a,0).TheequationofthetwocirclesareS=(x−a)2+y2

−r2=0andS1=(x+a)2+y2−R2=0.

Consider

∴RS±rS1=0Clearly,thecoefficientsoftheRandrintheseequationsarethesameandsotheyrepresentcircles.

ConsiderRS+rS1=0

Also,RS−rS=0hastheequation

Equations(5.52)and(5.53)canbewrittenas R=0and

Theconditionfororthogonalityis2gg1+2ff1=c+c1.

Therefore,thecircles areorthogonal.

Exercises(RadicalAxis)

1. Findtheradicalaxisofthecirclesx2+y2+2x+4y=0and2x2+2y2−7x−8y+1=0.Ans.:11x+16y−1=0

2. Findtheradicalaxisofthecirclesx2+y2−4x−2y−11=0andx2+y2−2x−6y+1=0andshowthattheradicalaxisisperpendiculartothelineofcentres.

Ans.:x−2y+5=0

3. Showthatthecirclesx2+y2−6x−9y+13=0andx2+y2−2x−16y=0toucheachother.Findthecoordinatesofpointofcontact.

Ans.:(5,1)

4. Findtheequationofthecommonchordofthecirclesx2+y2+2ax+2by+c=0andx2+y2+

2bx+2ay+c=0andalsoshowthatthecirclestouchif(a+b)2=2c.

5. Showthatthecirclesx2+y2+2x−8y+8=0andx2+y2+10x−2y+22=0toucheachotherandfindthepointofcontact.

Ans.:

6. Findtheequationofthecirclepassingthroughtheintersectionofthecirclesx2+y2=6andx2+

y2−6x+8=0andalsothroughthepoint(1,1).


7. Findtheequationofthecirclepassingthroughthepointofintersectionofthecirclesx2+y2−6x

+2y+4=0andx2+y2+2x−4y−6=0andwhoseradiusis3/2.

Ans.:5x2+5y2−18x+y+5=0

8. Ifthecirclesx2+y2+2gx+2fy=0andx2+y2+2g1x+2f1y=0toucheachotherthenshowthatfg1=f1g.

9. Findtheradicalcentreofthecirclesx2+y2+aix+biy+c=0,i=1,2,3.Ans.:(0,0)

10. Findtheradicalcentreofthecirclesx2+y2−x+3y−3=0,x2+y2−2x+2y+2=0andx2+

y2+2x+2y−9=0.Ans.:(2,1)

11. Theradicalcentreofthreecirclesisattheorigin.Theequationoftwoofthecirclesarex2+y2=1

andx2+y2+4x+4y−1=0.Findthegeneralformofthethirdcircle.Ifitpassesthrough(1,1)and(−2,1)thenfinditsequation.

Ans.:x2+y2+x−2y−1=0

12. Findtheradicalcentreofthecirclesx2+y2+x+2y+3=0,x2+y2+4x+7=0and2x2+2y2+3x+5y+9=0.

Ans.:(−2,−1)

13. Findtheequationofthecirclewhoseradiusis3andwhichtouchesthecirclex2+y2−4x−6y+2=0internallyatthepoint(−1,−1).

14. Showthattheradicalcentresofthreecirclesdescribedonthesidesofatriangleasdiameteristheorthocentreofthetriangle.

15. Findtheequationofthecirclewhichcutsorthogonallythethreecirclesx2+y2+y=0,x2+4y2+

4x+7=0,21x2+y2+3x+5y+9=0.

Ans.:x2+y2+4x+2y+1=0

16. AandBaretwofixedpointsandPmovessothatPA=n·PB.ShowthatthelocusofPisacircleandthatfordifferentvaluesofn,allthecircleshavethesameradicalaxis.

17. Findtheequationofcirclewhoseradiusis5andwhichtouchesthecirclex2+y2−2x−4y−20=0atthepoint(5,5).

18. Provethatthelengthofthecommonchordofthetwocircleswhoseequationsare(x−a)2+(y−

b)2=r2and(x−a)2+(y−b)2=c2is

19. Findtheequationtotwoequalcircleswithcentres(2,3)and(5,6)whichcutseachotherorthogonally.

20. IfthreecircleswithcentresA,BandCcuteachotherorthogonallyinpairsthenprovethatthepolarofAwithrespecttothecirclecentreBpassesthroughC.

21. Findthelocusofcentresofallthecircleswhichtouchthelinex=2aandcutthecirclex2+y2=

a2orthogonally.22. A,Barethepoints(a,0)and(−a,0).ShowthatifavariablecircleSisorthogonaltothecircleon

ABasdiameter,thepolarof(a,0)withrespecttothecircleSpassesthroughthefixedpoint(−a,0).

23. Ifacirclepassesthroughthepoint(a,b)andcutsthecirclex2+y2=k2orthogonallythenprove

thatthelocusofitscentresis2ax+2by−(a2+b2+k2)=0.

24. Showthatthecirclesx2+y2+10x+6y+14=0andx2+y2−4x+6y+8=0toucheachotheratthepoint(3,−1).

25. Showthatthecirclesx2+y2+2ax+4ay−3a2=0andx2+y2−8ax−6ay+7a2=0toucheachotheratthepoint(a,0).

26. Theequationofthreecirclesarex2+y2=1,x2+y2+8x+15=0andx2+y2+10y+24=0.Determinethecoordinateofthepointsuchthatthetangentsdrawnfromittothethreecirclesareequalinlength.

27. IfPandQbeapairofconjugatepointswithrespecttoacircleS=0thenprovethatthecircleonPQasdiametercutsthecircleS=0orthogonally.

28. Findtheequationofthecirclewhosediameteristhecommonchordofthecirclesx2+y2+2x+

3y+1=0andx2+y2+4x+3y+2=0.

5.6EXAMPLES(LIMITINGPOINTS)

Example5.6.1

IfA,BandCarethecentresofthreecoaxalcirclesandt1,t2andt3arethelengthsoftangentstothemfromanypointthenprovethat

Solution

Letthethreecirclesofcoaxalsystembe

ThecentresareA(−g1,0),B(−g2,0)andC(−g3,0)andBC=g3−g2,CA=g1−g3,

Then,

sinceΣ(g2−g3)=0andΣg1(g2−g3)=0.

Example5.6.2

Findtheequationsofthecircleswhichpassthroughthepointsofintersectionofx2+y2−2x+1=0andx2+y2−5x−6y−4=0andwhichtouchtheline2x−y+3=0.

Solution


Theequationofanycirclepassingthroughtheintersectionofthesetwocirclesisx2+y2−2x+1+λ(x+2y−1)=0.

Thecentreofthiscircleis andradius= Thecircle

(5.56)touchestheline2x−y+3=0.

Theequationofthecirclesarex2+y2−2x+1±2(x+2y−1)=0.(i.e.)x2+y2−2x+1+2x+4y−2=0andx2+y2−2x+1−2x−4y+2=0(i.e.)x2+y2+4y−1=0andx2+y2−4x+4y+3=0

Example5.6.3

Findtheequationofthecirclewhichpassesthroughtheintersectionofthe

circlesx2+y2=4andx2+y2−2x−4y+4=0andhasaradius

Solution

Theradicalaxisofthesetwocirclesis2x+4y−8=0.Anycirclepassingthroughtheintersectionofthesetwocirclesis

Centreis(−λ,−2λ).

Therefore,therequiredcirclesarex2+y2−4−2(2x+4y−8)=0and

(i.e.)x2+y2−4x−8y+12=0and5x2+5y2+4x+8y−36=0

Example5.6.4

Findtheequationofthecirclewhosediameteristhecommonchordofthecirclesx2+y2+2x+3y+1=0andx2+y2+4x+3y+2=0.

Solution

Theradicalaxisofthesetwocirclesis2x+1=0.Anycircleofthesystemisx2+y2+2x+3y+1+λ(2x+1)=0.

Centreis .

Sincetheradicalaxisisadiameter,centreliesontheradicalaxis.

Hence,theequationoftherequiredcircleis

Example5.6.5

Findtheequationofthecirclewhichtouchesx-axisandiscoaxalwiththecirclesx2+y2+12x+8y−33=0andx2+y2=5.

Solution


Anycircleofthecoaxalsystemisx2+y2−5+λ(6x+4y−14)=0.Centreis(−3λ,−2λ).

Thecircles(5.69)touchesx-axis(i.e.)y=0.

Therefore,thetwocirclesofthesystemtouchingx-axisare

Example5.6.6

Theline2x+3y=1cutsthecirclex2+y2=4inAandB.ShowthattheequationofthecircleonABasdiameteris13(x2+y2)−4x−6y−50=0.

Solution

Let

Anycirclepassingthroughtheintersectionofthecircleandthelineis

Centreis andradius=

IfABisadiameterofthecircle(5.72),theircentreshouldlieonAB.

Therefore,theequationofthecircleonABasdiameteris13(x2+y2−4)−2(2x+3y−1)=0.

∴13(x2+y2)−4x−6y+50=0

Example5.6.7

Apointmovessothattheratioofthelengthoftangentstothecirclesx2+y2+4x+3=0andx2+y2−6x+5=0is2:3.Showthatthelocusofthepointisacirclecoaxalwiththegivencircles.

Solution

ThelengthsoftangentsfromapointP(x1,y1)tothetwocirclesare

Giventhat,

Thelocusof

ThisisoftheformS1+λS2=0Hencethelocusofcircleisacirclecoaxalwiththetwogivencircles.

Example5.6.8

Findthelimitingpointsofthecoaxalsystemdeterminedbythecirclex2+y2+2x+4y+7=0andx2+y2+4x+2y+5=0.

Solution

Giventhat,

Theradicalaxisofthesetwocirclesis2x−2y−2=0.Anycircleofthecoaxalsystemisx2+y2+2x+4y+7+λ(2x−2y−2)=0.Centreis(−1−λ,−2+λ).

Radiusis

Limitingpointsarethecentresofcirclesofradiizero.Therefore,limitingpointsare(−2,−1)and(0,−3).

Example5.6.9

Thepoint(2,1)isalimitingpointofasystemofcoaxalcirclesofwhichx2+y2

−6x−4y−3=0isamember.Findtheequationtotheradialaxisandthecoordinatesoftheotherlimitingpoint.

Solution

Giventhat

x2+y2−6x−4y−3=0Since(2,1)isalimitpoint,thepointcirclecorrespondingtothecoaxalsystemis

Theradicalaxisofthesystemis

AnycircleofthecoaxalsystemisS+λL=0.

x2+y2−6x−4y−3+λ(2x+2y+8)=0

Centreis(3−λ,2−λ).

Radius

Forpointcircles,radius=0.

Therefore,thelimitingpointsarethecentresofpointcircleofthecoaxalsystem,thatis,(2,1)and(−5,−6).

Example5.6.10

Findtheequationofthecirclewhichpassesthoughtheoriginandbelongstothecoaxalsystemofwhichlimitingpointsare(1,2)and(4,3).

Solution

Since(1,2)and(4,3)arelimitingpointsoftwocirclesofthecoaxalsystemand(x−1)2+(y−2)2=0and(x−4)2+(y−3)2=0.

Radicalaxisis6x+2y−20=0.Anycircleofthesystemisx2+y2−2x−4y+5+λ(6x+2y−20)=0.Thispassesthroughtheorigin.

Hence,theequationofthesystemis

Example5.6.11

ApointPmovessothatitsdistancesfromtwofixedpointsareinaconstantratioλ.ProvethatthelocusofPisacircle.IfλvariesthenshowthatPgeneratesasystemofcoaxalcirclesofwhichthefixedpointsarethelimitingpoints.

Solution

LetP(x1,y1)beamovingpointandA(c,0)andB(0,−c)bethetwofixedpoints.Here,wehavechosenthefixedpointsonthex-axissuchthatPisitsmidpoint.Giventhat

ThisequationisoftheformS+λS′=0whichistheequationtoacoaxalsystemofcircles.Therefore,fordifferentvaluesofλ,Pgeneratesacoaxalsystemofcirclesof

which(x−a)2+y2=0and(x+a)2+y2=0aremembers.Theseequationsaretheequationofpointcircleswhosecentresare(a,0),(−a,0)whichisthefixedpoints.

Example5.6.12

Provethatthelimitingpointofthesystemx2+y2+2gx+c+λ(x2+y2+2fy+k)

=0subtendsarightangleattheoriginif

Solution

Thetwomembersofthesystemarex2+y2+2gx+c=0andx2+y2+2fy+k=0.Radicalaxisis2gx−2fy+c−k=0.Anycircleofthesystemisx2+y2+2gx+c+λ(2gx−2fy+c−k)=0.Centreis(−g−gλ,fλ).

Radius

Forpointcircle,radius=0.

Consideringthetwovaluesofλasλ1,λ2,

centresareA(−g(1+λ1),fλ1)andB(−g(1+λ2),fλ2)SinceOABisrightangledatO,OAisperpendiculartoOB.

Exercises

1. Findtheequationofthecirclepassingthroughtheintersectionofx2+y2−6=0andx2+y2+4y−1=0throughthepoint(−1,1).

Ans.:9x2+9y2+16y−34=0

2. Showthatthecirclesx2+y2=480andx2+y2−10x−24y+120=0toucheachotherandfindtheequation,ifathirdcirclewhichtouchesthecirclesattheirpointofintersectionandthex-axis

x2+y2−200x−400y+10000=0.

Ans.:5x2+5y2−40x+96y+30=0

3. Findtheequationofthecirclewhosecentreliesonthelinex+y−11=0andwhichpasses

throughtheintersectionofthecirclex2+y2−3x+2y−4=0withtheline2x+5y−2=0.

4. Findthelengthofthecommonchordofthecirclesx2+y2+4x−22y=0andx2+y2−10x+5y=0.

Ans.:40/7

5. Findthecoordinatesofthelimitingpointsofthecoaxalcirclesdeterminedbythetwocirclesx2+

y2−4x−6y−3=0andx2+y2−24x−26y+277=0.Ans.:(1,2),(3,1)

6. Findthecoordinatesofthelimitingpointsofthecoaxalsystemofcirclesofwhichtwomembers

arex2+y2+2x−6y=0and2x2+2y2−10y+5=0.Ans.:(1,2),(3,1)

7. Findthecoaxalsystemofcirclesifoneofwhosemembersisx2+y2+2x−6y=0andalimitingpointis(1,−2).

Ans.:x2+y2+2x+3y−7−λ(4x−y−12)=0

8. Findthelimitingpointofthecoaxalsystemdeterminedbythecirclesx2+y2−6x−6y+4=0

andx2+y2−2x−4y+3=0.

Ans.:

9. Findtheequationofthecoaxalsystemofcirclesoneofwhosemembersisx2+y2−4x−2y−5=0andthelimitingpointis(1,2).

Ans.:x2+y2−2x−4y+5+λ(x−y−5)=0

10. Iforiginisalimitingpointofasystemofcoaxalcirclesofwhichx2+y2+2gx+2fy+c=0isa

memberthenshowthattheotherlimitingpointsis

11. Showthattheequationofthecoaxalsystemwhoselimitingpointsare(0,0)and(a,b)isx2+y2+

k(2ax−2by−a2−b2)=0.

12. Theoriginisalimitingpointofasystemofcoaxalcirclesofwhichx2+y2+2gx+2fy+c=0isa

member.Showthattheequationofcirclesoftheorthogonalsystemis(x2+y2)(g+λf)+c(x−λy)=0fordifferentvaluesofx.

13. Showthatthecirclesx2+y2+2ax+2by+2λ(ax−by)=0whereλisaparameterfromacoaxalsystemandalsoshowthattheequationofthecommonradicalaxisandtheequationofcircles

whichareorthogonaltothissystemare

14. ApointPmovessuchthatthelengthoftangentstothecirclesx2+y2−2x−4y+5=0andx2+

y2+4x+6y−7=0areintheratio3:4.Showthatthelocusisacircle.

15. Showthatthelimitingpointsofthecirclex2+y2=a2andanequalcirclewithcentreontheline

lx+my+n=0beontheline(x2+y2)(lx+my+n)+a2(ln+mn)=0.

Chapter6

Parabola

6.1INTRODUCTION

Ifapointmovesinaplanesuchthatitsdistancefromafixedpointbearsaconstantratiotoitsperpendiculardistancefromafixedstraightlinethenthepathdescribedbythemovingpointiscalledaconic.Inotherwords,ifSisafixedpoint,lisafixedstraightlineandPisamovingpointandPMisthe

perpendiculardistancefromPonl,suchthat constant,thenthelocusofP

iscalledaconic.Thisconstantiscalledtheeccentricityoftheconicandisdenotedbye.

Ife=1,theconiciscalledaparabola.Ife<1,theconiciscalledanellipse.Ife>1,theconiciscalledahyperbola.ThefixedpointSiscalledthefocusoftheconic.Thefixedstraightlineis

calledthedirectrixoftheconic.Theproperty iscalledthefocus-directrix

propertyoftheconic.

6.2GENERALEQUATIONOFACONIC

Wecanshowthattheequationofaconicisaseconddegreeequationinxandy.Thisisderivedfromthefocus-directrixpropertyofaconic.LetS(x1,y1)bethefocusandP(x,y)beanypointontheconicandlx+my+n=0betheequationofthedirectrix.Thefocus-directrixpropertyoftheconicstates

(i.e.)

Thisequationcanbeexpressedintheformax2+2hxy+by2+2gx+2fy+c=0whichisaseconddegreeequationinxandy.

6.3EQUATIONOFAPARABOLA

LetSbethefocusandthelinelbethedirectrix.WehavetofindthelocusofapointPsuchthatitsdistancefromthefocusSisequaltoitsdistancefromthefixedlinel.

(i.e.) wherePMisperpendiculartothedirectrix.

DrawSXperpendiculartothedirectrixandbisectSX.LetAbethepointofbisectionandSA=AX=a.ThenthepointAisapointontheparabolasince

.TakeASasthex-axisandAYperpendiculartoASasthey-axis.Thenthe

coordinateofSare(a,0).Let(x,y)bethecoordinatesofthepointP.DrawPNperpendiculartothex-axis.

This,beingthelocusofthepointP,istheequationoftheparabola.Thisequationisthesimplestpossibleequationtoaparabolaandiscalledthestandardequationoftheparabola.

Note6.3.1:

1. ThelineAS(x-axis)iscalledtheaxisoftheparabola.2. ThepointAiscalledthevertexoftheparabola.3. AY(y-axis)iscalledthetangentatthevertex.4. Theperpendicularthroughthefocusiscalledthelatusrectum.5. Thedoubleordinatethroughthefocusiscalledthelengthofthelatusrectum.6. Theequationofthedirectrixisx+a=0.7. Theequationofthelatusrectumisx–a=0.

6.4LENGTHOFLATUSRECTUM

Tofindthelengthofthelatusrectum,drawLM′perpendiculartothedirectrix.

Then

6.4.1Tracingofthecurvey2=4ax

1. Ifx<0,yisimaginary.Therefore,thecurvedoesnotpassthroughtheleftsideofy-axis.2. Wheny=0,wegetx=0.Therefore,thecurvemeetsthey-axisatonlyonepoint,thatis,(0,0).

3. Whenx=0,y2=0,thatis,y=0.Hencethey-axismeetsthecurveattwocoincidentpoints(0,0).Hencethey-axisisatangenttothecurveat(0,0).

4. If(x,y)isapointontheparabolay2=4ax,(x,–y)isalsoapoint.Therefore,thecurveis

symmetricalaboutthex-axis.5. Asxincreasesindefinitely,thevaluesofyalsoincreasesindefinitely.Thereforethepointsofthe

curvelyingontheoppositesidesofx-axisextendtoinfinitytowardsthepositivesideofx-axis.

6.5DIFFERENTFORMSOFPARABOLA

1. Ifthefocusistakenatthepoint(–a,0)withthevertexattheoriginanditsaxisasx-axisthenits

equationisy2=–4ax.

2. Iftheaxisoftheparabolaisthey-axis,vertexattheoriginandthefocusat(0,a),theequationof

theparabolaisx2=4ay.

3. Ifthefocusisat(0,−a),vertex(0,0)andaxisasy-axis,thentheequationoftheparabolaisx2=–4ay.

ILLUSTRATIVEEXAMPLESBASEDONFOCUSDIRECTRIXPROPERTY

Example6.1

Findtheequationoftheparabolawiththefollowingfocianddirectrices:i. (1,2):x+y–2=0ii. (1,–1):x–y=0iii. (0,0):x–2y+2=0

Solution

i. LetP(x,y)beanypointontheparabola.DrawPMperpendiculartothedirectrix.Thenfromthe

definitionoftheparabola,

∴SP2=(x–1)2+(y–2)2PM=perpendiculardistancefrom(x,y)onx+y–2=0

Thisistheequationoftherequiredparabola.ii. ThepointSis(1,−1).Directrixisx–y=0

Fromanypointontheparabola,

iii. Sis(0,0).Directrixisx–2y+2=0

ForanypointPontheparabola,

Example6.2

Findthefoci,latusrectum,verticesanddirectricesofthefollowingparabolas:i. y2+4x–2y+3=0

ii. y2–4x+2y–3=0

iii. y2–8x–9=0

Solution

i.

Takex+ =X,y–1=Y.Shiftingtheorigintothepoint theequationoftheparabola

becomesY2=−4X.

∴Vertexis ,latusrectumis4,focusis andfootofthedirectrixis .The

equationofthedirectrixisx= or2x–1=0.

ii.

Shiftingtheorigintothepoint(–1,–1)bytakingx+1=Xandy+1=Ytheequationofthe

parabolabecomesY2=4X.∴Vertexis(−1,−1),latusrectum=4,focusis(0,−1)andfootofthedirectrixis(−2,−1).∴Theequationofthedirectrixisx+2=0.

y2–8x–9=0⇒y2=8x+9

Shifttheorigintothepoint andtake

∴TheequationoftheparabolabecomesY2=8X.Vertexis ,latusrectum=8andfocusis

Theequationofthedirectrixis

Exercises

1. Findtheequationoftheparabolawhosefocusis(2,1)anddirectrixis2x+y+1=0.

Ans.:x2–4xy+4y2–24x–12y+24=0

2. Findtheequationoftheparabolawhosefocusis(3,−4)andwhosedirectrixisx–y+5=0.

Ans.:x2+2xy+y2–16x–26y+25=0

3. Findthecoordinatesofthevertex,focusandtheequationofthedirectrixoftheparabola3y2=16x.Findalsothelengthofthelatusrectum.

Ans.:

4. Findthecoordinatesofthevertexandfocusoftheparabola2y2+3y+4x=2.Findalsothelengthofthelatusrectum.

Ans.:

5. Apointmovesinsuchawaythatthedistancefromthepoint(2,3)isequaltothedistancefromtheline4x+3y=5.Findtheequationofitspath.Whatisthenameofthiscurve?

Ans.:25[(x–2)2+(y–3)2]–(4x+3y–5)2

6.6CONDITIONFORTANGENCY

Findtheconditionforthestraightliney=mx+ctobeatangenttotheparabolay2=4axandfindthepointofcontact.

Solution

Theequationoftheparabolais

Theequationofthelineis

Solvingequations(6.1)and(6.2),wegettheirpointsofintersection.Thex-coordinatesofthepointsofintersectionaregivenby

Ify=mx+cisatangenttotheparabola,thentherootsofthisequationareequal.Theconditionforthisisthediscriminantisequaltozero.

Hence,theconditionfory=mx+ctobeatangenttotheparabolay2=4axisc=a/m.Substitutingc=a/minequation(6.3),weget

Therefore,thepointofcontactis

Note6.6.1:Anytangenttotheparabolais

6.7NUMBEROFTANGENTS

Showthattwotangentscanalwaysbedrawnfromapointtoaparabola.

Solution

Lettheequationtotheparabolabey2=4ax.Let(x1,y1)bethegivenpoint.Any

tangenttotheparabolais Ifthistangentpassesthrough(x1,y1),then

(1)Thisisaquadraticequationinm.

Therefore,therearetwovaluesofmandforeachvalueofmthereisatangent.Hence,therearetwotangentsfromagivenpointtotheparabola.

Note6.7.1:Ifm1,m2aretheslopesofthetwotangentsthentheyaretherootsofequation(6.3).

6.8PERPENDICULARTANGENTS

Showthatthelocusofthepointofintersectionofperpendiculartangentstoaparabolaisthedirectrix.

Solution

Lettheequationoftheparabolabey2=4ax.Let(x1,y1)bethepointofintersectionofthetwotangentstotheparabola.Anytangenttotheparabolais

Ifthistangentpassesthrough(x1,y1)then

Ifm1,m2aretheslopesofthetwotangentsfrom(x1,y1),thentheyaretherootsofequation(6.5).Sincethetangentsareperpendicular,

Therefore,thelocusof(x1,y1)isx+a=0,whichisthedirectrix.Showthatthelocusofthepointofintersectionoftwotangentstotheparabola

thatmakecomplementaryangleswiththeaxisisalinethroughthefocus.

Solution

Let(x1,y1)bethepointofintersectionoftangentstotheparabolay2=4ax.Any

tangenttotheparabolais Ifthislinepassesthrough(x1,y1),then

Ifm1,m2aretheslopesofthetwotangents,then

Ifthetangentsmakecomplementaryangleswiththeaxisoftheparabola,thenm1=tanθandm2=tan(90–θ).

Thelocusofthepointofintersectionofthetangentsisx–a=0,whichisastraightlinethroughtheorigin.

6.9EQUATIONOFTANGENT

Findtheequationofthetangentat(x1,y1)totheparabolay2=4ax.LetP(x1,y1)andQ(x2,y2)betwopointsontheparabolay2=4ax.Then

Theequationofthechordjoiningthepoints(x1,y1)and(x2,y2)is


Hence,theequationofthechordPQis

WhenthepointQ(x2,y2)tendstocoincidewithP(x1,y1),thechordPQbecomesthetangentatP.Hence,theequationofthetangentatPis

Aliter:Theequationoftheparabolaisy2=4ax.

Differentiatingthisequationwithrespecttox1,weget

Theequationofthetangentat(x1,y1)is

6.10EQUATIONOFNORMAL

Findtheequationofthenormalat(x1,y1)ontheparabolay2=4ax.

Solution

Theslopeofthetangentat(x1,y1)is

Therefore,theslopeofthenormalat(x1,y1)is

Theequationofthenormalat(x1,y1)is

6.11EQUATIONOFCHORDOFCONTACT

Findtheequationofthechordofcontactoftangentsfrom(x1,y1)totheparabolay2=4ax.

Solution

LetQRbethechordofcontactoftangentsfromP(x1,y1).LetQandRbethepoints(x2,y2)and(x3,y3),respectively.Then,theequationoftangentsatQandRare

ThesetwotangentspassthroughP(x1,y1).

Thesetwoequationsshowthatthepoints(x2,y2)and(x3,y3)lieonthelineyy1=2a(x+x1).Therefore,theequationofthechordofcontactoftangentsfromP(x1,y1)is

yy1=2a(x+x1).

6.12POLAROFAPOINT

Findthepolarofthepointwithrespecttotheparabolay2=4ax.

Definition6.12.1Thepolarofapointwithrespecttoaparabolaisdefinedasthelocusofthepointofintersectionofthetangentsattheextremitiesofachordpassingthroughthatpoint.

Solution

LetP(x1,y1)bethegivenpoint.LetQRbeavariablechordpassingthroughP.LetthetangentsatQandRintersectat(h,k).Thentheequationofthechordofcontactoftangentsfrom(h,k)isyk=2a(x+h).ThischordpassesthroughP(x1,y1).

∴y1k=2a(x1+h)Thenthelocusof(h,k)isyy1=2a(x+x1)Hence,thepolarof(x1,y1)withrespecttoy2=4axis

yy1=2a(x+x1)

Note6.12.1:PointPisthepoleofthelineyy1=2a(x+x1).

Note6.12.2:Findthepoleofthelinelx+my+n=0withrespecttotheparabolay2=4ax.Let(x1,y1)bethepole.Thenthepolarof(x1,y1)is

Butthepolarof(x1,y1)isgivenbylx+my+n=0(6.13)

Equations(6.12)and(6.13)representthesameline.Then,identifyingthesetwoequations,weget

Hence,thepoleofthelineis

6.13CONJUGATELINES

Definition6.13.1Twolinesaresaidtobeconjugatetoeachotherifthepoleofeachliesontheother.Findtheconditionforthelineslx+my+n=0andl1x+m1y+n1=0tobeconjugatelineswithrespecttotheparabolay2=4ax.

Solution

Let(x1,y1)bethepoleofthelinesl1x+m1y+n1=0withrespecttotheparabola.Thepolarof(x1,y1)withrespecttothepolary2=4axislx+my+n=0.Theequationofthepolarof(x1,y1)withrespecttotheparabolay2=4axis

Butthepolarof(x1,y1)isgivenbylx+my+n=0(6.15)

Equations(6.14)and(6.15)representthesameline.Identifyingthesetwoequations,weget

Thepoleofthelinelx+my+n=0is

Sincethelineslx+my+n=0andl1x+m1y+n1=0areconjugatetoeachother,thepoleoflx+my+n=0willlieonl1x+m1y+n1=0.


6.14PAIROFTANGENTS

Findtheequationofpairoftangentsfrom(x1,y1)totheparabolay2=4ax.

Solution

Theequationofalinethrough(x1,y1)is

Anypointonthislineis(x1+rcosθ,y1+rsinθ).Thepointsofintersectionofthelineandtheparabolaaregivenby

Thetwovaluesofrofthisequationarethedistancesofpoint(x,y)tothepoint(x1,y1).Ifline(6.16)isatangenttotheparabola,thenthetwovaluesofrmustbeequalandtheconditionforthisisthediscriminantofquadratic(6.17)iszero.

∴4(y1sinθ–2acosθ)2=4sin2θ(y2–4ax1)Eliminatingθinthisequationwiththehelpof(6.16),weget

Therefore,theequationofpairoftangentsfrom(x1,y1)is

6.15CHORDINTERMSOFMIDPOINT

Findtheequationofachordoftheparabolaintermsofitsmiddlepoint(x1,y1).

Solution


Anypointonthislineis(x1+rcosθ,y1+rsinθ).Whenthechordmeetstheparabolay2=4ax,thispointliesonthecurve.

ThetwovaluesofrarethedistancesRPandRQ,whichareequalinmagnitudebutoppositeinsign.Theconditionforthisisthecoefficientofrisequaltozero.

Thisistherequiredequationofthechord.

6.16PARAMETRICREPRESENTATION

x=at2,y=2atsatisfytheequationy2=4ax.Thismeans(at2,2at)isapointontheparabola.Thispointisdenotedby‘t’andtiscalledaparameter.

6.17CHORDJOININGTWOPOINTS

Findtheequationofthechordjoiningthepoints onthe

parabolay2=4ax.

Solution

Theequationofthechordjoiningthepointsis

Note6.17.1:Thechordbecomesthetangentat‘t’ift1=t2=t.Therefore,theequationofthetangentattis

y(2t)=2x+2at2oryt=x+at2

6.18EQUATIONSOFTANGENTANDNORMAL

Findtheequationofthetangentandnormalat‘t’ontheparabolay2=4ax.

Solution

Theequationoftheparabolaisy2=4ax.Differentiatingwithrespecttox,

Theequationofthetangentattis

Theslopeofthenormalattis−t.Theequationofthenormalat‘t’is

6.19POINTOFINTERSECTIONOFTANGENTS

Findthepointofintersectionoftangentsatt1andt2ontheparabolay2=4ax.

Solution

Theequationoftangentsatt1andt2are

Hence,thepointofintersectionis[at1t2,a(t1+t2)].

6.20POINTOFINTERSECTIONOFNORMALS

Findthepointofintersectionofnormalsatt1andt2.

Solution

6.21NUMBEROFNORMALSFROMAPOINT

Showthatthreenormalscanalwaysbedrawnfromagivenpointtoaparabola.

Solution

Lettheequationoftheparabolabey2=4ax.Theequationofthenormalattis

y+xt=2at+at3Ifthispassesthrough(x1,y1)then

Thisbeingacubicequationint,therearethreevaluesfort.Foreachvalueoftthereisanormalfrom(x1,y1)totheparabolay2=4ax.

Note6.21.1:Ift1,t2,t3aretherootsofequation(6.18),then

Note6.21.2:From(6.18),2at1+2at2+2at3=0Therefore,thesumofthecoordinatesofthefeetofthenormalisalwayszero.

6.22INTERSECTIONOFAPARABOLAANDACIRCLE

Provethatacircleandaparabolameetatfourpointsandshowthatthesumoftheordinatesofthefourpointsofintersectioniszero.

Solution


Lettheequationoftheparabolabe


Anypointontheparabolais(at2,2at).Whenthecircleandtheparabolaintersect,thispointliesonthecircle,

Thisbeingafourthdegreeequationint,therearefourvaluesoft.Foreachvalueoftthereisapointofintersection.Hence,therearefourpointsofintersectionofacircleandaparabola.Ift1,t2,t3,t4bethefourrootsofequation(6.24),then

Multiplyingequation(6.25)by2a,weget

2at1+2at2+2at3+2at4=0Therefore,thesumoftheordinatesofthefourpointsofintersectioniszero.

ILLUSTRATIVEEXAMPLESBASEDONTANGENTSANDNORMALS

Example6.3

Findtheequationsofthetangentandnormaltotheparabolay2=4(x–1)at(5,4).

Solution

y2=4(x–1)Differentiatingwithrespecttox,

∴Theequationofthetangentat(5,4)isy–4= (x–5).

2y–8=x–5orx–2y+3=0.Theslopeofthenormalat(5,4)is−2.∴Theequationofnormalat(5,4)isy–4=−2(x–5)or2x+y=14.

Example6.4

Findtheconditionthatthestraightlinelx+my+n=0isatangenttotheparabola.

Solution

Anystraightlinetangenttotheparabolay2=4axisoftheformy=mx+cif

Considerthelinelx+my+n=0(i.e.)my=−lx−n

Ifthisisatangenttotheparabola,y2=4axthen

Example6.5

Acommontangentisdrawntothecirclex2+y2=r2andtheparabolay2=4ax.Showthattheangleθwhichitmakeswiththeaxisoftheparabolaisgivenby

Solution

Lety=mx+cbeacommontangenttotheparabola

andthecircle

Ify=mx+cistangenttotheparabola(6.27)then

Ify=mx+cisatangenttothecircle(6.29)then

Equations(6.29)and(6.30)representthesamestraightline.Identifyingweget

Sincem2hastobepositive,

Example6.6

Astraightlinetouchesthecirclex2+y2=2a2andtheparabolay2=8ax.Showthatitsequationisy=±(x+2a).

Solution


Theequationoftheparabolais

Atangenttotheparabola(6.32)is

Atangenttothecircle(6.33)is

Equations(6.33)and(6.34)representthesamestraightline.Identifyingweget,

m2=1or−2;m2=−2isimpossible.

∴m2=1orm=±1∴Theequationofthecommontangentisy=±x±2a.

∴y=±(x+2a)

Example6.7

Showthatforallvaluesofm,theliney=m(x+a)+ willtouchtheparabolay2

=4a(x+a).Henceshowthatthelocusofapoint,thetwotangentsformwhichtotheparabolasy2=4a(x+a)andy2=4b(x+b)onetoeachareatrightangles,isthelinex+a+b=0.

Solution

Solving(6.35)and(6.36),wegettheirpointsofintersection.Thex-coordinatesoftheirpointsofintersectionaregivenby,

∴Thetwovaluesofxandhenceofyofthepointsofintersectionarethesame.

Hence, isatangenttotheparabolay2=4a(x+a).Let(x1,y1)be

thepointofintersectionofthetwotangentstotheparabolay2=4a(x+a),y2=

4a(y+b).Thetangentsare and Sincetheypass

through(x1,y1),wehave

Sincethetangentsareatrightangles,m1m2=−1.Subtracting(6.38)from(6.37),weget

since

Cancelling

Thelocusof(x1,y1)isx+a+b=0.

Example6.8

Provethatthelocusofthepointofintersectionoftwotangentstotheparabolay2

=4ax,whichmakesanangleofαwithx-axis,isy2–4ax=(x+a)2tan2α.Determinethelocusofpointofintersectionofperpendiculartangents.

Solution

Let(x1,y1)bethepointofintersectionoftangents.Anytangenttotheparabola

isy=mx+ .Ifthispassesthrough(x1,y1)theny=mx1+ .

∴Thelocusof(x1,y1)isy2–4ax=(x+a)2tan2α.Ifthetangentsareperpendicular,tanα=tan90°=∞∴Thelocusofperpendiculartangentsisdirectrix.

Example6.9

Provethatiftwotangentstoaparabolaintersectonthelatusrectumproducedthentheyareinclinedtotheaxisoftheparabolaatcomplementaryangles.

Solution

Let(x1,y1)theequationoftheparabolabey2=4ax.Lety=mx+ beany

tangenttotheparabola.Letthetwotangentsintersectat(a,y1),apointonthe

latusrectum.Then(a1,y1)lies,ony=mx+ .

Ifm1andm2aretheslopesofthetwotangentstotheparabolathenm1m2=1.(i.e.)tanθ·tan(90–θ)=1.(i.e.)Thetangentmakescomplementaryanglestotheaxisoftheparabola.

Example6.10

Provethatthelocusofpolesofthechordsoftheparabolay2=4axwhichsubtendsaconstantangleαatthevertexisthecurve(x+4a)2tan2α=4(y2–4ax).

Solution

Let(x1,y1)bethepoleofachordoftheparabola.Thenthepolarof(x1,y1)is

whichisthechordofcontactfrom(x1,y1).ThecombinedequationofthelinesAQandARisgotbyhomogenizationoftheequationoftheparabolay2=4axwiththehelpof(6.40).∴Thecombinedequationofthelinesis

Thelocusof(x1,y1)is(x+4a)2tan2α=4(y2–4ax).

Note6.10.1:Ifα=90°,thelocusof(x1,y1)isx+4a=0.

Example6.11

Iftwotangentsaredrawntoaparabolamakingcomplementaryangleswiththeaxisoftheparabola,provethatthechordofcontactpassesthroughthepointwheretheaxiscutsthedirectrix.

Solution

Lety=mx+ beatangentfromapoint(x1,y1)totheparabolay2=4ax.Then

y1=mx1+ orm2x1–my1+a=0.Ifthetangentsmakecomplementaryangles

withtheaxisoftheparabolathenm1m2=1or

Theequationofthechordofcontactfrom(a,y1)totheparabolaisyy1=2a(x+a).Whenthechordofcontactmeetsthex-axis,y=0.

∴x+a=0orx=−a.∴Thechordofcontactpassesthroughthepoint(−a,0)wheretheaxiscutsthedirectrix.

Example6.12

Findthelocusofpolesoftangentstotheparabolay2=4axwithrespecttotheparabolax2=4by.

Solution

Let(x1,y1)bethepolewithrespecttotheparabolax2=4by.Thenthepolarof

(x1,y1)isxx1=2b(y+y1), .Thisisatangenttotheparabolay2=4ax.

Theconditionfortangentisc= (i.e.) orx1y1+2ab=0.

Thelocusof(x1,y1)isthestraightlinexy+2ab=0.

Example6.13

Fromavariablepointonthetangentatthevertexofaparabola,theperpendicularisdrawntoitspolar.Showthattheperpendicularpassesthroughafixedpointontheaxisoftheparabola.

Solution

Theequationofthetangentatthevertexisx=0.Anypointonthislineis(0,y1).Thepolarof(0,y1)withrespecttotheparabolay2=4axisyy1=2ax.Theequationoftheperpendiculartothepolarof(x1,y1)isy1x+2ay=k.This

passesthrough(0,y1).

∴k=2ay1.∴Theequationoftheperpendiculartothepolarfrom(0,y1)isy1x+2ay=2ay1,whenthislinemeetsthex-axis,y1x=2ay1orx=2aHence,theperpendicularpassesthroughthepoint(2a,0),afixedpointonthe

axisoftheparabola.

Example6.14

Thepolarofanypointwithrespecttothecirclex2+y2=a2touchestheparabolay2=4ax.Showthatthepointliesontheparabolay2=–ax.

Solution

Thepolarofthepoint(x1,y1)withrespecttothecirclex2+y2=a2isxx1+yy1=a2

Thisisatangenttotheparabolay2=4ax.Theconditionforthatisc= .

Thelocusof(x1,y1)isy2=–axwhichisaparabola.

Example6.15

Findthelocusofpoleswithrespecttotheparabolay2=4axoftangentstothecirclex2+y2=c2.

Solution

Letthepolewithrespecttotheparabolay2=4axbe(x1,y1).Thenthepolarof(x1,y1)is

yy1=2a(x+x1)

(i.e.) .Thisisatangenttothecirclex2+y2=a2.Theconditionfor

thisis‘c2=a2(1+m2).

Thelocusof(x1,y1)is4a2x2=c2(y2+4a2).

Example6.16

ApointPmovessuchthatthelinethroughitperpendiculartoitspolarwithrespecttotheparabolay2=4axtouchestheparabolax2=4by.ShowthatthelocusofPis2ax+by+4a2=0.

Solution

LetPbethepoint(x1,y1).ThepolarofPwithrespecttoy2=4axis

Theequationoftheperpendicularto(6.41)isy1x+2ay+k=0.Thispassesthrough(x1,y1).

Hence,theequationoftheperpendicularisy1x+2ay–(x1y1+2ay1)=0.

(i.e.)

Thisisatangenttotheparabolax2=4by.

∴Thecondition

∴Thelocusof(x1,y1)is2ax+by+4a2=0.

Example6.17

IfthepolarofthepointPwithrespecttoaparabolapassesthroughQthenshowthatthepolarofQpassesthroughP.

Solution

Lettheequationoftheparabolabey2=4ax.LetPandQbethepoints(x1,y1)and(x2,y2).ThenthepolarofPisyy1=2a(x+x1).SincethispassesthroughQ(x2,y2),wegety1y2=2a(x1+x2).Thiscondition

showsthatthepoint(x1,y1)liesonthelineyy2=2a(x+x2).

∴ThepolarofQpassesthroughthepointP(x1,y1).

Example6.18

Pisavariablepointonthetangentatthevertexoftheparabolay2=4ax.ProvethatthelocusofthefootoftheperpendicularfromPonitspolarwithrespecttotheparabolaisthecirclex2+y2–2ax=0.

Solution

Pisthevariablepointonthetangentatthevertexoftheparabolay2=4ax.Theequationofthetangentatthevertexisx=0.AnypointonthetangentatthevertexisP(0,y1).Thepolarof(0,y)is

Theequationoftheperpendiculartothispolaris

Thispassesthrough(0,y1).

∴2ay1=k.∴TheequationoftheperpendicularfromPtoitspolaris

Let(l,m)bethepointofintersectionof(6.42)and(6.43).Then2al–my1=0.

y1l+2am–2ay1=0Solving

Now,

∴Thelocusof(l,m)isx2+y2–2ax=0.

Example6.19

Iffromthevertexoftheparabolay2=4ax,apairofchordscanbedrawnatrightanglestooneanotherandwiththesechordsasadjacentsidesarectanglebemade,provethatthelocusoffurtherangleoftherectangleistheparabolay2=4a(x–8a).

Solution

LetAPandAQbethechordsoftheparabolasuchthat .Completethe

rectangleAPRQ.ThenthemidpointsofARandPQarethesame.LettheequationsofAPbey=mx.Solvingy=mxandy2=4ax,wegetm2x2=4axor

∴ThepointPis .SinceAQisperpendiculartoAP,slopeofAQis .

Hence,thepointQis(4am2,−4am).

Let(x1,y1)bethepointR.ThemidpointofARis ThemidpointofPQis

SincethemidpointofARisthesameasthatofPQ,

Hence,thelocusof(x1,y1)isy2=4a(x–8a).

Example6.20

Showthatifr1andr2bethelengthsofperpendicularchordsofaparaboladrawn

throughthevertexthen

Solution

ThecoordinatesofPare(r1cosθ,r1sinθ).ThecoordinatesofQare(r2sinθ,r2cosθ).SincePliesontheparabolay2=4ax,

Similarly,

Also

Fromequations(6.45)and(6.46),

Example6.21

Showthatthelatusrectumofaparabolabisectstheanglebetweenthetangents

Showthatthelatusrectumofaparabolabisectstheanglebetweenthetangentsandnormalateitherextremity.

Solution

LetLSL′bethelatusrectumoftheparabolay2=4ax.ThecoordinatesofLare(a,2a).TheequationoftangentatLisy·2a=2a(x+a)

Theslopeofthetangentis1.∴TheslopeofthenormalatLis−1.LSisperpendiculartox-axis.

∴LatusrectumbisectstheanglebetweenthetangentsandnormalatL.

Example6.22

Showthatthelocusofthepointsofintersectionoftangentstoy2=4axwhichinterceptaconstantlengthdonthedirectrixis(y2–4ax)(x+a)2=d2x2.

Solution

LetP(x1,y1)bethepointofintersectiontangenttotheparabola.ThentheequationofthepairoftangentsPQandPRisT2=SS1.(i.e.)[yy1–2a(x+x1]2=(y2–4ax)(y2–4ax1).Whentheselinesmeetthe

directrixx=−a,wehave

Ify1andy2aretheordinatesofthepointofintersectionoftangentswiththedirectrixx+a=0,then

Then

∴Thelocusof(x1,y1)is(y2–4ax)(x+a)2=d2x2.

Example6.23

Showthatthelocusofmidpointsofchordsofaparabolawhichsubtendarightangleatthevertexisanotherparabolawhoselatusrectumishalfthelatusrectumoftheparabola.

Solution


Let(x1,y1)bethemidpointofthechordPQ.ThentheequationofPQisT=S1

ThecombinedequationofthelinesAPandAQisgotbyhomogenizationofequation(6.48)withthehelpof(6.49).

∴ThecombinedequationofOPandOQis

Thelocusof(x1,y1)isy2=2a(x–4a)whichisaparabolawhoselatusrectumishalfthelatusrectumofthegivenparabola.

Example6.24

Showthatthelocusofmidpointsofchordsoftheparabolaofconstantlength2lis(y2–4ax)(y2+4a2)+4a2l2=0.

Solution

Let(x1,y1)bethemidpointofachordoftheparabola


Anypointonthislineis(x1+rcosθ,y1+rsinθ).Thispointliesontheparabolay2=4ax.

ThetwovaluesofrarethedistancesRPandRQwhichareequalinmagnitudebutoppositeinsign.Theconditionforthisisthecoefficientofr=0.(i.e.)y1sinθ–2acosθ=0.

ThenfromEquation(6.50),

Thelocusof(x1,y1)is(y2–4ax)(y2+4a2)+4a2l2=0.(sincer=l)

Example6.25

Showthatthelocusofthemidpointsoffocalchordsofaparabolaisanotherparabolawhosevertexisatthefocusofthegivenparabola.

Solution

Letthegivenparabolabe

Let(x1,y1)bethemidpointofachordofthisparabola.Thenitsequationis

Ifthisisafocalchordthenthispassesthrough(a,0).

Thelocusof(x1,y1)isy2=2a(x–a)whichisaparabolawhosevertexisatthefocusofthegivenparabola.

Example6.26

Fromapointcommontangentsaredrawntothecircle andthe

parabolay2=4ax.Findtheareaofthequadrilateralformedbythecommontangents,thechordofcontactofthecircleandthechordofcontactoftheparabola.

Solution

Anytangenttotheparabolay2=4axis

Ifthisisalsoatangenttothecircle

∴m2(1+m2)=2orm4+m2–2=0or(m2–1)(m2+2)=0⇒m2=1or−2.Butm2=−2isinadmissiblesincem2hastobepositive.∴m

=±1.Hencethecommontangentsarey=±(x+a).ThetwotangentsmeetatP(−a,

0).

Theequationofthechordofcontactfrom(−a,0)tothecircle is

Theequationofthechordofcontactfrom(−a,0)totheparabolay2=4axis0=2a(x–a)orx–a=0.Whenx=a,y=±2a.HenceNandQare(a,2a)and(a,−2a).

When

∴AreaofquadrilateralLMQN=AreaoftrapeziumLMQN

Example6.27

ThepolarofapointPwithrespecttotheparabolay2=4axmeetsthecurveinQandR.ShowthatifPliesonthelinelx+my+n=0thenthelocusofthemiddlepointoftheQRisl(y2–4ax)+2a(lx+my+n)=0.

Solution

LetPbethepoint(h,k).ThepolarofP(h,k)withrespecttotheparabolay2=4axis

ThepolarofPmeetstheparabolay2=4axatQandR.LetP(x1,y1)bethemidpointofQR.Itsequationis

Equations(6.55)and(6.56)representthesameline.∴Identifyingequations(6.55)and(6.56)weget,

Sincethepoint(h,k)liesonlx+my+n=0,lh+mk+n=0.

Usingequation(6.56),

Example6.28

Provethatareaofthetriangleinscribedintheparabolay2=4axis

wherey1,y2andy3aretheordinatesoftheverticesofthe

triangle.

Solution

Let(x1,y1),(x2,y2)and(x3,y3)betheverticesofthetriangleinscribedinthe

parabolay2=4ax.Thentheverticesare .Theareaof

thetriangleis

Example6.29

Anequilateraltriangleisinscribedintheparabolay2=4axoneofwhoseverticesisatthevertexoftheparabola.Finditsside.

Solution

ThecoordinatesofBareB(rcos30°,rsin30°),

Sincethispointliesontheparabolay2=4ax,then

Exercises

1. Showthattwotangentscanbedrawnfromagivenpointtoaparabola.Ifthetangentsmakeanglesθ1andθ2withxaxissuchthat

i. tanθ1+tanθ2isaconstantshowthatthelocusofpointofintersectionoftangentsisastraightlinethroughthevertexofaparabola.

ii. iftanθ1·tanθ2isaconstantshowthatthelocusofthepointofintersectingisastraightline.

iii. ifθ1+θ2isaconstantshowthatthelocusofthepointofintersectionoftangentsisastraightlinethroughthefocus.

iv. ifθ1andθ2arecomplementaryanglesthenthelocusofpointofintersectionisthestraightlinex=a.

2. Findthelocusofpointofintersectionoftangentstotheparabolay2=4axwhichincludesanangle

of .

Ans.:3(x+a)2=y2–4ax

3. Showthatthelocusofthepolesofchordsoftheparabolay2=4axwhichsubtendsanangleof45°

atthevertexisthecurve(x+a)2=4(y2–4ax).

4. Showthatthelocusofpolesofalltangentstotheparabolay2=4axwithrespecttotheparabolay2

=4bxistheparabolaay2=4b2x.5. Showthatthelocusofpolesofchordsoftheparabolawhichsubtendsarightangleatthevertexisx+4a=0.

6. Showthatiftangentsbedrawntotheparabolay2=4axfromanypointonthestraightlinex+4a=0,thechordofcontactsubtendsarightangleatthevertexoftheparabola.

7. Perpendicularsaredrawnfrompointsonthetangentatthevertexontheirpolarswithrespectto

theparabolay2=4ax.Showthatthelocusofthefootoftheperpendicularisacirclecentreat(a,0)andradiusa.

8. Showthatthelocusofpoleswithrespecttotheparabolay2=4axoftangentstothecirclex2+y2

=4a2isx2–y2=4a2.9. ApointPmovessuchthatthelinethroughtheperpendiculartoitspolarwithrespecttothe

parabolay2=4axtouchestheparabolax2=4by.ShowthatthelocusofPis2ax+by+4a2x=0.

10. Ifachordoftheparabolay2=4axsubtendsarightangleatitsfocus,showthatthelocusofthe

poleofthischordwithrespecttothegivenparabolaisx2+y2+6ax+a2=0.

11. Showthatthelocusofpolesofallchordsoftheparabolay2=4axwhichareataconstantdistance

dfromthevertexisd2y2+4a2(d2–x2)=0.

12. Showthatthelocusofpolesofthefocalchordsoftheparabolay2=4axisx+a=0.

13. Iftwotangentstotheparabolay2=4axmakeequalangleswithafixedlineshowthatthechordofcontactpassesthroughafixedpoint.

14. Provethatthepolarofanypointonthecirclex2+y2–2ax–3a2=0withrespecttothecirclex2

+y2+2ax–3a2=0touchestheparabolay2=4ax.

15. Showthatthelocusofthepoleswithrespecttotheparabolay2=4axofthetangentstothecurve

x2–y2=aistheellipse4x2+y2=4ax.

16. Pisavariablepointontheliney=b,provethatthepolarofPwithrespecttotheparabolay2=4axisafixeddirectrix.

17. TheperpendicularfromapointOonitspolarwithrespecttoaparabolameetthepolarinthepointsMandcutstheaxisinG.Thepolarmeetsx-axisinTandtheordinatethroughOintersectsthecurveinPandP′.ShowthatthepointsG,M,P,P′andTlieonacirclewhosecentreisatthefocusS.

18. Tangentsaredrawntotheparabolay2=4axfromapoint(h,k).Showthattheareaofthetriangle

formedbythetangentsandthechordofcontactis

19. Provethatthelengthofthechordofcontactofthetangentsdrawnfromthepoint(x1,y1)tothe

parabolay2=4axis Henceshowthatoneofthetrianglesformedbythese

tangentsandtheirchordofcontactis

20. TangentsaredrawnfromavariablepointPtotheparabolay2=4axsuchthattheyformatriangle

ofconstantareawiththetangentatthevertex.ShowthatthelocusofPis(y2–4ax)x2=4c2.21. Provethatthetangenttoaparabolaandtheperpendiculartoitfromitsfocusmeetonthetangent

atthevertex.22. Showthataportionofatangenttoaparabolainterceptedbetweendirectrixandthecurvesubtends

arightangleatthefocus.

23. Thetangenttotheparabolay2=4axmakeanglesθ1andθ2withtheaxis.Showthatthelocusofthepointofintersectionsuchthatcotθ1+cotθ2=cisy=ac.

24. Ifperpendicularsbedrawnfromanytwofixedpointsontheaxisofaparabolaequidistantfromthefocusonanytangenttoit,showthatthedifferenceoftheirsquaresisaconstant.

25. Provethattheequationoftheparabolawhosevertexandfocusonx-axisatdistances4aand5a

fromtheoriginrespectively(a>0)isy2=4a(x–4a).Alsoobtaintheequationtothetangenttothiscurveattheendoflatusrectuminthefirstquadrant.

Ans.:y=x–a

26. Chordsofaparabolaaredrawnthroughafixedpoint.Showthatthelocusofthemiddlepointsisanotherparabola.

27. Findthelocusofthemiddlepointsofchordsoftheparabolay2=2xwhichtouchesthecirclex2+

y2–2x–4=0.

28. Atangenttotheparabolay2+4bx=0meetstheparabolay2=4axatPandQ.Showthatthelocus

ofthemiddlepointofPQisy2(2a+b)=4a2x.

29. Througheachpointofthestraightlinex–my=hisdrawnachordoftheparabolay2=4axwhich

isbisectedatthepoint.Provethatitalwaystouchestheparabola(y+2am)2=8axh.

30. Twolinesaredrawnatrightangles,onebeingatangenttotheparabolay2=4axandtheotherto

y2=4by.Showthatthelocusoftheirpointofintersectionisthecurve(ax+by)(x2+y2)=(bx–

ay)2.

31. Acirclecutstheparabolay2=4axatrightanglesandpassesthroughthefocus.Showthatthe

centreofthecircleliesonthecurvey2(a+x)=a(a+bx)2.32. Twotangentsdrawnfromapointtotheparabolamakeanglesθ1andθ2withthex-axis.Showthat

thelocusoftheirpointofintersectioniftan2θ1+tan2θ2=cisy

2–cx2=2ax.33. IfatrianglePQRisinscribedinaparabolasothatthefocusSistheorthocentreandthesidesmeet

theaxesinpointsK,LandMthenprovethatSK·SL·SM–4SA2=0whereAisthevertexoftheparabola.

34. Chordsoftheparabolay2=4axaredrawnthroughafixedpoint(h,k).Showthatthelocusofthe

midpointisaparabolawhosevertexis andlatusrectumis2a.

35. Showthatthelocusofthemiddlepointsofasystemofparallelchordsofaparabolaisalinewhichisparalleltotheaxisoftheparabola.

36. Showthatthelocusofthemidpointsofchordsoftheparabolawhichsubtendsaconstantangleα

atthevertexis(y2–2ax–8a2)2tan2α=16α2(4ax–y2).

ILLUSTRATIVEEXAMPLESBASEDONPARAMETERS

Example6.30

Provethatperpendiculartangentstotheparabolawillintersectonthedirectrix.

Solution

Letthetangentsatt1andt2intersectatP.Theequationoftangentsatt1andt2

are

Theslopesofthetangentsare .Sincethetangentsareperpendicular,

∴t1t2=−1

Thepointofintersectionofthetangentsatt1andt2isP(at1t2,a(t1+t2))(i.e.)

(−a,a(t1+t2)).Thispointliesonthelinex+a=0.∴Perpendiculartangentsintersectonthedirectrix.

Example6.31

Provethatthetangentsattheextremitiesofafocalchordintersectatrightanglesonthedirectrix.

Solution

Lett1andt2betheextremitiesofafocalchord.Thentheequationofthechordisy(t1+t2)=2x+2at1t2.Thispassesthroughthefocus(a,0).

∴Tangentsatt1andt2areperpendicular.Thepointofintersectionoftangentsatt1andt2is[at1t2,a(t1+t2)](i.e.)(−a,

a(t1+t2)).Thispointliesonthedirectrix.Hencethetangentsattheextremitiesofafocalchordintersectatrightanglesonthedirectrix.

Example6.32

Provethatanytangenttoaparabolaandperpendicularonitfromthefocusmeetonthetangentatthevertex.

Solution

Lettheequationoftheparabolay2=4ax.Theequationofthetangentattis

Theslopeofthetangentis .Theslopeoftheperpendiculartoitis–t.Hence

theequationoftheperpendicularlinepassingthroughfocus(a,0)is

Multiplyingequation(6.60)byt,weget

Equation(6.59)–equation(6.61)givesx(1+t2)=0orx=0.

∴y=at

Hence,thepointofintersectionof(6.59)and(6.60)is(0,at)andthispointliesony-axis.

Example6.33

Showthattheorthocentreofthetriangleformedbythetangentsatthreepointsonaparabolaliesonthedirectrix.

Solution

Lett1,t2andt3bepointsofcontactofthetangentsatthepointsA,BandC,respectivelyontheparabolay2=4ax,formingatrianglePQR.TheequationofQRis

Pisthepointofintersectionoftangentsatt1andt2.Thispointis[at1t2,a(t1+t2)].TheslopeofPL,perpendiculartoQRis−t3.∴TheequationofPLisy–a(t1+t2)=–t3[x–at1t2]

(i.e.)

ThentheequationofQMperpendicularfromQonPRis

Equation(6.63)–equation(3)givesx(t3–t2)=a(t2–t3)orx=−a.Thispointliesonthedirectrixx+a=0.Hencetheorthocentreliesonthe

directrix.

Example6.34

Thecoordinatesoftheendsofafocalchordoftheparabolay2=4axare(x1,y1)and(x2,y2).Provethatx1x2=a2andy1y2=−4a2.

Solution

Lett1andt2betheendsofafocalchord.Thentheequationofthefocalchordisy(t1+t2)=2x+at1t2.Sincethispassesthroughthefocus(a,0),0=2a+at1t2ort1t2=−1.

Example6.35

Aquadrilateralisinscribedinaparabolaandthreeofitssidespassthroughfixedpointsontheaxis.Showthatthefourthsidealsopassesthroughafixedpointontheaxisoftheparabola.

Solution

Lett1,t2,t3andt4berespectivelyverticesA,B,CandDofthequadrilateralinscribedintheparabolay2=4ax.TheequationofchordABis

Whenthismeetsthex-axisy=0(i.e.)x=−at1t2=k1.SinceABmeetsthex-axisatafixedpoint,

Similarly,

Multiplyingthese,weget

Hence,thefourthsideofthequadrilateralalsopassesthroughafixedpoint.

Example6.36

Tangentstotheparabolay2=4axaredrawnatpointswhoseabscissaeareintheratiok:1.Provethatthelocusoftheirpointofintersectionisthecurvey2=(k1/4

+k–1/4)2x2.

Solution

Letthetangentsatt1andt2intersectatP(x1,y1)

Giventhat

Thepointofintersectionofthetangentsatt1andt2isx1=at1t2andy1=a(t1+t2).

∴Thelocusof(x1,y1)isy2=(k1/4+k−1/4)2x.

Example6.37

Showthatthelocusofthemiddlepointofalltangentsfrompointsonthedirectrixtotheparabolay2=4axisy2(2x+a)=a(x+3a)2.

Solution

Let(−a,y1)beapointonthedirectrix.Lettbethepointofcontactoftangentsfrom(−a,y1)totheparabolay2=4ax.Theequationofthetangentattis

Sincethispassesthrough(−a,y1),

∴Thepointonthedirectrixis

Let(x1,y1)bethemidpointoftheportionoftangentbetweenthedirectrixandthepointofcontact.Then

and2y1t=2at2–a+at2

Thelocusof(x1,y1)isy2(2x+a)=a(3x+a)2.

Example6.38

TangentsaredrawnfromavariablepointPtotheparabolay2=4ax,suchthattheyformatriangleofconstantareac2withthetangentatthevertex.ShowthatthelocusofPis(y2–4ax)x2=4c4.

Solution

LetP(x1,y1)bethepointofintersectionoftangentsatt1andt2.Theequationof

thetangentatt1is ThismeetsthetangentatthevertexatQ.∴Qis

(0,at1).Similarly,Ris(0,at2).Pisthepointofintersectionoftangentsatt1andt2andthepointisP(at1t2,a(t1+t2)).TheareaofΔPQRisgivenasc2.

Therefore,thelocusof(x1,y1)isx2(y2–4ax)=4c4.

Example6.39

Provethatthedistanceofthefocusfromtheintersectionoftwotangentstoaparabolaisameanproportionaltothefocalradiiofthepointofconstant.

Solution

Letthetangentsat intersectatP.Thenthe

coordinatesofthepointPare(at1t2,a(t1+t2)).Sisthepoint(a,0).

Example6.40

Provethatthelocusofthepointofintersectionofnormalsattheendsofafocalchordofaparabolaisanotherparabolawhoselatusrectumisonefourthofthatofthegivenparabola.

Solution


Lett1andt2betheendsofafocalchordoftheparabola.Forafocalchordt1t2=−1.

Theequationofthenormalatt1andt2are

If(x1,y1)isapointofintersectionofthenormalsatt1andt2then

Thelocusof(x1,y1)isy2=a(x–3a)whichisaparabolawhoselatusrectumisonefourthofthelatusrectumoftheoriginalparabola.

Example6.41

Ifthenormalatthepointt1ontheparabolay2=4axmeetsthecurveagainatt2

provethat

Solution

Theequationofthenormalatt1is

Theequationofthechordjoiningthepointst1andt2is

Equations(6.73)and(6.74)representthesamelines.Thereforeidentifyingwe

get

Example6.42

Ifthenormalsattwopointst1,t2ontheparabolay2=4axintersectagainatapointonthecurveshowthatt1+t2+t3=0andt1t2=2andtheproductofordinatesofthetwopointsis8a2.

Solution

Thenormalst1andt2meetatt3.

Subtracting Sincet1–t2≠0,t1t2=2.

Solvingequations(6.75)and(6.76),weget

Example6.43

Findtheconditionthatthelinelx+my+n=0isanormaltotheparabolaisy2=4ax.

Solution

Letthelinelx+my+n=0beanormalat‘t’.Theparabolaisy2=4ax.Theequationofthenormalattis

Buttheequationofthenormalisgivenas

Identifyingequations(6.77)and(6.78),weget and

(i.e.)al3+2alm2+m2n=0.

Example6.44

Showthatthelocusofpolesofnormalchordsoftheparabolaisy2=4axis(x+2a)y2+4a3=0.

Solution

Let(x1,y1)bethepoleofanormalchordnormalatt.Theequationofthepolarof(x1,y1)is

Theequationofthenormalattis

Equations(6.79)and(6.80representthesameline.∴Identifyingequations(6.79)and(6.80),weget

Example6.45

Intheparabolay2=4axthetangentatthepointPwhoseabscissaisequaltothelatusrectummeetstheaxisonTandthenormalatPcutsthecurveagaininQ.ProvethatPT:TQ=4:5.

Solution

LetPandQbethepointst1andt2respectively.Giventhat

Theequationofthetangentatt1is

whenthismeetsthex-axis,y=0.

HenceTisthepoint(−4a,0).Alsoasthenormalatt1meetsthecurveatt2,

Example6.46

Showthatthelocusofapointsuchthattwoofthethreenormaldrawnfromittotheparabolay2=4axcoincideis27ay2=4(x–2a)3.

Solution

Let(x1,y1)beagivenpointandtbefootofthenormalfrom(x1,y1)totheparabolay2=4ax.Theequationofthenormalat‘t’is

Sincethispassesthrough(x1,y1)wehave

Ift1,t2andt3befeetofthenormalsfrom(x1,y1)totheparabolathent1,t2andt3aretherootsofequation(6.84).

Iftwoofthethreenormalscoincidethent1=t2.

Fromequations(6.88)and(6.89), Sincet1isarootof

equation(6.84)

Example6.47

Ifthenormalsfromapointtotheparabolay2=4axcuttheaxisinpointswhosedistancesfromthevertexareinAPthenshowthatthepointliesonthecurve

27ay2=2(x–2a)3.

Solution

Let(x1,y1)beagivenpointandtbethefootofanormalfrom(x1,y1).Theequationofthenormalattis

Sincethispassesthrough(x1,y1),y1+x1t=2at+at3.

Ift1,t2andt3bethefeetofthenormalsfrom(x1,y1)then

Whenthenormalattmeetsthex-axis,y=0,from(6.91)wegetxt=2at+at3

orx=2a+at2.Thenthex-coordinatesofthepointswherethenormalmeetsthex-axisare

givenby GiventheseareinAP.

areinAP.

Fromequation(6.95),or

Sincet2isarootofequation(6.91),

∴Thelocusof(x1,y1)is27ay3=2(x–2a)3.

Example6.48

Showthatthelocusofthepointofintersectionoftwonormalstotheparabolawhichareatrightanglesisy2=a(x–3a).

Solution

If(x1,y1)isthepointofintersectionoftwonormalstotheparabolay2=4axthen

Ift1,t2andt3bethefeetofthethreenormalsfrom(x1,y1)then

Sincetwoofthenormalsareperpendicularthent1t2=−1

Sincet3isarootofequation(6.97),

Example6.49

Provethatanormalchordofaparabolawhichsubtendsarightangleatthe

vertexmakesanangle withthex-axis.

Solution



ThecombinedequationofthelinesAPandAQis

Sincethetwolinesareatrightangles,coefficientofx2+coefficientofy2=0.

∴t=0ort2=2.t=0correspondstothenormaljoiningthroughthevertex.

∴Thenormalsmakeanangle withthex-axis.

Example6.50

Provethattheareaofthetriangleformedbythenormalstotheparabolay2=4ax

atthepointst1,t2andt3is

Solution

Theequationsofthenormalsatt1,t2,t3are

Solvingtheseequationspairwisewegettheverticesofthetriangle.Hencethe

verticesare andtwoothersimilarpoints.

Example6.51

ProvethatthelengthoftheinterceptsonthenormalatthepointP(at2,2at)totheparabolay2=4axmadebythecircledescribedonthelinejoiningthefocusand

Pasdiameteris

Solution

TheequationofthenormalatPisy+xt=2at+at3.

LetthecircleonPSasdiametercutthenormalatPatRandthex-axisatT.

Example6.52

NormalsatthreepointsP,QandRoftheparabolay2=4axmeetin(h,k).Prove

thatthecentroidofΔPQRliesontheaxisatadistance fromthevertex.

Solution

Lettbeafootofanormalfrom(h,k).Theequationofthenormalattis

Thispassesthrough(h,k).

⇒at3+t(2a–h)–k=0.Ift1,t2andt3arethefeetofthenormalsfrom(h,k)thent1+t2+t3=0,

ThecentroidoftheΔPQRis

Sincethecentroidliesonthex-axis,

Thex-coordinatesofthecentroidis

∴Centroidisatadistance fromthevertexoftheparabola.

Example6.53

ThenormalsatthreepointsP,QandRonaparabolameetatTandSbethefocusoftheparabola.ProvethatSP·SQ·SR=aTS2.

Solution

LetTbethepoint(h,k).ThenP,QandRarethefeetofthenormalsfromT(h,k).Theequationofthenormalattisy+xt=2at+at3.Ift1,t2andt3bethefeetofthenormalsfromTthent1+t2+t3=0.

Sisthepoint(a,0).

Example6.54

TheequationofachordPQoftheparabolay2=4axislx+my=1.Showthat

thenormalsatP,Qmeetonthenormalat

Solution

LetPandQbethepointst1andt2.ThenormalsatPandQmeetatR.Ift3isthefootofthenormalofthe3rdpointthen

TheequationofthechordPQislx+my=1.SincePandQarethepointst1andt2,

Example6.55

IfthenormalatPtotheparabolay2=4axmeetsthecurveatQandmakeanangleθwiththeaxisshowthat

i. itwillcuttheparabolaatθatanangle and

ii. PQ=4asecθcosec2θ.

Solution

LetPbethepoint(at2,2at).Theequationofthenormalattisy+xt=2at+at3.

Thenormalattmeetsthecurveat

LetɸbetheanglebetweenthenormalandthetangentatQ.Theslopeofthe

tangentatQis

Slopeofthenormalattis–t.

sincetanθ=−t

Example6.56

Provethatthecirclepassingthroughthefeetofthethreenormalstoaparaboladrawnfromanypointintheplanepassesthroughthevertexoftheparabola.Alsofindtheequationofthecirclepassingthroughthefeetofthenormals.

Solution



LetP,QandRbethefeetofthenormalstoy2=4axfromagivenpoint(h,k).Thenwehaveat3+(2a–h)t–k=0.

Ift1,t2andt3bethefeetofthenormalsatP,QandRthent1+t2+t3=0.Weknowthatthecircle(6.111)andtheparabola(6.110)cutatfourpointsandift1,t2,t3andt4arethefourpointsofintersectionofthecircleandtheparabolathentheyaretherootsoftheequation,

Ift1,t2andt3correspondtothefeetofthenormalsfrom(h,k)then

Fromequations(6.113)and(6.114),t4=1.

Butt4isthepoint whichisthevertexoftheparabola.Hence

thecirclepassingthroughthefeetofthenormalsfromagivenpointalsopassesthroughthevertexoftheparabola.Henceequation(6.112)becomes

sincec=0Equations(6.111)and(6.115)arethesame.Bycomparingthecoefficients,

weget

∴Theequationofthecirclepassingthroughthefeetofthenormalis

Exercises

1. Showthattheportionofthetangentinterceptedbetweenthepointofcontactandthedirectrix

subtendsarightangleatthefocus.2. IfthetangentatapointPontheparabolameetstheaxisatTandPNistheordinateatPthenshow

thatAN=AT.3. IfthetangentatPmeetsthetangentatthevertexinYthenshowthatSYisperpendiculartoTPand

SY2=ASSP.4. IfA,BandC,arethreepointsonaparabolawhoseordinatesareinGPthenprovethatthetangents

atAandCmeetontheordinatesofB.5. Provethatthemiddlepointoftheinterceptsmadeonatangenttoaparabolabythetangentsattwo

pointsPandQliesonthetangentwhichisparalleltoPQ.

6. Ifpoints(at2,2at)isoneextremityofafocalchordoftheparabolay2=4ax,showthatthelength

ofthefocalchordis

7. Showthatthetangentsatoneextremityofafocalchordofaparabolaisparalleltothenormalattheotherextremity.

8. Ifthetangentsatthreepointsontheparabolay2=4axmakeangles60°,45°and30°withtheaxisoftheparabola,showthattheabscissaeandordinatesofthethreepointsareinGP.

9. Showthatthecircledescribedonthefocalchordofaparabolaasdiametertouchesthedirectrix.10. Showthatthetangentatoneextremityofafocalchordofaparabolaisparalleltothenormalatthe

otherextremity.11. Provethatthesemilatusrectumofaparabolaistheharmonicmeanofthesegmentsofafocal

chord.12. Provethatthecircledescribedonfocalradiiasdiametertouchesthetangentsatthevertexofa

parabola.

13. Threenormalstoaparabolay2=4xaredrawnthroughthepoint(15,12).Showthattheequationsare3x–y–33=0,4x+y–72=0andx–y–3=0.

14. ThenormalsattwopointsPandQofaparabolay2=4axmeetatthepoint(x1,y1)ontheparabola.ShowthatPQ=(x1+4a)(x1–8a).

15. Showthatthecoordinatesofthefeetofthenormalsoftheparabolay2=4axdrawnfromthepoint(6a,0)are(0,0),(4a,4a)and(4a,–4a).

16. ThenormalatPtotheparabolay2=4axmakesanangleawiththeaxis.Showthattheareaofthetriangle,formedbyitisthetangentsatitsextremitiesisaconstant.

17. IfP,QandRarethepointst1,t2andt3ontheparabolay2=4ax,suchthatthenormalatQandR

meetatPthenshowthat:i. thelinePQispassesthroughafixedpointontheaxis.ii. thelocusofthepoleofPQisx=a.

iii. thelocusofthemidpointofPQisy2=2a(x+2a).

iv. theordinatesofPandQaretherootsoftheequationy2+xy+8a2=0wheret3istheordinateofthepointofintersectionofthenormalsatPandQ.

18. IfacirclecutsaparabolaatP,Q,RandSshowthatPQandRSareequallyinclinedtotheaxis.

19. ThenormalsatthepointsPandRontheparabolay2=4axmeetontheparabolaatthepointP.

ShowthatthelocusoftheorthocentreofΔPQRisy2=a(x+6a)andthelocusofthecircumcentre

ofΔPQRistheparabola2y=x(x–a).20. Provethattheareaofthetriangleinscribedinaparabolaistwicetheareaofthetriangleformedby

thetangentsatthevertices.21. ProvethatanythreetangentstoaparabolawhoseslopesareinHPenclosesatriangleofconstant

area.22. Provethatthecircumcircleofatrianglecircumscribingaparabolapassesthroughthefocus.

23. IfthenormalsatanypointPoftheparabolay2=4axmeettheaxisatGandthetangentatvertexatHandifAbethevertexoftheparabolaandtherectangleAGQHbecompleted,provethatthe

equationtothelocusofQisx2=2ax+ay2.24. ThenormalatapointPofaparabolameetsthecurveagainatQandTisthepoleofPQ.Show

thatTliesonthedirectrixpassingthroughPandthatPTisbisectedbythedirectrix.

25. Iffromthevertexoftheparabolay2=4ax,apairofchordsbedrawnatrightanglestooneanotherandwiththesechordsasadjacentsidesarectanglebemadethenshowthatthelocusoffurther

angleoftherectangleistheparabolay2=4a(x–8a).

26. Thenormaltotheparabolay2=4axatapointPonitmeetstheaxisinG.ShowthatPandGareequidistantfromthefocusoftheparabola.

27. Twoperpendicularstraightlinesthroughthefocusoftheparabolay2=4axmeetitsdirectrixinTandT′respectively.ShowthatthetangentstotheparabolatotheperpendicularlinesintersectatthemidpointofTT′.

28. IfthenormalsatanypointP(18,12)totheparabolay2=8xcutsthecurveagainatQshowthat9·

PQ=80

29. IfthenormalatPtotheparabolay2=4axmeetsthecurveagainatQandifPQandthenormalat

Qmakeanglesθandɸ,respectivelywiththeaxis,provethattanθtan2ɸ+tan2θ+2=0.30. PQisafocalchordofaparabola.PP′andQQ′arethenormalsatPandQcuttingthecurveagain

atP′andQ′.ShowthatP′Q′isparalleltoPQandisthreetimesPQ.

31. IfPQbeanormalchordoftheparabola.y2=4axandifSbethefocus,showthatthelocusofthe

centroidofthetriangleSPQisy2(ay2+180a2–108ax)+128a4=0.32. IfthetangentsatPandQmeetatTandtheorthocenteroftheΔPTQliesontheparabola,show

thateithertheorthocentreisatthevertexorthechordPQisnormaltotheparabola.

33. Ifthreenormalsfromapointtotheparabolay2=4axcutstheaxisinpoints,whosedistancesfrom

thevertexareinAP,showthatthepointonthecurve27ay2=2(x–a)3.34. Tangentsaredrawntoaparabolafromanypointonthedirectrix.Showthatthenormalsatthe

pointsofcontactareperpendiculartoeachotherandthattheyintersectonanotherparabola.

35. Showthatiftwotangentstoaparabolay2=4axinterceptaconstantlengthonanyfixedtangent,thelocusoftheirpointofintersectionisanotherequalparabola.

36. Showthattheequationofthecircledescribedonthechordinterceptedbytheparabolay2=4axon

theliney=mx+casdiameterism2(x2+y2)+2(mc–2a)x–4ay+c(4am+c)=0.37. Circlesaredescribedonanytwocommonchordsofaparabolaasdiameter.Provethattheir

commonchordpassesthroughthevertexoftheparabola.

38. IfP(h,k)isafixedpointintheplaneofaparabolay2=4ax.ThroughPavariablesecantisdrawn

tocuttheparabolainQandR.TisapointonQRsuchthat

i. PQ·PR=PT2.ShowthatthelocusofTis(y–k)2=k2–4ah.

ii. PQ+PR=PT.ShowthatthelocusofTisy2–k2=4a(x–h).

39. Showthatthelocusofthepointofintersectionoftangents,totheparabolay2=4axatpoints

whoseordinatesareintheratio

40. Showthatthelocusofthemiddlepointsofasystemofparallelchordsofaparabolaisalinewhichisparalleltotheaxisoftheparabola.

41. P,QandRarethreepointsonaparabolaandthechordPQmeetsthediameterthroughRinT.

OrdinatesPMandQNaredrawntothisdiameter.ShowthatRMRN=RT2.

Chapter7

Ellipse

7.1STANDARDEQUATION

Aconicisdefinedasthelocusofapointsuchthatitsdistancefromafixedpointbearsaconstantratiotoitsdistancefromafixedline.Thefixedpointiscalledthefocusandthefixedstraightlineiscalledthedirectrix.Theconstantratioiscalledtheeccentricityoftheconic.Iftheeccentricityislessthanunitytheconiciscalledanellipse.Letusnowderivethestandardequationofanellipseusingtheabovepropertycalledfocus-directrixproperty.

7.2STANDARDEQUATIONOFANELLIPSE

LetSbethefocusandlinelbethedirectrix.DrawSXperpendiculartothedirectrix.DivideSXinternallyandexternallyintheratioe:1(e<1).LetAandA′

bethepointsofdivision.Since and ,fromthedefinitionofellipse,

thepointsAandA′lieontheellipse.LetAA′=2aandCbeitsmiddlepoint.

Addingequations(7.1)and(7.2),wegetSA+SA′=e(AX+A′X).

Subtractingequations(7.1)from(7.2),wegetSA′−SA=e(CX′−CX)

TakeCSasthex-axisandCMperpendiculartoCS,asy-axis.LetP(x,y)beanypointontheellipse.DrawPMperpendiculartothe

directrix.ThenthecoordinatesofSare(ae,0).Fromthefocus-directrixproperty

oftheellipse,

Thisiscalledthestandardequationofanellipse.

Note7.2.1:

1. Equation(7.5)canbewrittenas:

2. AA′iscalledthemajoraxisoftheellipse.3. BB′iscalledtheminoraxisoftheellipse.4. Ciscalledthecentreoftheellipse.5. Thecurvemeetsthex-axisatthepointA(a,0)andA′(−a,0).6. Thecurvemeetsthey-axisatthepointsB(0,b)andB′(0,−b).7. Thecurveissymmetricalaboutboththeaxes.If(x,y)isapointonthecurve,then(x,−y)and(−x,y)arealsothepointsonthecurve.

8. Fromtheequationoftheellipse,weget

Therefore,foranypoint(x,y)onthecurve,−a≤x≤aand−b≤y≤b.9. Thedoubleordinatethroughthefocusiscalledthelatusrectumoftheellipse.

(i.e.)LSL′isthelatusrectum.

10. Secondfocusandseconddirectrix:Onthenegativesideoftheorigin,takeapointS′suchthatCS=CS′andanotherpointX′suchthatCX=CX′=a.

DrawX′M′perpendiculartoAA′andPM′perpendiculartoX′M′.Thenwecanshowthat

givesthelocusofPas HereS′iscalledthesecondfocusandX′M′isthesecond

directrix.11.

i. ShiftingtheorigintothefocusS,theequationoftheellipseis

ii. ShiftingtheorigintoA,theequationoftheellipseis

iii. ShiftingtheorigintoX,theequationofthefocusis

12. Theequationofanellipseiseasilydeterminedifwearegiventhefocusandtheequationofthedirectrix.

7.3FOCALDISTANCE

Thesumofthefocaldistancesofanypointontheellipseisequaltothelengthofthemajoraxis.

Intheabovefigure,(section2.2)

Note7.3.1:

7.4POSITIONOFAPOINT

Apoint(x1,y1)liesinside,onoroutsideoftheellipseaccordingas −1

isnegative,zeroorpositive.

LetQ(x1,y1)beapointontheordinatePNwherePisapointontheellipse

Then,

Similarly,iftheQ′(x′,y′)isapointoutsidetheellipse,

EvidentlyifQ(x′,y′)isapointontheellipse,

7.5AUXILIARYCIRCLE

Thecircledescribedonthemajoraxisasdiameteriscalledtheauxiliarycircle.LetPbeanypointontheellipse.LettheordinatethroughPmeettheauxiliary

circleatP′.Since wehavethegeometricalrelation,P′N2=AN·A′N.

ThepointP′wheretheordinatePNmeetstheauxiliarycircleiscalledthecorrespondingpointofP.Therefore,theordinateofanypointontheellipsetothatofcorrespondingpointontheellipseareintheratiosoflengthsofsemi-minoraxisandsemi-majoraxis.Thisratiogivesanotherdefinitiontoanellipse.Consideracircleandfromeachpointonit,drawperpendiculartoadiameter.Thelocusofthesepointsdividingtheseperpendicularsinagivenratioisan

ellipseandforthisellipsethegivencircleistheauxiliarycircle.

ILLUSTRATIVEEXAMPLESBASEDONFOCUS-DIRECTRIXPROPERTY

Example7.1

Findtheequationoftheellipsewhosefoci,directrixandeccentricityaregivenbelow:

i. Focusis(1,2),directrixis2x−3y+6=0andeccentricityis2/3ii. Focusis(0,0),directrixis3x+4y−1=0andeccentricityis5/6

iii. Focusis(1,–2),directrixis3x−2y+1=0andeccentricityis1/

Solution

i. LetP(x1,y1)beapointontheellipse.Then

Therefore,thelocusof(x1,y1)istheellipse101x2+81y2+48x−330x−324y+441=0.

ii.

Therefore,thelocusof(x1,y1)istheellipse27x2+20y2−24xy+6x+8y−1=0.

iii.

Therefore,thelocusof(x1,y1)istheellipse17x2+22y2+12xy−58x+108y+129=0.

Example7.2

Findtheequationoftheellipsewhose

i. Fociare(4,0)and(−4,0)and

ii. Fociare(3,0)and(−3,0)and

Solution

i. Ifthefociare(ae,0)and(−ae,0)thentheequationoftheellipseis Here,ae=4and

∴Theequationoftheellipseis

ii. Ifthefociare(ae,0)and(−ae,0)theequationoftheellipseis

Here,ae=3and a2e2=9and

Therefore,theequationoftheellipseis

Example7.3

Findtheeccentricity,fociandthelengthofthelatusrectumoftheellipse.

i. 9x2+4y2=36

ii. 3x2+4y2−12x−8y+4=0

iii. 25x2+9y2−150x−90y+225=0.

Solution

i. 9x2+4y2=36Dividingby36,weget

Thisisanellipsewhosemajoraxisisthey-axisandminoraxisisthex-axisandcentreattheorigin.

Therefore,eccentricity=

Therefore,fociare

Therefore,latusrectum=

ii.

Shifttheorigintothepoint(2,1).Therefore,centreis(2,1).


Therefore,fociare(3,1)and(1,1)withrespecttooldaxes.

Lengthofthelatusrectum

iii.

Shifttheorigintothepoint(3,5).


Therefore,centreis(3,5).Thisisanellipsewithy-axisonthemajoraxisandx-axisastheminoraxis.

Therefore,focilieonthelinex=3.

Therefore,fociare(3,9)and(3,1)andLeangthofthelatusrectum=

Exercises

1. Findthecentre,fociandlatusrectumoftheellipse:

i. 3x2+4y2+12x+8y−32=0Ans.:(−2,−1);(0,−1);(−4,−1);6

ii. 9x2+25y2=225

Ans.:

iii. x2+9y2=9

Ans.:

iv. 2x2+3y2−4x+6y+4=0

Ans.:

2. Findtheequationoftheellipsewhosefociare(0,±2)andthelengthofmajoraxisis2

Ans.:5x2+y2=5

3. Findtheequationoftheellipsewhosefociis(3,1),eccentricity anddirectrixisx−y+6=0.

Ans.:7x2+2xy+7y2−60x−20y+44=0

4. Findtheequationofellipsewhosecentreisattheorigin,onefocusis(0,3)andthelengthofsemi-majoraxisis5.

Ans.:

5. Findtheequationofellipsewhosefocusis(1,−1),eccentricityis anddirectrixisx−y+3=0.

Ans.:7x2+2xy+7y2−22x+22y+7=0

6. Findtheequationoftheellipsewhosecentreis(2,−3),onefocusat(3,−3)andonevertexat(4,−3).

Ans.:3x2+4y2−12x+24y+36=0

7. Findthecoordinatesofthecentre,eccentricityandfocioftheellipse8x2+6y2−6x+12y+13=0

Ans.:

8. Findtheequationoftheellipsewithfociat(0,1)and(0,−1)andminoraxisoflength1.

Ans.:2x2+4y2=5

9. Anellipseisdescribedbyusingoneendlessstringwhichispassedthroughtwopoints.Iftheaxesare6and4unitsfindthenecessarylengthandthedistancebetweenthepoints.

Ans.:

7.6CONDITIONFORTANGENCY

Tofindtheconditionthatthestraightliney=mx+cmaybeatangenttotheellipse:

Lettheequationoftheellipsebe


Solvingequations(7.6)and(7.7),wegettheirpointsofintersection;thex-

coordinatesofthepointsofintersectionaregivenby

Ify=mx+cisatangenttotheellipsethenthetwovaluesofxofthisequationareequal.Theconditionforthatisthediscriminantofthequadraticequationiszero.

Thisistherequiredconditionfortheliney=mx+ctobeatangenttothegivenellipse.

Note7.6.1:Theequationofanytangenttotheellipseisgivenby

7.7DIRECTORCIRCLEOFANELLIPSE

Toshowthatalwaystwotangentscanbedrawnfromagivenpointtoanellipseandthelocusofpointofintersectionofperpendiculartangentsisacircle:


Anytangenttothisellipseis

Ifthistangentpassesthroughthepoint(x1,y1)theny1=

Thisisaquadraticequationinmandhencetherearetwovaluesform.Foreachvalueofm,thereisatangent(realorimaginary)andhencetherearetwo

tangentsfromagivenpointtoanellipse.Ifm1andm2aretherootsofthe

equation(7.11),then

Ifthetwotangentsareperpendicularthenm1m2=−1.

Thelocusof(x1,y1)isx2+y2=a2+b2whichisacircle,centreat(0,0)and

radius

Note7.7.1:Thiscircleiscalledthedirectorcircleoftheellipse.

7.8EQUATIONOFTHETANGENT

Tofindtheequationofthechordjoiningthepoints(x1,y1)and(x2,y2)andfindtheequationofthetangentat(x1,y1)totheellipse:LetP(x1,y1)andQ(x2,y2)betwopointsontheellipse.Lettheequationof

ellipsebe

Then

and

Subtracting,

Fromequation(7.15),wegettheequationofthechordjoiningthepoints(x1,y1)and(x2,y2)as:

Thischordbecomesthetangentat(x1,y1)ifQtendstoPandcoincideswithP.Hence,byputtingx2=x1andy2=y1inequation(7.16),wegettheequationofthetangentat(x1,y1).Therefore,theequationofthetangentat(x1,y1)is:

Dividingbya2b2,weget

However, since(x1,y1)liesontheellipse.

Therefore,fromequation(7.17),theequationofthetangentat(x1,y1)is

7.9EQUATIONOFTANGENTANDNORMAL

Tofindtheequationoftangentandnormalat(x1,y1)totheellipse

Theequationoftheellipseis

Differentiatingwithrespecttox,weget

However, =slopeofthetangentat(x1,y1).Therefore,theequationof

thetangentat(x1,y1)is,

Dividingbya2b2,weget

Slopeofthenormalat(x1,y1)is

Theequationofthenormalat(x1,y1)totheellipseis

Dividingbyx1,y1,weget,

Therefore,theequationofnormalat(x1,y1)totheellipse is

7.10EQUATIONTOTHECHORDOFCONTACT

Tofindtheequationtothechordofcontactoftangentsdrawnfrom(x1,y1)

totheellipse


LetQRbethechordofcontactoftangentsfromP(x1,y1).LetQandRbethepoints(x2,y2)and(x3,y3),respectively.ThentheequationoftangentsatQandR

are:

ThesetwotangentspassthroughP(x1,y1).

Therefore, and

Theabovetwoequationsshowthatthepoints(x2,y2)and(x3,y3)lieontheline

Hence,theequationofthechordofcontactis

7.11EQUATIONOFTHEPOLAR

TofindtheequationofthepolarofthepointP(x1,y1)ontheellipse

LetP(x1,y1)bethegivenpoint.LetQRbeavariablechordthroughthepointP(x1,y1).LetthetangentsatQandRmeetatT(h,k).TheequationofthechordcontactfromT(h,k)is:

Thischordofcontactpassesthrough(x1,y1).

ThelocusofT(h,k)isthepolarofthepoint(x1,y1).

Therefore,thepolarof(x1,y1)is

Note7.11.1:

1. Whenthepoint(x1,y1)liesontheellipse,thepolarof(x1,y1)isthetangentat(x1,y1).Whenthepoint(x1,y1)liesinsidetheellipsethepolarof(x1,y1)isthechordofcontactoftangentsfrom(x1,y1).

2. Theline iscalledthepolarofthepoint(x1,y1)and(x1,y1)iscalledthepoleofthe

line

7.12CONDITIONFORCONJUGATELINES

Tofindthepoleofthelinelx+my+n=0withrespecttotheellipse

anddeducetheconditionforthelineslx+my+n=0andl1x+m1y

+n1=0tobeconjugatelines:

Let(x1,y1)bethepoleoftheline

withrespecttotheellipse

Thenthepolarof(x1,y1)is:

Thentheequations(7.24)and(7.26)representthesameline.∴Identifyingequations(7.24)and(7.26),weget

Hence,thepoleofthelinelx+my+n=0is Twolinesaresaidto

beconjugateifthepoleoftheeachliesontheother.

∴Thepoint liesonthelinel1x+m1y+n1=0.

Thisistherequiredconditionforthelineslx+my+n=0andl1x+m1y+n1=0tobeconjugatelines.

ILLUSTRATIVEEXAMPLESBASEDONTANGENTS,NORMALS,POLE-POLARANDCHORD

Example7.4

Findtheequationofthetangenttotheellipsex2+2y2=6at(2,−1).

Solution

Theequationoftheellipseisx2+2y2=6.


Therefore,theequationofthetangentat(2,−1)is

(i.e.)2x−2y=6⇒x−y=3

Example7.5

Findtheequationofthenormaltotheellipse3x2+2y2=5at(−1,1).

Solution

Therefore,theequationofthenormaltotheellipse3x2+2y2=5is2x+3y=1.

Example7.6

IfBandB′aretheendsoftheminoraxisofanellipsethenprovethatSB=S′B′=awhereSandS′arethefocianda′isthesemi-majoraxis.Showalsothat

SBS′B′isarhombuswhoseareais2abe.

Solution

Sis(ae,0);S′is(−ae,0)Bis(0,b);B′is(0,−b)

Inthefigure,SBS′B′thediagonalsSS′andBB′areatrightangles.Therefore,SBS′B′isarhombus.

Example7.7

IfthetangentatPoftheellipse meetsthemajoraxisatTandPNisthe

ordinateofP,thenprovethatCN·CT=a2whereCisthecentreoftheellipse.

Solution

LetPbethepoint(x1,y1)

Theequationoftangentat(x1,y1)is

Whenthetangentmeetsthex-axis,y=0

Example7.8

ThetangentatanypointPontheellipse meetsthetangentsatAandA′

(extremitiesofmajoraxis)inLandM,respectively.ProvethatAL·A′M=b2.

Solution

LettheequationofthetangentatPbe Theequationofthe

tangentatthepointAisx=a.Solvingthesetwoequations,weget

Example7.9

IfSYandS′Y′beperpendicularsfromthefociuponthetangentsatanypointof

theellipse ,thenprovethatY,Y′lieonthecirclex2+y2=a2andthatSY

·S′Y′=b2.

Solution

TheequationofthetangentatanypointPis

Theslopeofthetangentism.

Therefore,theslopeoftheperpendicularlineSYis Sisthepoint(ae,0).

∴TheequationofSYisy= (x−ae).

LetY,thefootoftheperpendicular,be(x1,y1)Thenfromequation(7.27),weget

Fromequation(7.28),weget

Addingequations(7.29)and(7.30),weget

Cancelling,

Thelocusof(x1,y1)isx2+y2=a2.Similarly,wecanprovethatthelocusofY′isalsothiscircle.Hence,YandY′lieonthiscircle.

Note7.12.1:Thiscircleiscalledtheauxiliarycircle(x2+y2=a2).Thisisthecircledescribedonthemajoraxisasdiameter.

Example7.10

IfnormalatapointPontheellipse meetsthemajoraxisatGthen

provethat:

i. CG=e2CN,whereCisthecentreoftheellipseandNisthefootoftheperpendicularfromPtothemajoraxis.

ii. SG=eSPwhereSisthefocusoftheellipse.

Solution

i. LetPbethepoint(x1,y1).


Whenthismeetsthex-axis,y=0.

ii.

Example7.11

Inanellipse,provethatthetangentandnormalatanypointParetheexternalandinternalbisectorsoftheangleSPS′whereSandS′arethefoci.

Solution

LetP(x1,y1)beanypointontheellipse.

LetthenormalatPmeetthex-axisatL.TheequationofthenormalatPis

Wheny=0,x=e2x1∴Lis(e2x1,0).



Therefore,thenormalPListheinternalbisectorof SincethetangentatPis

perpendiculartothenormalatP,thetangentPistheexternalbisector.

Example7.12

Findtheanglesubtendedbyafocalchordoftheellipse passing

throughanendoftheminoraxisatthecentreoftheellipse.

Solution


Theequationofthefocalchordis

ThecombinedequationofthelinesCBandCQisgotbyhomogenizationoftheequationoftheellipsewiththehelpofstraightline(7.37).

TheanglebetweenthelinesCBandCQisgivenby

SincetheangleBCQisobtuse,θ=tan−1

Example7.13

Abarofgivenlengthmoveswithitsextremitiesontwofixedstraightlinesat

Abarofgivenlengthmoveswithitsextremitiesontwofixedstraightlinesatrightangles.Provethatanypointoftheroddescribesanellipse.

Solution

LetOAandOBbethetwoperpendicularlinesandABbetherodoffixedlength.LetP(x1,y1)beanypointoftherod.LettherodbeinclinedatanangleθwithOX.

(i.e.)

TakePA=aandPB=b.Thenx=OQ=RP=bcosθ,y=QP=bsinθ,

Hence,

Therefore,thelocusofPisanellipse.

Example7.14

Theequation25(x2−6x+9)+16y2=400representsanellipse.Findthecentreandfocioftheellipse.Howshouldtheaxisbetransformedsothattheellipseis

representedbytheequation

Solution

25(x2−6x+9)+16y2=40025(x−3)2+16y2=400

Dividingby400, Takex−3=X,y=Y.

Then

ThemajoraxisofthisellipseistheY-axis.

Centreis(3,0).Fociare(3,±ae)(i.e.) (i.e.)(3,±3).Nowshiftorigin

tothepoint(3,0)andthenrotatetheaxesthroughrightangles.Thenthe

equationoftheellipsebecomes

Example7.15

Showthatifs,s′arethelengthsoftheperpendicularonatangentfromthefoci,a,a′thosefromtheverlicesandethatfromthecentrethens,s′−e2=e2(aa′−c2)whereeistheeccentricity.

Solution


FociareS(ae,0)andS′(−ae,0).VerticesareA(a,0)andA′(−a,0),centreis(0,

0).Anytangenttotheellipse(7.38)is

TheperpendiculardistancefromS(ae,0)is=

Fromequations(7.39)and(7.40),wegetss′−c2=e2(aa′−c2).

Example7.16

Acircleofradiusrisconcentricwiththeellipse .Provethateach

commontangentisinclinedtotheaxisatanangletan−1 andtowardsits

length.

Solution


Theequationofthecircleconcentricwiththeellipseis

Anytangenttotheellipseis

Anytangenttothecircleis

Ifthetangentisacommontangentthen

Therefore,theinclinationtothemajoraxisisθ=tan−1

Example7.17

Provethatthesumofthesquaresoftheperpendicularsofanytangentofan

ellipse fromtwopointsontheminoraxis,eachdistance from

thecentreis2a2.

Solution

Theequationoftheellipseis .

Anytangenttotheellipseis .Theperpendiculardistancefrom

tothetangentis

Theperpendiculardistancefrom isgivenby

Example7.18

Letdbetheperpendiculardistancefromthecentreoftheellipse tothe

tangentdrawnatapointPontheellipse.IfF1andF2arethetwofociofthe

ellipsethenshowthat .

Solution

Theequationoftheellipseis .LetP(x1,y1)beanypointonit.

Theequationofthetangentat(x1,y1)is .

TheperpendiculardistancefromConthistangentis

WeknowthatPF1=a−ex1,PF2=a+ex1


Example7.19

Showthatthelocusofthemiddlepointsoftheportionofatangenttotheellipse

includedbetweentheaxesisthecurve

Solution

Anytangenttotheellipse is

Whenthetangentmeetsthex-axis,y=0.

Whenitmeetsthey-axis,x=0.

Therefore,thepointsofintersectionofthetangentswiththeaxesare

and Let(x1,y1)bethemidpointoflineAB.

Therefore,thelocusofP(x1,y1)is

Example7.20

Provethatthetangenttotheellipse meetstheellipse in

points,tangentsatwhichareatrightangles.

Solution

Anytangenttotheellipse is

AtQandRletthetangentsmeettheellipse

LetL(x1,y1)bethepointofintersectionoftangentsatQandR.ThenQRisthechordofcontactformL.Itsequationis

Equations(7.48)and(7.49)representthesameline.Identifyingequations(7.48)

and(7.49),weget

Therefore, Thelocusof(x1,y1)istheequationofthedirector

circleoftheellipse(7.49).However,directorcircleistheintersectionofperpendiculartangents.Hence,thetangentsatQandRareatrightangles.

Example7.21

AchordPQofanellipsesubtendsarightangleatthecentreoftheellipse

ShowthatthelocusoftheintersectionofthetangentsatQandRisthe

ellipse

Solution

Theequationofellipseis

LetR(x1,y1)bethepointofintersectionoftangentsatPandQ.TheequationofthechordofcontactofPQis

ThecombinedequationofCPandCQisgotbyhomogenizationofequation(7.51)withthehelpofequation(7.52).


ThelocusofP(x1,y1)is

Example7.22

Showthatthelocusofpoleswithrespecttotheellipse ofanytangentto

theauxiliarycircleis

Solution

Let(x1,y1)bethepolewithrespecttotheellipse .

Thepolarof(x1,y1)is

Thisisatangenttotheauxiliarycirclex2+y2=a2.Theconditionforthatisc2=a2(1+m2).

Dividingby

Thelocusof(x1,y1)is

Example7.23

Showthatthelocusofpolesoftangentstothecircle(x−h)2+(y−k)2=r2with

respecttotheellipse is

Solution

Let(x1,y1)bethepolewithrespecttotheellipse .Thenthepolarof(x1,

y1)is Thislineisatangenttothecircle(x−h)2+(y−k)2=r2.The

conditionforthisisthattheradiusofthecircleshouldbeequaltotheperpendiculardistancefromthecentreonthetangents.


Example7.24

Findthelocusofthepoleswithrespecttotheellipseofthetangentstotheparabolay2=4px.

Solution

Let(x1,y1)bethepolewithrespecttotheellipse Thepolarof(x1,y1)is

Thisisatangenttotheparabolay2=4px.

∴Theconditionis

Therefore,thelocusof(x1,y1)isa2py2+b4x=0.

Example7.25

AnytangenttoanellipseiscutbythetangentsattheextremitiesofthemajoraxisinthepointTandT′.ProvethatthecircledrawnonTT′asdiameterpassesthroughthefoci.

Solution


TheendsofmajoraxisareA(a,0)andA′(−a,0).Anytangenttotheellipseis

ThistangentmeetsthetangentsatA,A′atTandT′,respectively.Thenthe

coordinatesofTandT′areT T′ Theequation

ofthecircleonTT′asdiameteris

ThiscirclepassesthroughthefociS(ae,0)andS′(−ae,0).

Example7.26

TheordinateNPofapointPontheellipseisproducedtomeetthetangentatoneendofthelatusrectumthroughthefocusSinQ.ProvethatQN=SP.

Solution

LetLSL′bethelatusrectumthroughthefocusS.TheequationoftangentatLis

LetPbethepoint(x1,y1).TheequationoftheordinateatPis

WhenthetangentatLmeetstheordinateatPinQ,thecoordinatesofQaregivenbysolvingequations(7.56)and(7.57).

or

y1=a−ex1∴QN=a−ex1

WeknowthatSP=a−ex1.Therefore,QN=SP.

Example7.27

ThetangentandnormalatapointPontheellipsemeettheminoraxisinTandQ.ProvethatTQsubtendsarightangleateachofthefoci.

Solution

Theequationofellipseis

TheequationofthetangentandnormalatP(x1,y1)is

WhenthetangentandnormalmeetintheminoraxisinTandQ,respectively,

thecoordinatesofTandQareT and

ThecoordinatesofSare(ae,0).

Slopeof

Slopeof

Now,m1m2=−1.Therefore,TQsubtendsarightangleatthefocusS.ThecoordinatesofS′are

(−ae,0).HenceitisprovedthatTQsubtendsarightangleatS′.

Example7.28

IfSandS′bethefocioftheellipse andebeitseccentricitythenprove

thattan wherePisanypointontheellipse.

Solution

Theequationofellipseis .

ThecoordinatesofSare(ae,0)andS′are(−ae,0).

∴SS′=2ae

InanyΔABC,weknowthattan wheresisthesemiperimeterof

ΔABC.

Let

Then

Hence,

Example7.29

AvariablepointPontheellipseofeccentricityeisjoinedtoitsfociSandS′.ProvethatthelocusoftheincentreoftheΔPSS′isanellipsewhoseeccentricity

is

Solution

Lettheequationoftheellipsebe .

ThecoordinatesofthefociareS(ae,0)andS′(−ae,0).LetP(h,k)beanypointontheellipse.ThenSP+S′P=2a.AlsoSS′=2ae.AlsoSP=a−eh,S′P=a+ek.Letthecoordinatesoftheincentrebe(x1,y1).Then

Since(h,k)liesontheellipse

ThelocusofP(x1,y1)is whichisanellipsewhoseeccentricitye1

isgivenby,

Therefore,thelocusoftheincentreoftheΔPSS′isanellipsewhoseeccentricity

e1is

Exercises

1. Findtheequationofthetangenttotheellipsewhichmakesequalinterceptsontheaxes.

Ans.:

2. Findthelengthoflatusrectum,eccentricity,equationofthedirectrixandfocioftheellipse25x2+

16y2=400.

Ans.:

3. Theequationtotheellipseis2x2+y2−8x−2y+1=0.Findthelengthofitssemiaxescoordinatesofthefoci,lengthoflatusrectumandequationofthedirectrix.

Ans.:2,2 ,(2,–1),(2,3),2 ,x−2=0,y−1=0

4. Provethat touchestheellipse andfindthecoordinatesofthepointof

contact.

Ans.:

5. IfpbethelengthoftheperpendicularfromthefocusSoftheellipse onthetangentsat

Pthenshowthat

6. IfSTbetheperpendicularfromthefocusSonthetangentatanypointPontheellipse

thenshowthatTliesontheauxiliarycircleoftheellipse.

7. Thelinexcosα+ysinα=pinterceptedbytheellipse subtendsarightangleatits

centreprovethatthevalueofpis

8. Ifthechordofcontactofthetangentsdrawnfromthepoint(α,β)totheellipse

touchesthecirclex2+y2=c2provethatthepoint(α,β)liesontheellipse

9. Pisapointontheellipse andQ,thecorrespondingpointontheauxiliarycircle.Ifthe

tangentatPtotheellipsecutstheminoraxisinT,thenprovethatthelineQTtouchestheauxiliarycircle.

10. Tangentstotheellipse makeanglesθ1andθ2withthemajoraxis.Findtheequation

ofthelocusoftheirintersectionwhentan(θ1+θ2)isaconstant.11. Showthatthelocusofthepointofintersectionoftwoperpendiculartangentstoanellipseisa

circle.12. Provethatachordofanellipseisdividedharmonicallybyanypointonitanditspolewithrespect

totheellipse.13. IfthepolarofPwithrespecttoanellipsepassesthroughthepointQ,showthatpolarofQpasses

throughP.14. Findtheconditionforthepoleofthestraightlinelx+my=1withrespecttotheellipse

maylieontheellipse

Ans.:a2l2+b2m2=4

15. Chordsoftheellipse touchthecirclex2+y2=r2.Findthelocusoftheirpoles.

16. Chordsoftheellipse alwaystouchtheellipse .Showthatthelocusofthe

polesis

17. ProvethattheperpendicularfromthefocusofanellipsewhosecentreisConanypolarofPwillmeetCPonthedirectrix.

18. Showthatthefocusofanellipseisthepoleofthecorrespondingdirectrix.

19. Atangenttotheellipse meetstheellipse atQandR.Showthatthe

locusofthepoleofQRwithrespecttothelatterisx2+y2=a2+b2.20. Ifthemidpointofachordliesonafixedlinelx+my+n=0,showthatthelocusofpoleofthe

chordistheellipse

21. Findthelocusofthepolesofchordsoftheellipse whichtouchtheparabolaay2=

−2b2x.

22. Theperpendicularfromthecentreoftheellipse onthepolarofapointwithrespectto

theellipseisequaltoc.Provethatthelocusofthepointistheellipse,

23. Showthatthelocusofthepoleswithrespecttoanellipseofastraightlinewhichtouchesthecircledescribedontheminoraxisoftheellipseasdiameter.

24. Showthatthelocusofpolesoftangentstotheellipse withrespecttox2+y2=abis

anequalellipse.25. Provethatthetangentsattheextremitiesoflatusrectumofanellipseintersectonthe

correspondingdirectrix.

26. Findthecoordinatesofallpointsofintersectionoftheellipse andthecirclex2+y2=

6.Writedowntheequationofthetangentstotheellipseandcircleatthepointofintersectionandfindtheanglebetweenthem.

27. Tangentsaredrawnfromanypointontheellipse tothecirclex2+y2=a2.Provethat

theirchordofcontacttouchestheellipsea2x2+b2y2=r4.

28. Provethattheanglebetweenthetangentstotheellipse andthecirclex2+y2=abat

theirpointofintersectionistan−1

29. Provethatthesumofthereciprocalsofthesquaresofanytwodiametersofanellipsewhichareatrightanglestooneanotherisaconstant.

30. Anellipseslidesbetweentwostraightlinesatrightanglestoeachother.Showthatthelocusofitscentreisacircle.

31. Showthatthelocusofthefootofperpendicularsdrawnfromthecentreoftheellipse

onanytangenttoitis(x2+y2)2=a2x2+b2y2.32. Twotangentstoanellipseinterestatrightangles.Provethatthesumofthesquaresofthechords

whichtheauxiliarycircleinterceptsonthemisconstantandequaltothesquareofthelinejoiningthefoci.

33. Showthattheconjugatelinesthroughafocusofanellipseareatrightangles.34. Anarchwayisintheformofasemi-ellipse,themajoraxisofwhichcoincideswiththeroadlevel.

Ifthebreadthoftheroadis34feetandamanwhois6feethigh,justreachesthetopwhen2feetfromasideoftheroad,findthegreatestheightofthearch.

35. IfthepoleofthenormaltoanellipseatPliesonthenormalatQthenshowthatthepoleofthenormalatQliesonthenormalatP.

36. PQ,PRisapairofperpendiculartangentstotheellipse .ProvethatQRalwaystouches

theellipse

37. Showthatthepoints(xr,yr),r=1,2,3arecollineariftheirpolarswithrespecttotheellipse

areconcurrent.

38. Ifl1andl2bethelengthoftwotangentstotheellipse atrightanglestooneanother,

provethat

39. IfRPandRQaretangentsfromanexternalpointR(x1,y1)totheellipse andSbethe

focusthenshowthat

7.13ECCENTRICANGLE

LetPbeapointontheellipseandP′bethecorrespondingpointontheauxiliarycircle.TheangleCP′makeswiththepositivedirectionofx-axisiscalledtheeccentricangleofthepointPontheellipse.Ifthisangleisdenotedbyθ,thenCN=acosθandNP′=asinθ.

Weknowthat

ThenthecoordinatesofanypointPare(CN,NP).

(i.e.)(acosθ,bsinθ)∴‘θ’iscalledtheeccentricangleanditisalsocalledtheparameterofthepointP.

7.14EQUATIONOFTHECHORDJOININGTHEPOINTS

Tofindtheequationofthechordjoiningthepointswhoseeccentricanglesare‘θ’and‘ϕ’:Thetwogivenpointsare(acosθ,bsinθ)and(acosϕ,bsinϕ).Theequation

ofthechordjoiningthetwopointsis

Dividingbyab,

Therefore,theequationofthechordjoiningthepointswhoseeccentricangles

are‘θ’‘ϕ’is

Note7.14.1:Thischordbecomesthetangentat‘θ’ifϕ=θ

∴Theequationofthetangentat‘θ’is

7.15EQUATIONOFTANGENTAT‘Θ’ONTHEELLIPSE


Differentiatingwithrespecttox,weget,

Theequationofthetangentat‘θ’is,

Dividingbyab,

Theslopeofthenormalat

Therefore,theequationofthenormalatθis:

Dividingbysinθcosθ,weget,

Therefore,equationofnormalat‘θ’ontheellipse is

7.16CONORMALPOINTS

Ingeneral,fournormalscanbedrawnfromagivenpointtoanellipse.Ifα,β,γ,andδbetheeccentricanglesofthesefourconormalpointsthenα+β+γ+δisanoddmultipleofπ.Let(h,k)beagivenpoint.LetP(acosθ,bsinθ)beanypointontheellipse

.

Theequationofthenormalatθis

Ifthenormalpassesthrough(h,k)then

Thisisafourthdegreeequationintandhencetherearefourvaluesfort.Foreachvalueoft,thereisavalueofθandhencetherearefourvaluesofθsayα,β,γ,andδ.Hence,therearefournormalsfromagivenpointtoanellipse.

Hence, aretherootsoftheequation(7.63).

7.17CONCYCLICPOINTS

Acircleandanellipsewillcutfourpointsandthatthesumoftheeccentricanglesofthefourpointsofintersectionisanevenmultipleofπ.Lettheequationoftheellipsebe


Anypointontheellipseis(acosb,asinθ).Whenthecircleandtheellipseintersect,thispointliesonthecircle.

Substitutingthesevaluesinequation(7.67),weget

Equation(7.68)isafourthdegreeequationintandhencetherearefourvaluesfort,realorimaginary.Foreachvalueofttherecorrespondsavalueofθ.Henceingeneraltherearefourpointsofintersectionofacircleandanellipsewitheccentricanglesθ1,θ2,θ3,andθ4.Weknowthat,

7.18EQUATIONOFACHORDINTERMSOFITSMIDDLEPOINT

Tofindtheequationofachordintermofitsmiddlepoint:


LetR(x1,y1)bethemidpointofachordPQofthisellipse.LettheequationofchordPQbe

Anypointonthislineis(x1+rcosθ,y1+rsinθ).Whenthechordmeetstheellipsethispointliesontheellipse(7.69).

IfR(x1,y1)isthemidpointofthechordPQthenthetwovaluesofrarethedistancesPRandRQwhichareequalinmagnitudebutoppositeinsign.

Theconditionforthisisthecoefficientofr=0.

Substituting inequation(7.72),weget

Hence,theequationofchordintermsofitsmiddlepointisT=S1

where

7.19COMBINEDEQUATIONOFPAIROFTANGENTS

Tofindthecombinedequationofpairoftangentsfrom(x1,y1)totheellipse

Lettheequationofthechordthrough(x1,y1)be

Anypointonthislineis(x1+rcosθ,y1+rsinθ)

Ifthispointliesontheellipse ,

Thetwovaluesofrarethedistancesofthepointofintersectionofthechordandtheellipsefrom(x1,y1).Thelinewillbecomeatangentifthetwovaluesofrareequal.Theconditionforthisisthediscriminantofthequadraticequationiszero.

Usingthevaluesofcosθandsinθfromequation(7.70a),

Thisisthecombinedequationofthepairoftangentsfrom(x1,y1).

Note7.19.1:Thecombinedequationofthepairoftangentsfromthepoint(x1,y1)is

Ifthetwotangentsareperpendicularthencoefficientofx2+coefficientofy2=0.

Thelocusof(x1,y1)isx2+y2=a2+b2.Therefore,thelocusofthepointofintersectionofperpendiculartangentsisacircle.Thisequationiscalledthedirectrixofthecircle.

7.20CONJUGATEDIAMETERS

Example7.30

Findtheconditionthatthelinelx+my+n=0maybeatangenttotheellipse

.

Solution

Letlx+my+n=0beatangenttotheellipse .

Letthelinebetangentat‘θ’.Theequationofthetangentatθis

However,theequationoftangentisgivenas

Identifyingequations(7.71a)and(7.72a),weget

Squaringandadding,weget


Example7.31

Findtheconditionforthelinelx+my+n=0tobeanormaltotheellipse

.

Solution


Theequationofnormalis

Letthisequationbenormalat‘θ’.Theequationofthenormalat‘θ’is

Theequations(7.73)and(7.74)representthesameline.Therefore,identifyingequations(7.73)and(7.74),weget

Squaringandaddingequations(7.78)and(7.79),weget

Thisisrequiredcondition.

Example7.32

Showthatthelocusofthepointofintersectionoftangentstoanellipseatthepointswhoseeccentricanglesdifferbyaconstantisanellipse.

Solution

LettheeccentricanglesofPandQbeα+βandα−β.∴(α+β)−(α−β)=2β=2k;aconstant∴β=k.

TheequationoftangentsatPandQare

Let(x1,y1)betheirpointofintersection.Then



Sinceβ=k,thelocusof(x1,y1)is whichisanellipse.

Example7.33

Showthatthelocusofpolesofnormalchordsoftheellipse is

Solution

Let(x1,y1)bethepoleofthenormalchordoftheellipse

Thenthepolarof(x1,y1)withrespecttoellipseis

Letthisbenormalat‘θ’ontheellipseofequation(7.84).Thentheequationofthenormalat‘θ’is

Equations(7.80)and(7.81)representthesameline.Therefore,identifyingequations(7.80)and(7.81),weget

Squaringandaddingweget,


Example7.34

Findthelocusofmidpointsofthenormalchordsoftheellipse .

Solution

Let(x1,y1)bethemidpointofachordoftheellipsewhichisnormalatθ.Theequationofthechordintermsofitsmiddlepointis

Theequationofthenormalat‘θ’is

Equations(7.82)and(7.83)representthesameline.Therefore,identifyingequations(7.87)and(7.88)weget,




Example7.35

Ifthechordjoiningtwopoints,whoseeccentricanglesareαandβontheellipse

cutsthemajoraxisatadistancedfromthecentre,showthattan

Solution

Theequationofthechordjoiningthepointswhoseeccentricanglesareαand

Thislinemeetsthemajoraxisatthepoint(d,0).

Example7.36

Thetangentatthepointαontheellipsemeetauxiliarycircleontwopointswhichsubtendarightangleatthecentre.Showthattheeccentricityoftheellipseis(1+sin2α)–1/2.

Solution


Theequationoftheauxiliarycircleis

Theequationofthetangentat .Thislinemeetstheauxiliary

circleatPandQ.ThenthecombinedequationofthelinesCQand

(i.e.)

since coefficientofx2+coefficientofy2=0

Example7.37

Ifthenormalattheendofalatusrectumofanellipsepassesthroughoneextremityoftheminoraxis,showthattheeccentricityofthecurveisgivenbye4

+e2−1=0.

Solution

Lettheequationoftheellipsebe .

ThecoordinatesoftheendLofthelatusrectumare Theequationofthe

normalatLis

Thisnormalpassesatthepoint

Thislinepassesthroughthepoint(0,−b).

Example7.38

Provethatthetangentandnormalatapointontheellipsebisecttheanglebetweenthefocalradiiofthatpoint.

Solution


LetPTandPQbethetangentandnormalatanypointPontheellipse.The

equationofthenormalat(x1,y1)is

Whenthismeetthemajoraxis,y=0.

SinceSP′=a+ex1andSP=a−ex1.

Hence,PGbisectsinternally SincethetangentPTisperpendiculartoSG,

PTistheexternalbisectorof Therefore,thetangentandnormalatParethe

bisectorsoftheanglesbetweenthefocalradiithroughthatpoint.

Example7.39

Showthatthelocusofthemiddlepointofchordoftheellipse which

subtendsarightangleatthecentreis

Solution


Let(x1,y1)bethemidpointofachordoftheellipseofequation(7.87).

Thenitsequationis

IfCisthecentreoftheellipse,thecombinedequationofthelinesCPandCQis


Example7.40

Provethattheportionofthetangenttotheellipseinterceptedbetweenthecurveandthedirectrixsubtendsarightangleatthecorrespondingfocus.

Solution

LetPbethepoint(acosθ,bsinθ)ontheellipse .Theequationofthe

tangentatθis

Theequationofthecorrespondingdirectrixis

Solvingequations(7.88)and(7.89),wegetT,thepointofintersection.

TheslopeofSPis

TheslopeofSTis

Example7.41

Anormalinclinedat45°tothex-axisoftheellipse isdrawn.Itmeets

themajorandminoraxisinPandQrespectively.IfCisthecentreoftheellipse,

showthattheareaof∆CPQis sq.units.

Solution


Whenthismeetsx-axis,y=0.

Therefore,Pis

Whenitmeetsy-axis,x=0.

Therefore,Qis

Cis(0,0).

Slopeofthenormal=

Example7.42

Ifα−βisaconstant,provethatthechordjoiningthepoints,‘α’and‘β’touchesafixedellipse.

Solution

Theequationofthechordjoiningthepointsαandβis

Take thentheaboveequationbecomes =cosk.

Thislineisatangenttotheellipse

Example7.43

Ifthechordjoiningthevariablepointsatθandϕontheellipse

subtendsarightangleatthepoint(a,0)thenshowthat

Solution

Pisthepoint(acosθ,bsinθ).

Qisthepoint(acosϕ,bsinϕ).

SlopeofAPis

SlopeofAQis

SinceAPisperpendiculartoAQ,

Example7.44

Ifthenormaltotheellipse atthepointαcutsthecurvejoinin2αshow

thatcos

Solution


a2=14,b2=5

Theequationofthenormalat‘α’is

Example7.45

IfthenormalatanypointPtotheellipse meetsthemajorandminor

axesinGandgandifCFbetheperpendicularuponthisnormal,whereCisthecentreoftheellipse,thenprovethatPF·Pg=a2andPF·PG=b2.

Solution

LetP(acosθ,bsinθ)beanypointontheellipse.LetthenormalatPmeetthemajoraxisinGandminoraxising.LetCFbetheperpendicularfromCtothenormalatP.TheequationsofthetangentandnormalatPare

ThenthecoordinatesofGandgare and

PF=CLwhereCListheperpendicularonthetangent.

Example7.46

Showthattheconditionforthenormalsatthepoints(xi,yi),i=1,2,3onthe

ellipse tobeconcurrentis

Solution

Let(h,k)bethepointofconcurrenceofthenormal.Theequationofthenormal

at

Sincethisnormalpassesthrough(h,k),

Similarly,

Eliminatinghandkfromequations(7.94),(7.95)and(7.96),weget

Example7.47

Showthattheareaofthetriangleinscribedinanellipseis

aretheeccentricanglesofthevertices

andhencefindtheconditionthattheareaofthetriangleinscribedinanellipseismaximum.

Solution

LetΔABCbeinscribedintheellipse

LetA,BandCbethepoints(acosα,bsinα),(acosβ,bsinβ)and(acosγ,bsinγ),respectively.Thentheareaofthe∆ABCisgivenby,

IfA′,B′andC′arethecorrespondingpointsontheauxiliarycirclethen

AreaofΔABCisthegreatestwhentheareaofΔA′B′C′isthegreatest.However,theareaofA′B′C′isthegreatestwhenthetriangleisequilateral.In

thiscasetheeccentricanglesofthepointsP,QandRare

(i.e.)TheeccentricanglesofthepointsP,QandRdifferby

Example7.48

Ifthreeofthesidesofaquadrilateralinscribedinanellipseareinafixeddirection,showthatthefourthsideofthequadrilateralisalsoinafixeddirection.

Solution

Letα,β,γandδbetheeccentricanglesoftheverticesofthequadrilateralABCDinscribedintheellipse

ThentheequationofthechordPQis

TheslopeofthechordPQis .

SincethedirectionofPQisfixed, constant.

Similarly,

Therefore,thedirectionofPSisalsofixed.

Example7.49

Provethattheareaofthetriangleformedbythetangentsatthepointsα,βandγis

Solution

Theequationoftangentsatαandβare

Solvingequations(7.104)and(7.105)weget,

Therefore,thepointofintersectionoftangentsatPis

Hence,theareaofthetriangleformedbythetangentsatα,βandγis

Exercises

1. Ifαandβbetheeccentricanglesattheextremitiesofachordofanellipseofeccentricitye,prove

thatcos

2. LetPandQbetwopointsonthemajoraxisofanellipse equidistantfromthecentre.

ChordsaredrawnthroughPandQmeetingtheellipseatpointswhoseeccentricanglesareα,β,g

andδ.Thenprovethattan

3. Provethatthechordjoiningthepointsontheellipse whoseeccentricanglesdifferby

touchesanotherellipsewhosesemi-axesarehalfthoseofthefirst.

4. PSP′andQSQ′aretwofocalchordsofellipse suchthatPQisadiameter.Provethat

P′Q′passesthroughafixedpointonthemajoraxisoftheellipse.Findalsoitsequation.

Ans.:

5. PandP′arethecorrespondingpointsonanellipseanditsauxiliarycircle.ProvethatthetangentsatPandP′intersectonthemajoraxis.

6. ThetangentatoneendPofadiameterPP′ofanellipseandanychordP′QthroughtheotherendmeetatR.ProvethatthetangentatQbisectsPR.

7. Provethatthethreeellipses willhaveacommontangentif

8. AnytangenttotheellipseiscutbythetangentsattheendsofthemajoraxisinTandT′.ProvethatthecircleonTT′asdiameterwillpassthroughthefoci.

9. Findthecoordinatesofthepointsontheellipse ,thetangentsatwhichwillmakeequal

angleswiththeaxis.Alsoprovethatthelengthoftheperpendicularfromthecentreoneitherof

theseis

Ans.:

10. Findtheconditionforthelinexcosα+ysinα=pisatangenttotheellipse

Ans.:αcos2α+bsin2α=p2

11. Ifthetangenttotheellipse ,interceptslengthsαandβonthecoordinateaxesthen

showthat

12. Ifxcosα+ysinα−p=0beatangenttotheellipse ,provethatp2=a2cos2α+b2

sinα.IfPbethepointofcontactofthetangentxcosα+ysinα=pandN,thefootofthe

perpendicularonit,fromthecentreoftheellipse,provethat

13. ThetangentatoneendofPofadiameterOP′ofanellipseandanychordP′QthroughtheotherendmeetinR.ProvethatthetangentatQbisectsOR.

14. PandP′arecorrespondingpointsonanellipseandtheauxiliarycircle.ProvethatthetangentsatPandP′intersectonthemajoraxis.

15. IfthenormalatapointPontheellipseofsemi-axesa,bandcentreCcutsthemajorandminor

axesatGandg,showthata2Cg2+b2CG2=(a2−b2)2.AlsoprovethatPG=e·GN,wherePNistheordinateofP.

16. ThetangentsandnormalatapointPontheellipse meetthemajoraxisinTandT′so

thatTT′=a.ProvethattheeccentricangleofPisgivenbye2cos2θ+cosθ−1=0.17. Provethat,thelinejoiningtheextremitiesofanytwoperpendiculardiametersofanellipsealways

touchesaconcentriccircle.18. Showthatthelocusofthefootoftheperpendiculardrawnfromthecentreoftheellipse

onanytangenttoitis(x2+y2)2=(a2x2+b2y2)2.

19. IfPisanypointontheellipse whoseordinateisy′,provethattheanglebetweenthe

tangentatPandthefixeddistanceofPis

20. Showthatthefeetofthenormalsthatcanbedrawnfromthepoint(h,k)totheellipse

lieonthecurveb2(k−y)+a2y(x−h)=0.

21. Ifthenormalsatthefourpoints(xi,yi),i=1,2,3,4ontheellipse areconcurrent

showthat:

22. Ifthenormalsatthefourpointsθi,i=1,2,3,4areconcurrent,provethat(Σcosθi)(Σsecθi)=4.

Showthatthemeanpositionofthesefourpointsis where(h,k)isthepoint

ofconcurrency.

23. Ifthenormalsatthepointsα,βandγontheellipse areconcurrentthenprovethat

24. Ifα,β,γandδaretheeccentricanglesofthefourcornerpointsontheellipse then

provethat:(i)Σcos(α+β)=0and(ii)Σsin(α+β)=0.25. IfthepoleofthenormaltoanellipseatPliesonthenormalatQ,showthatthepoleofthenormal

atQliesonthenormalatP.26. FindthelocusofthemiddlepointsofthechordsofellipsewhosedistancefromthecentreCis

constantc.

27. Findthelocusofthemidpointofchordsoftheellipseofconstantlength2l.

28. Showthatthelocusofmidpointsofchordsoftheellipse ,tangentsattheendsof

whichintersectonthecirclex2+y2=a2is

29. Ifthemidpointofachordliesonafixedlinelx+my+n=0showthatthelocusofthepoleofthe

chordistheellipse

30. Showthatthelocusofmiddlepointsofthechordsoftheellipsethatpassthroughafixedpoint(h,

k)istheellipse

31. Provethatthelocusofthepointofintersectionoftangentstoanellipseattwopointswhoseeccentricanglesdifferbyaconstantisanellipse.Ifthesumoftheeccentricanglesbeconstantthenprovethatthelocusisastraightline.

32. TPandTQarethetangentsdrawntoanellipsefromapointTandCisitscentre.Provethatthe

areaofthequadrilateralCPTQisabtan whereθandϕaretheeccentricanglesofPandQ.

33. TheeccentricanglesoftwopointsPandQontheellipse areθandϕ.Provethatthe

areaofthisparallelogramformedbythetangentsattheendsofthediametersthroughPandQis4abcosec(θ−ϕ).

34. Chordsoftheellipse passthroughafixedpoint(h,k).Showthatthelocusoftheir

middlepointsistheellipse

35. IfPisanypointonthedirectorcircle,showthatthelocusofthemiddlepointsofthechordin

whichthepolarofPcutstheellipseis

36. Showthatthelocusofmidpointsofthechordsoftheellipse touchingtheellipse

37. IfthenormalstoanellipseatPi,i=1,2,3,4areconcurrentthenthecirclethroughP1,P2andP3meetstheellipseagaininapointP4whichistheotherendofthediameterthroughP4.

38. Findthecentreofthecirclepassingthroughthethreepoints,ontheellipsewhoseeccentricanglesareα,βandγ.

39. IfABCbeamaximumtriangleinscribedinanellipsethenshowthattheeccentricanglesofthe

verticesdifferby andthenormalsA,BandCareconcurrent.

40. Thetangentandnormaltotheellipsex2+4y2=2,atthepointPmeetthemajoraxisinQandR,

respectivelyandQR=2.ShowthattheeccentricangleofPiscos−1

41. Iftwoconcentricellipsesbesuchthatthefociofonelieontheotherthenprovethattheangle

betweentheiraxesis wheree1ande2aretheireccentricities.

42. Showthatthelengthofthefocalchordoftheellipse whichmakesanangleθwiththe

majoraxisis .

43. Ifthenormalsaredrawnattheextremitiesofafocalchordofanellipse,provethatalinethroughtheirpointofintersectionparalleltothemajoraxiswillbisectthechord.

44. Iftangentsfromthepointtotheellipse cutoffalengthequaltothemajoraxisfrom

thetangentat(a,0),provethatTliesonaparabola.45. IfthenormalatanypointPonanellipsecutsthemajoraxisatG,provethatthelocusofthe

middlepointofPQisanellipse.

46. Showthatthelocusoftheintersectionoftwonormalstotheellipse whichare

perpendiculartoeachotheris

47. Iftheanglebetweenthediameterofanypointoftheellipse andthenormalatthat

pointisθ,provethatthegreatestvalueof

48. Pisanypointonanellipse.ProvethatthelocusofthecentroidGofthepointPandthetwofocioftheellipseisaconcentricellipseofthesameeccentricity.

49. IfP,Q,RandSareconormalpointsonanellipse,showthatthecirclepassingthroughPandRwillcuttheellipseatapointS′whereSandS′aretheendsofadiameteroftheellipse.

50. Showthatthelocusofpoleofanytangenttotheellipsewithrespecttotheauxiliarycircleisasimilarconcentricellipsewhosemajoraxisisatrightanglestothatoftheoriginalellipse.

51. Thenormalsoffourpointsofanellipsemeetat(h,k).Iftwoofthepointslieon

provethattheothertwopointslieon

52. Ifthenormalstotheellipse attheendsofthechordslx+my=1andl1x+m1y=1be

concurrentthenshowthata2ll1=b2mm1=−1.

53. Provethattwostraightlinesthroughthepointsofintersectionofanellipsewithanycirclemakeequalangleswiththeaxesoftheellipse.

54. Showthattheequationofapairofstraightlineswhichareatrightanglesandeachofwhichpasses

throughthepoleoftheothermaybewrittenaslx+my+n=0andn(mx−ny)+lm(a2−b2)=0.Alsoprovethattheproductofthedistancesofsuchpairoflinesfromthecentrecommonly

exceeds

55. ShowthattherectangleundertheperpendiculardrawntothenormalatapointofanellipsefromthecentreandfromthepoleofthenormalisequaltotherectangleunderthefocaldistancesofP.

56. ProvethatifP,Q,RandSarethefeetofthenormalstotheellipse andthecoordinates

(x1,y1),(x2,y2),arethepolesofPQandRSthentheyareconnectedbytherelations

57. Ifthenormalsatfourpointsoftheellipse areconcurrentandiftwopointslieonthe

linelx+my=1,showthattheothertwopointslieontheline .Henceshowthatifthe

feetofthetwonormalsfromapointPtothisellipsearecoincidentthenthelocusofthemidpoints

ofthechordsjoiningthefeetoftheothernormalsis

7.20.1LocusofMidpoint

Locusofmidpointofaseriesofparallelchordsoftheellipse:Let(x1,y1)bethemidpointofachordparalleltotheliney=mx.ThentheequationofthechordisT=S1.

Itsslopeis

Sincethischordisparalleltoy=mx,

Thelocusof(x1,y1)is whichisastraightlinepassingthroughthecentre

oftheellipse.Ify=m1xbisectallchordsparalleltoy=mxthen

Bysymmetryofthisresult,weseethatthediametery=mxbisectallthechordsparalleltoy=m1x.

Definition7.20.1Twodiametersaresaidtobeconjugatetoeachotherifchordsparalleltooneisbisectedbytheother.Therefore,theconditionforthediameter

y=mxandy=m1xtobeconjugatediametersis

7.20.2Property:TheEccentricAnglesoftheExtremitiesofaPairofSemi-conjugateDiameterDifferbyaRightAngle

LetPCP′andDCD′beapairofconjugatediameters.LetPbethepoints(acosθ,

bsinθ)andDbethepoints(acosϕ,bsinϕ).ThentheslopeofCPis

TheslopeofCDis

SinceCPandCDaresemi-conjugatediameters

Therefore,theeccentricanglesofapairofsemi-conjugatediametersdifferbyarightangle.

Note7.20.1:ThecoordinatesofDare

(i.e.)(−asinθ,bcosθ)

Therefore,ifthecoordinatesofPare(acosθ,bsinθ)thenthecoordinatesofDare(−asinθ,bcosθ).ThecoordinatesofP′are(−acosθ,−bsinθ).ThecoordinatesofD′are(asinθ,−bcosθ).

7.20.3Property:IfCPandCDareaPairofSemi-conjugateDiametersthenCD2

+CP2isaConstant

ThecoordinatesofC,PandDareC(0,0)

P(acosθ,bsinθ)andD(−asinθ,bcosθ).

Then

7.20.4Property:TheTangentsattheExtremitiesofaPairofConjugateDiametersofanEllipseEnclosesaParallelogramWhoseAreaIsConstant

LetPCP′andDCD′beapairofconjugatediameters.LetPbethepoint(acosθ,

bsinθ).ThenDisthepoint

(i.e.)(−asinθ,bcosθ).

TheequationofthetangentatPis

TheslopeofthetangentatPis

TheslopeofCDis

Sincethetwoslopesareequal,thetangentsatPisparalleltoDCD′.Similarly,wecanshowthatthetangentatP′isparalleltoDCD′.Therefore,thetangentat

PandP′areparallel.Similarly,thetangentDandD′areparallel.Hence,thetangentsatP,P′,D,D′fromaparallelogramEFGH.TheareaoftheparallelogramEFGH

7.20.5Property:TheProductoftheFocalDistancesofaPointonanEllipseIsEqualtotheSquareoftheSemi-diameterWhichIsConjugatetotheDiameter

ThroughthePoint

LetSandS′bethefociofellipse .LetPbeanypointontheellipseand

drawMPM′perpendiculartothedirectrix.

Then,

7.20.6Property:IfPCP′andDCD′areConjugateDiameterthenTheyarealsoConjugateLines

Weknowthatthepolarofapointandthechordofcontactoftangentsfromittotheellipsearethesame.Therefore,thepoleofthediameterPCP′willbethepointofintersectionofthetangentsatPandP′whichareparallel.Therefore,thepoleofPCP′liesatinfinityontheconjugatediameterDCD′.Hence,PCP′andDCD′areconjugatelines.

Note7.20.2:Conjugatediameterisaspecialcaseofconjugatelines.

7.21EQUI-CONJUGATEDIAMETERS

Definition7.21.1Twodiametersofanellipsearesaidtobeequiconjugatediametersiftheyareofequallength.

7.21.1Property:Equi-conjugateDiametersofanEllipseLiealongtheDiagonalsoftheRectangleFormedbytheTangentattheEndsofitsAxes

LetPCP′andDCD′betwoconjugatediametersoftheellipse .Letthe

coordinatesofPbe(acosθ,bsinθ).ThenthecoordinatesofDare(−asinθ,bcosθ).CisthepointC(0,0).

IfCPandCDareequi-conjugatediametersthenCP2=CD2.

When theequationofthediameteris

When equationsofthesetwoconjugatediametersare Therefore,

theequi-conjugatediametersare whicharetheequationsofthe

diagonalsformedbythetangentsatthefourverticesoftheellipse.

ILLUSTRATIVEEXAMPLESBASEDONCONJUGATEDIAMETERS

Example7.50

Showthatthelocusofthepointofintersectionoftangentsattheextremitiesofa

pairofconjugatediametersoftheellipse istheellipse

Solution

LetPCP′andDCD′beapairofconjugatediametersoftheellipse .

LetPbethepoint(acosθbsinθ).ThenD′isthepoint

Let(x1,y1)bethepointofintersectionofthetangentsatPandD.TheequationsofthetangentsatPandDare

Sincethesetwotangentsmeetat(x1,y1),

and

Squaringandaddingfromequations(7.103)and(7.104),weget


Example7.51

IfPandDaretheextremitiesofapairofconjugatediameteroftheellipse

showthatthelocusofthemidpointofPDis

Solution

LetPbethepoint(acosθ,bsinθ).ThenthecoordinatesofDare(−asinθ,bcosθ).Let(x1,y1)bethemidpointofPD.

Squaringandaddingfromequations(7.105)and(7.106),weget

Therefore,thelocusof(x1,y1)is whichisaconcentricellipse.

Example7.52

IfCPandCDaretwoconjugatesemi-diametersofanellipse then

provethatthelinePDtouchestheellipse

Solution

LettheeccentricangleofPbeθ.ThentheeccentricangleofDis The

equationofthechordPDis

(i.e)

where

Thisstraightlinetouchestheellipse

Example7.53

FindtheconditionthatthetwostraightlinesrepresentedbyAx2+2Hxy+By2=

0maybeapairofconjugatediametersoftheellipse .

Solution

LetthetwostraightlinesrepresentedbyAx2+2Hxy+By2=0bey=m1xandy

=m2x.Then

Theconditionforthelinestobeconjugatediametersis


Example7.54

IfPandDbetheendsofconjugatesemi-diametersoftheellipsethenshowthatthelocusofthefootoftheperpendicularfromthecentreonthelinePDis2(x2+y2)2=a2x2+b2y2.

Solution

LettheeccentricangleofPbeθ.ThentheeccentricangleofDisθ+ .The

equationofPDis

Theequationofthelineperpendiculartothisandpassingthroughthecentre(0,0)is

Let(x1,y1)bethefootoftheperpendicularfrom(0,0)onPD.Then(x1,y1)liesontheabovetwolines.

Solvingfor and ,weget

Substitutingfor and inequation(7.107),weget

Therefore,thelocusof(x1,y1)is2(x2+y2)=a2x2+b2y2.

Example7.55

CPandCDaresemi-conjugatediametersoftheellipse .Ifthecircleson

CPandCDasdiametersintersectinRthenprovethatthelocusofthepointRis2(x2+y2)2=a2x2+b2y2.

Solution

LetPbethepoint(acosθ,bsinθ).ThenDisthepoint(−asinθ,bcosθ).Cisthepoint(0,0).TheequationsofthecirclesonCPandCDasdiametersarex(x−acosθ)+

y(y−bsinθ)=0andx(x+asinθ)+y(y−bcosθ)=0.

(i.e.)x2+y2=axcosθ+bysinθandx2+y2=−axsinθ+bycosθ.Let(x1,y1)beapointofintersectionofthesetwocircles.Then

Bysquaringandaddingequations(7.111)and(7.112),weget

Therefore,thelocusof(x1,y1)is(x2+y2)2=a2x2+b2y2.

Example7.56

Ifthepointsofintersectionoftheellipses and bethepointsof

conjugatediametersoftheformerprovethat

Solution

Anyconicpassingthroughthepointofintersectionoftheellipses

is

whereλ=−1,equation(7.118)reducesto

Thisbeingahomogeneousequationofseconddegreeinxandyrepresentsapairofstraightlines,thatis,equation(7.116)representsapairofstraightlines

passingthroughtheorigin.

or

Example7.57

Ifαandβbetheanglessubtendedbythemajoraxisofanellipseattheextremitiesofapairofconjugatediametersthenshowthatcos2α+cos2βisaconstant.

Solution

Letequationoftheellipsebe .

LetPbethepoint(acosα,bsinβ).ThenDisthepoint

TheslopeofAPis

TheslopeofA′Pis

Changingaintoα+ ,


Example7.58

Ifxcosα+ysinα=pisachordjoiningtheendsPandDofconjugatesemi-diametersoftheellipsethenprovethata2cos2α+b2sin2α=2p2.

Solution

Lettheequationoftheellipsebe .LetPCP′andDCD′beapairof

conjugatediameters.LetPbethepoint(acosθ,bsinθ)thenDisthepoint

TheequationofPDis

However,theequationofPDisgivenas

Equations(7.119)and(7.120)representthesameline.Identifyingequations(7.119)and(7.120),weget

Example7.59

CPandCDareconjugatediametersoftheellipse .Atangentisdrawn

paralleltoPDmeetingCPandCDinRandSrespectively.ProvethatRandSlie

ontheellipse

Solution

LetCPandCDbeapairofconjugatediametersoftheellipse

LetPbethepoint(acosθ,bsinθ).ThenDisthepoint(−asinθ,bcosθ).SlopeofPDis

LettheequationofthetangentparalleltoPDbe

LetRbethepoint(h,k).Since(h,k)liesonthistangent,

Inaddition,theequationofCPis

Sincethispassesthrough(h,k),

Substitutinginequation(7.122),weget

Eliminatingmfromequations(7.123)and(7.124),

Dividingbya2b2, Thelocusof(h,k)is Similarly,thepointS

alsoliesontheaboveellipse.

Example7.60

Atangenttotheellipse cutsthecirclex2+y2=a2+b2inPandQ.

ProvethatCPandCQarealongconjugatessemi-diametersoftheellipsewhereCisthecentreofthecircle.

Solution


Theequationofthecircleisx2+y2=a2+b2.(7.125)Theequationofthetangentatθontheellipseis

ThismeetsthecircleinPandQ.ThecombinedequationCPandCQisgotbyhomogenizationofequation(7.125)withthehelpofequation(7.126),

∴CPandCQareconjugatesemi-diametersoftheellipse.

Example7.61

Provethattheacuteanglebetweentwoconjugatediametersisleastwhentheyareofequallength.

Solution

LetPCP′andDCD′betheconjugatediameters.

Fromequations(7.127)and(7.128),

RHSisleastwhenthedenominatoristhelargest.Thishappenswhen

Therefore,theacuteanglebetweenthediametersisminimumwhentheconjugatediametersareofequallengthandtheleastacuteangleisgivenby

Example7.62

Findthelocusofthepointofintersectionofnormalsattwopointsonanellipsewhichareextremitiesofconjugatediameters.

Solution



LetPandDbetheextremitiesofapairofconjugatediametersoftheellipse(7.129).LetPandDbethepointsP(acosθ,bsinθ)andD(−asinθ,bcosθ).TheequationsofthenormalatPandDare

Solvingequations(7.130)and(7.131),weget,


Therefore,thelocusofthepointofintersectionofthesetwonormalsis(a2x2+b2y2)3=(a2−b2)2(a2x2−b2y2)2.

Example7.63

Ifthepointofintersectionoftheellipses and beatthe

extremitiesoftheconjugatediametersoftheformerthenprovethat

Solution

Thegivenellipsesare

Solvingequations(7.134)and(7.135)wegettheirpointofintersections.

Equation(7.134)−(7.135)gives

Thisisapairofstraightlinespassingthroughtheorigin.Ify=mxisoneofthe

linesthen

Thisisaquadraticequationinm.Ifm1andm2aretheslopesofthetwostraightlinesthroughtheoriginthen

Ifm1andm2aretheslopesofthepairofconjugatediametersthen


Example7.64

LetPandQbetheextremitiesoftwoconjugatediametersoftheellipse

andSbethefocus.ThenprovethatPQ2−(SP−SQ)2=2b2.

Solution

LetSbe(ae,0)andPbe(acosθ,bsinθ).ThenSP=a−aecosθ,SQ=a+aesinθ.

Example7.65

IfCPandCDaresemi-conjugatediametersoftheellipse ,provethat

thelotusoftheorthocentreofΔPCDis2(b2y2+a2x2)3=(a2−b2)2(a2x2−b2y2)2.

Solution

LetPbethepoint(asinθ,bcosθ).ThenDis(−asinθ,bcosθ).TangentatPisparalleltoCD.TangentatDisparalleltoCP.Therefore,thealtitudesthroughPandDarethenormalsatPandD,

respectively.Let(x1,y1)betheorthocentre.TheequationofthenormalatPis

TheequationofthenormalatQis

Solvingequations(7.139)and(7.140)wegetthecoordinatesoftheorthocentre.


Therefore,thelocusoftheorthocentreis2(a2x2+b2y2)3=(a2−b2)2(a2x2−b2y2)2.

Exercises

1. LetCPandCQbeapairofconjugatediametersofanellipseandletthetangentsatPandQmeetatR.ShowthatCRandPQbisecteachother.

2. Findtheconditionthatforthediametersof throughitspointsofintersectionwiththe

linelx+my+n=0tobeconjugate.

Ans.:l2+m2=a2l2+b2m2

3. Provethatb2x2+2hxy−a2y2=0representsconjugatediametersoftheellipse forall

valuesofh.

4. Provethata2x2+2hxy−b2y2=0representsconjugatediametersoftheellipseax2+by2=1forallvaluesofh.

5. Findthecoordinatesoftheendsofthediameteroftheellipse16x2+25y2=400whichisconjugateto5y=4x.

Ans.:

6. Findthelengthofsemi-diameterconjugatetothediameterwhoseequationisy=3x.7. Throughthefociofanellipse,perpendicularsaredrawntoapairofconjugatediameters.Prove

thattheymeetonaconcentricellipse.

8. Adiameteroftheellipse meetsonelatusrectuminPandtheconjugatediameter

meetstheotherlatusrectuminQ.ProvethatPQtouches

9. IfPP′isadiameterandQisanypointontheellipse,provethatQPandQP′areparalleltoapairofconjugatediametersoftheellipse.

10. Ifα+β=γ(aconstant)thenprovethatthetangentsataandbontheellipse .intersect

onthediameterthroughγ.11. Showthatthelinejoiningtheextremitiesofanytwodiametersofanellipsewhichareatright

anglestooneanotherwillalwaystouchafixedcircle.12. Showthatthesumofthereciprocalsofthesquareofanytwodiametersofanellipsewhichareat

rightanglestooneanotherisaconstant.

13. PandQareextremitiesoftwoconjugatediametersoftheellipse .andSisafocus.

ProvethatPQ2+(SP−SQ)2=2b2.14. Ifthedistancebetweenthetwofociofanellipsesubtendsangles2αand2βattheendsofapairof

conjugatediameters.Showthattan2α+tan2βisaconstant.15. Showthatthesumofthesquaresofthenormalattheextremitiesofconjugatesemi-diametersand

terminatedbymajoraxisisa2(1−e2)(2−e2).16. IfPandQaretwopointsonanellipsesuchthatCPisconjugatetothenormalatQ,provethatCQ

isconjugatetothenormalatP.

17. Twoconjugatediametersoftheellipse centreatCmeetthetangentatanypointPisE

andF.ProvethatPE·PF=CD2.

18. IfCPandCDareconjugatesemi-diametersoftheellipse ,thenormalatPcutsthe

majoraxisatGandthelineDCinFthenprovethatPG:CD=b:a.

19. ThenormalatavariablepointPofanellipse cutsthediameterCDconjugatetoPin

Q.ProvethattheequationofthelocusofQis

20. Showthatforaparallelograminscribedinanellipse,thesumofthesquaresofthesidesisconstant.

21. Showthatthemaximumvalueofthesmalleroftwoanglesbetweentwoconjugatediametersofan

ellipseis andtheminimumvalueofthisangleis whereaandbareitssemi-

majorandsemi-minoraxes,respectively.

22. IfPCP′andDCD′aretwoconjugatediametersoftheellipse .andQisanypointonthe

circlex2+y2=c2thenprovethatPQ2+DQ2+P2Q2+PQ2=2(a2+b2+2c2).

23. Twoconjugatediametersoftheellipse cutthecirclex2+y2=r2atPandQ.Show

thatthelocusofthemidpointofPQisa2[(x2+y2)2−r2x2]+b2[(x2+y2)2−r2y2]=0.24. Inanellipsewhosesemi-axesareaandb,provethattheacute-anglebetweentwoconjugate

diameterscannotbelessthan

25. IfCPandCDareconjugatediametersofanellipseshowthat4(CP2−CD2)=(SP−S′P)2−(SD

−S′D)2.26. Twoconjugatesemi-diametersofanellipseareinclinedatanglesαandβtothemajoraxis.Show

thattheirlengthscanddareconnectingtherelationc2sin2α+d2sin2β=0.27. Findtheconditionforthelinesl1x+m1y=0andl2x+m2y=0tobeconjugatediametersof

.

Ans.:a2ll1+b2mm1=0

28. Showthatax2+2hxy−by2=0representsconjugatediametersoftheellipseax2+by2=1forallvaluesofa.

29. Provethatax2+2hxy−by2=0representsconjugatediametersoftheellipseax2+by2=1forallvaluesofh.

30. CPandCQareconjugatesemi-diametersoftheellipse .AtangentparalleltoPQ

meetsCPandCQinRandS,respectively.ShowthatRandSlieontheellipse

31. IftwoconjugatediametersCPandCQofanellipsecutthedirectorcircleinLandM,provethatLMtouchestheellipse.

32. Twoconjugatediametersoftheellipse cutsthecirclex2+y2=ratPandQ.Show

thatthelocusofthemidpointofPQisa2[(x2+y2)2−r2x2]+b2[(x2+y2)2−r2y2]=0.

33. TheeccentricanglesoftwopointsPandQontheellipse areαandβ.Provethatthe

areaoftheparallelogramformedbythetangentsattheendsofthediametersthroughPandQis

andhenceshowthatitisleastwhenPandQaretheextremitiesofapairofconjugate

diameters.

34. LetPCP′beadiameteroftheellipse .IfthenormalatPmeetstheordinateatP′inT,

showthatthelocusofTis

35. IftwoconjugatediametersCPandCQofanellipsecutthedirectorcircleinLandM,provethatLMtouchestheellipse.

36. Inanellipse,apairofconjugatediametersisproducedtomeetadirectrix.Showthattheorthocentreofthetrianglessoformedisafocus.

37. ThroughafixedpointP,apairoflinesisdrawnparalleltoavariablepairofconjugatediametersofagivenellipse.ThelinesmeettheprincipalaxesinQandR,respectively.ShowthatthemidpointofQRliesonafixedline.

38. PerpendicularsPMandPNaredrawnfromanypointPofanellipseontheequi-conjugatediameteroftheellipse.ProvethattheperpendicularsfromPtoitspolarbisectMN.

39. Intheellipse3x2+7y2=21,findtheequationsoftheequi-conjugatediametersandtheirlengths.

Ans.:

40. Provethatthetangentstotheellipse atthepointswhoseeccentricanglesareθand

meetononeoftheequi-conjugatediameters.

41. Fromapointononeoftheequi-conjugatediametersofanellipsetangentsaredrawntotheellipse.

Showthatthesumoftheeccentricanglesofthepointofcontactisanoddmultipleof .

42. Tangentsaredrawnfromanypointontheellipse tothecirclex2+y2=r2.Provethat

thechordsofcontactaretangentstotheellipsea2x2+b2y2=r4.If ,provethatthe

lineandthecentretothepointsofcontactwiththecircleareconjugatediametersofthesecondellipse.

43. AnytangenttoanellipsemeetsthedirectorcircleinPandD.ProvethatCPandCDareinthedirectionsofconjugatediametersoftheellipse.

44. IfCPisconjugatetothenormalatQ,provethatCQisconjugatetothenormalatP.45. Provethatthestraightlinesjoiningthecentretotheintersectionofthestraightline

withtheellipseareconjugatediameters.

Chapter8

Hyperbola

8.1DEFINITION

Ahyperbolaisdefinedasthelocusofapointthatmovesinaplanesuchthatitsdistancefromafixedpointisalwaysetimes(e>1)itsdistancefromafixedline.Thefixedpointiscalledthefocusofthehyperbola.Thefixedstraightlineiscalledthedirectrixandtheconstanteiscalledtheeccentricityofthehyperbola.

8.2STANDARDEQUATION

LetSbethefocusandthelinelbethedirectrix.DrawSXperpendiculartothedirectrix.DivideSXinternallyandexternallyintheratioe:1(e>1).LetAand

A′bethepointofdivision.Since and thepointsAandA′lieonthe

curve.LetAA′=2aandCbeitsmiddlepoint.


Subtractingequation(8.1)fromequation(8.2),weget

TakeCSasthex-axisandCYperpendiculartoCXasthey-axis.Then,thecoordinatesofSare(ae,0).LetP(x,y)beanypointonthecurve.DrawPMperpendiculartothedirectrixandPNperpendiculartox-axis.

Fromthefocusdirectrixpropertyofhyperbola,

Dividingbya2(e2−1),weget

Thisiscalledthestandardequationofhyperbola.

Note8.2.1:

1. Thecurvemeetsthex-axisatpoints(a,0)and(−a,0).

2. Whenx=0,y2=−a2.Therefore,thecurvemeetsthey-axisonlyatimaginarypoints,thatis,therearenorealpointsofintersectionofthecurveandy-axis.

3. If(x,y)isapointonthecurve,(x,−y)and(−x,y)arealsopointsonthecurve.Thisshowsthatthecurveissymmetricalaboutboththeaxes.

4. Foranyvalueofy,therearetwovaluesofx;asyincreases,xincreasesandwheny→∞,xalso→∞.Thecurveconsistsoftwosymmetricalbranches,eachextendingtoinfinityinboththedirections.

5. AA′iscalledthetransverseaxisanditslengthis2a.6. BB′iscalledtheconjugateaxisanditslengthis2b.

7. Ahyperbolainwhicha=biscalledarectangularhyperbola.Itsequationisx2−y2=a2.Its

eccentricityis

8. ThedoubleordinatethroughthefocusSiscalledlatusrectumanditslengthis

9. ThereisasecondfocusS′andaseconddirectrixl′tothehyperbola.

8.3IMPORTANTPROPERTYOFHYPERBOLA

Thedifferenceofthefocaldistancesofanypointonthehyperbolaisequaltothelengthoftransverseaxis.

8.4EQUATIONOFHYPERBOLAINPARAMETRICFORM

(asecθ,btanθ)isapointonthehyperbola forallvaluesofθ,θis

calledaparameterandisdenotedby‘θ’.Theparametricequationsofhyperbolaarex=asecθ,y=btanθ.

8.5RECTANGULARHYPERBOLA

Ahyperbolainwhichb=aiscalledarectangularhyperbola.Thestandardequationoftherectangularhyperbolaisx2−y2=a2.

8.6CONJUGATEHYPERBOLA

ThefociareS(ae,0)andS′(−ae,0)andtheequationsofthedirectricesare

Bythesymmetryofthehyperbola,ifwetakethetransverseaxisasthey-

axisandtheconjugateaxisasx-axis,thentheequationofthehyperbolais

Thishyperbolaiscalledtheconjugatehyperbola.Here,thecoordinatesofthe

fociareS(0,be)andS′(0,−be).Theequationsofthedirectricesare

Thelengthofthetransverseaxisis2b.Thelengthoftheconjugateaxisis2a.

Thelengthofthelatusrectumis

Thefollowingaresomeofthestandardresultsofthehyperbolawhose

equationis

1. Theequationofthetangentat(x1,y1)is

2. Theequationofthenormalat(x1,y1)is

3. Theequationofthechordofcontactoftangentsfrom(x1,y1)is

4. Thepolarof(x1,y1)is

5. Theconditionthatthestraightliney=mx+cisatangenttothehyperbolaisc2=a2m2−b2and

istheequationofatangent.

6. Theequationofthechordofthehyperbolahaving(x1,y1)asthemidpointis

7. Theequationofthepairoftangentsfrom(x1,y1)isT2=SS1

8. Parametricrepresentation:x=asecθ,y=btanθisapointonthehyperbolaandthispointisdenotedbyθ.θiscalledaparameterofthehyperbola.

Theequationofthetangentat

Theequationofthenormalat

9. Thecircledescribedonthetransverseaxisasdiameteriscalledtheauxiliarycircleandits

equationisx2+y2=a2.

10. Theequationofthedirectorcircle(thelocusofthepointofintersectionofperpendiculartangents)

isx2+y2=a2−b2.

Example8.6.1

Findtheequationofthehyperbolawhosefocusis(1,2),directrix2x+y=1and

eccentricity

Solution

LetP(x1,y1)beanypointonthehyperbola.Then

Hence,theequationofthehyperbolawhichisthelocusof

(x1,y1)is7x2+12xy−2y2−2x+14y−22=0.

Example8.6.2

Showthattheequationofthehyperbolahavingfocus(2,0),eccentricity2anddirectrixx−y=0isx2+y2−4xy+4x−4=0.

Solution

Sis(2,0):e=2andequationofthedirectrixisx−y=0.LetP(x,y)beany

pointonthehyperbola.Then,

Hence,theequationofthehyperbolaisx2+y2−4xy+4x−4=0.

Example8.6.3

Findtheequationofthehyperbolawhosefocusis(2,2),eccentricity and

directrix

3x−4y=1

Solution

Sis(2,2): anddirectrix3x−4y=1.LetP(x,y)beanypointonthe

hyperbola.

Hence,theequationofthehyperbolais19x2+216xy−44y2−346x−472y−791=0.

Example8.6.4

Findtheequationofthehyperbolawhosefocusis(0,0),eccentricity and

directrixxcosα+ysinα=p

Solution

Foranypointonthehyperbola,

Hence,theequationofthehyperbolais16(x2+y2)−25(xcosα+ysinα−p)2

=0.

Example8.6.5

Findtheequationofthehyperbolawhosefociare(6,4)and(−4,4)andeccentricity2.

Solution

Sis(6,4)andS′(−4,4),andCisthemidpointofSS′

Hence,theequationofthehyperbolais

Example8.6.6

Findtheequationofthehyperbolawhosecenteris(−3,2),oneendofthe

transverseaxisis(−3,4)andeccentricityis

Solution

Centreis(−3,4)Ais(−3,4)∴A′is(−3,6);a=2


(sincethelineparalleltoy-axisisthetransverseaxis)

Example8.6.7

Findtheequationofthehyperbolawhosecentreis(1,0),onefocusis(6,0),andlengthoftransverseaxisis6.

Solution

Hence,theequationofthehyperbolais (i.e.)16x2−9y2−32x−

128=0.

Example8.6.8

Findtheequationofthehyperbolawhosecentreis(3,2),onefocusis(5,2)andonevertexis(4,2).

Solution

Cis(3,2),Ais(4,2)andSis(5,2).Hence,CA=1andthetransverseaxisisparalleltox-axis.

∴a=1Alsoae=2.Sincea=1ande=2,b2=a2(e2−1)=1(4−1)=3.


Example8.6.9

Findtheequationofthehyperbolawhosecentreis(6,2),onefocusis(4,2)ande=2.

Solution

Transverseaxisisparalleltox-axisandCS=2unitsinmagnitude.


Example8.6.10

Findthecentre,eccentricityandfociofhyperbola9x2−16y2=144.

Solution

Dividingby144,weget

Hence,thecentreofthehyperbolais(0,0)

Hence,thefociare(5,0)and(−5,0).

Example8.6.11

Findthecentre,fociandeccentricityof12x2−4y2−24x+32y−127=0

Solution

Hence,centreis(1,4).

Hence,thefociare(6,4)and(−4,4).

Example8.6.12

Findthecentreandeccentricityofthehyperbola9x2−4y2+18x+16y−43=0.

Solution

Hence,centreis(−1,2),a2=4andb2=9.

Example8.6.13

IffromthecentreCofthehyperbolax2−y2=a2,CMisdrawnperpendiculartothetangentatanypointofthecurvemeetingthetangentatMandthecurveatN,showthatCM·CN=a2.

Solution

TheequationofthetangentatP(x1,y1)inx2−y2=a2isxx1−yy1=a2.

TheequationofthelineCNisxy1+yx1=0ThenCM,whichisperpendicularfromConthetangent,isgivenby

Solvingx2−y2=a2andxy1+yx1=0wegetthecoordinatesofN

Example8.6.14

Tangentstothehyperbolamakeanglesθ1,θ2withthetransverseaxis.Findtheequationofthelocusofpointofintersectionsuchthattan(θ1+θ2)isaconstant.

Solution

Lettheequationofthehyperbolabe .Then,theequationofthetangent

tothehyperbolais

Ifthistangentpassesthrough(x1,y1),then

Ifm1andm2aretheslopesofthetwotangents,then

Itisgiventhattan(θ1+θ2)=k.

Hence,thelocusof(x1,y1)isk(x2+y2−a2−b2)2xy=0.

Example8.6.15

Provethattwotangentsthatcanbedrawnfromanypointonthehyperbolax2−

y2=a2−b2totheellipse whichmakecomplementaryangleswiththe

axes.

Solution

Thetangentdrawnfromanypointtotheellipse is

Sincethispassesthrough(x1,y1)

Ifm1andm2aretheslopesofthetangents,then

Since(x1,y1)liesonx2−y2=a2−b2,wehave

Hence,thetwotangentsmakecomplementaryangleswiththeaxes.

Example8.6.16

Chordsofthehyperbola areataconstantdistancefromthecentre.Find

thelocusoftheirpoles.

Solution

Let(x1,y1)bethepolewithrespecttohyperbola .Thepolarof(x1,y1)

is Theperpendiculardistancefromthecentreonthepolaris

(aconstant)(i.e.)

Hence,thelocusof(x1,y1)is

Example8.6.17

Findtheequationofcommontangentstothehyperbolas and

Solution

Thetwogivenhyperbolasare and

Theconditionsfory=mx+ctobeatangenttothehyperbolasare

c2=a2m2−b2and

Hence,therearetwocommontangentswhoseequationsare

Example8.6.18

Showthatthelocusofmidpointsofnormalchordsofthehyperbolax2−y2=a2

is

(y2−x2)3=4a2x2y2.

Solution

Theequationofthehyperbolaisx2−y2=a2.Let(x1,y1)bethemidpointofanormalchordofthehyperbola.Theequation

ofthenormalis andtheequationofthechordintermsofthe

middlepointis Boththeseequationsrepresentthesameline.

Hence,identifyingthem,weget

Squaringandsubtracting,weget

Thefocusof(x1,y1)is(y2−x2)3=4a2x2y2.

Example8.6.19

Provethatthelocusofmiddlepointsofchordsofthehyperbola

passingthroughafixedpoint(h,k)isahyperbolawhosecentreis

Solution

Theequationofthehyperbolais .

Theequationofthechordofthehyperbolaintermsofitsmiddlepointis

Sincethischordpassesthroughthefixedpoint(h,k),

Thelocusof(x1,y1)is ,whichisahyperbolawhose

centreis

Example8.6.20

Showthatthelocusofthefootoftheperpendicularfromthecentreuponany

normaltothehyperbola

Solution

LetP(asecθ,btanθ)beapointonthehyperbola.Letm(x1,y1)bethefootoftheperpendicularfromthecentrewithnormalatP.Theequationofthenormal

atPis

TheequationoftheperpendicularfromC(0,0)onthisnormalis

Thesetwolinesintersectat(x1,y1)

Solvingthesetwoequationforx1andy1weget

Hence,thelocusof(x1,y1)is

Example8.6.21

Chordsofthecurvex2+y2=a2touchthehyperbola .Provethattheir

middlepointslieonthecurve(x2+y2)2=a2x2−b2y2.

Solution

Let(x1,y1)bethemidpointofthechordofthecircle.Itsequationis

Thisisatangenttothehyperbola .

Hence,theconditionis

Hence,thelocusof(x1,y1)is(x2+y2)2=(a2x2−b2y2).

Example8.6.22

Showthatthelocusofmidpointsofnormalchordsofthehyperbolax2−y2=a2

is(y2−x2)2=4a2xy.

Solution

Let(x1,y1)bethemidpointofthenormalchordofthehyperbolax2−y2=a2.Then,theequationofthechordis


Thesetwoequationsrepresentthesameline.Identifying,weget

Thelocusof(x1,y1)is(y2−x2)3=4a2x2y2.

Example8.6.23

Anormaltothehyperbola meetstheaxesatQandRandlinesQLand

RLaredrawnatrightanglestotheaxesandmeetatL.ProvethatthelocusofthepointListhehyperbola(a2x2−b2y2)=(a2+b2)2.ProvefurtherthatthelocusofthemiddlepointofQRis4(a2x2−b2y2)=(a2+b2)2.

Solution

LetP(h,k)bethepointonthehyperbola .Theequationofthenormal

at(h,k)is Whenthislinemeetsthex-axisy=0

Therefore,thecoordinatesofQare .ThecoordinatesofRare

.Let(x1,y1)bethecoordinatesofL.Then,

since(h,k)liesonthe

hyperbola.Thelocusof(x1,y1)isa2x2-b2y2=(a2+b2)2.

Let(α,β)bethemidpointofQR.Then,

since(h,k)lieson

thehyperbola.Thelocusof(α,β)isis4(a2x2-b2y2)=(a2+b2)2.

Example8.6.24

Thechordsofthehyperbolax2−y2=a2touchtheparabolay2=4ax.Provethatthelocusoftheirmidpointisthecurvey2(a−y)=x3.

Solution

Let(x1,y1)bethemidpointofthechordofthehyperbola .Itsequation

is

Thelineisatangenttotheparabolay2=4ax.Theconditionis

Thelocusof(x1,y1)isx(x2−y2)=ay2(i.e.)y2(a−y)=x8.

Example8.6.25

Avariabletangenttothehyperbola meetsthetransverseaxisatQand

thetangentatthevertexatR.ShowthatthelocusofthemidpointQRisx(4y2+b2)=ab2.

Solution

Theequationofthetangentat'θ'is

Whenthislinemeetsthetransverseaxis,y=0andx=acosθ.HereQis(acosθ,0).Whenitmeetsthelinex=a,

Let(h,k)bethemidpointofQR.Then,

Hence,thelocusof(h,k)isb2x+4xy2=ab2orx(b2+4y2)=ab2.

Example8.6.26

Showthatthelocusofthemidpointsofthechordsofthehyperbola that

subtendsarightangleatthecentreis

Solution

LetP(x1,y1)bethemidpointofachordofthehyperbols .

Then,theequationofthechordisT=S1

Thechordsubtendsarightangleatthecentreofthehyperbola.Hence,thecombinedequationofthelinesCPandCQis

Since∠QCR=90°,coefficientofx2+coefficientofy2=0.

Example8.6.27

Frompointsonthecirclex2−y2=a2tangentsaredrawntothehyperbolax2−y2

=a2.Provethatthelocusofthemiddlepointsofthechordsofcontactisthecurve(x2−y2)=a2(x2+y2).

Solution

LetP(x1,y1)beapointonthecirclex2+y2=a2.

Let(h,k)bethemidpointofthechordofcontactQRofthetangentsfromPtothehyperbolax2−y2=a2.Thentheequationofchordofcontacttothehyperbolais

xx1−yy1=a2

Theequationofthechordintermsofthemiddlepoint(h,k)is

xh−yk=h2−k

Thesetwoequationsrepresentthesameline.Identifyingthem,weget

Hence,thelocusof(h,k)is(x2−y2)2=a2(x2+y2).

Example8.6.28

Ifthetangentandnormalatanypointofthehyperbola meetonthe

conjugateaxisatQandR,showthatthecircledescribedwithQRasthediameterpassesthroughthefociofthehyperbola.

Solution

Theequationofthetangentandnormalat(x1,y1)onthehyperbola are

ThesetwolinesmeettheconjugateaxisatQandR.Thereforesubstitutex=0in

equations(8.3)and(8.4).ThecoordinatesofQare ThecoordinatesofR

are

TheequationofthecirclewithQRasdiameteris

Substitutingx=±aeandy=0

(i.e.)a2e2−(a2+b2)=0(i.e.)a2e2−a2e2=0whichistrue.Hence,thecirclewithQRasdiameterpassesthroughthefoci.

Exercises

1. Findtheequationofthehyperbolawhosefocusis(1,2),directrix2x+y=1andeccentricity .

Ans.:7x2+12xy−2y2−2x+14y−22=0

2. Showthattheequationofthehyperbolahavingfocus(2,0),eccentricity2anddirectrixx−y=0

isx2+y2−4xy+4=0.

3. Findtheequationofthehyperbolawhosefocusis(2,2),eccentricity anddirectrix3x−4y=1.

Ans.:19x2+44y2−216xy−346x+472y−791=0

4. Findtheequationofthehyperbolawhosefocusis(0,0),eccentricity anddirectrixxcosα+ysin

α=p.

Ans.:16(x2+y2)−25(xcosα+ysinα−p)2=0

5. Findtheequationofthehyperbolawhosecentreis(−3,2)andoneendofthetransverseaxisis

(−3,4)andeccentricityis .

Ans.:4x2−21y2+24x+84y+36=0

6. Findtheequationofthehyperbolawhosefociare(6,4)and(−4,4)andeccentricity2.

Ans.:

7. Findtheequationofthehyperbolawhosecentreis(1,0),onefocusis(6,0)andlengthoftransverseaxisis6.

Ans.:16x2−9y2−32x−128=0

8. Findtheequationofthehyperbolawhosecentreis(3,2),onefocusis(5,2)andonevertexis(4,

2).

Ans.:3x2−y2−18x−4y+20=0

9. Findtheequationofthehyperbolawhosecentreis(6,2),onefocusis(4,2)andeccentricity2.

Ans.:

10. Findthecentre,eccentricityandfociofhyperbola9x2−16y2=144.

Ans.:

11. Findthecentre,fociandeccentricityof12x2−4y2−24x+32y−127=0.Ans.:(1,4),(6,4)and(−4,4)

12. Findthecentre,fociandeccentricityofthehyperbola9x2−4y2−18x+16y−43=0.

Ans.:

13. IfSandS′arethefociofahyperbolaandpisanypointonthehyperbola,showthatS′P−SP=2a.

14. Findthelatusofthehyperbola .

Ans.:

15. Findtheequationofthehyperbolareferredtoitsaxisastheaxisofcoordinateiflengthoftransverseaxisis5andconjugateaxisis4.

Ans.:

16. Findthelatusrectumofthehyperbola4x−9y2=36.

Ans.:

17. Findthecentre,eccentricityandfociofthehyperbolax2−2y2−2x+8y−1=0.

Ans.:

18. Findthecentre,eccentricity,focianddirectrixofthehyperbola16x2−9y2+32x+36y−164=0.

Ans.:

19. Thehyperbola passesthroughtheintersectionofthelines7x+13y−87=0and5x−

8y+7=0anditslatusrectumis Findaandb.

Ans.:

20. Tangentsaredrawntothehyperbola3x2−2y2=6fromthepointPandmakeθ1,θ2withx-axis.Ifthetanθ1tanθ2isaconstant,provethatlocusofPis

2x2−y2=7.

21. Findtheequationoftangentstothehyperbola3x2−4y2=15whichareparalleltoy=2x+k.Findthecoordinatesofthepointofcontact.

Ans.:

22. Tangentsaredrawntothehyperbolax2−y2=c2areinclinedatanangleof45°,showthatthe

locusoftheirintersectionis(x2+y2)2+4a2(x2−y2)=4a4.

23. Provethatthepolarofanypointontheellipse withrespecttothe will

touchtheellipseattheotherendoftheordinatethroughthepoint.

24. Ifthepolarofpoints(x1,y1)and(x2,y2)withrespecttohyperbolaareatrightanglesthenshow

thatb4x1x2+a4y1y2=0.

25. Findthelocusofpolesofnormalchordsofthehyperbola .

26. Chordsofthehyperbola subtendarightangleatoneofthevertices.Showthatthe

locusofpolesofallsuchchordsisthestraightlinex(a2+b2)=a(a2−b2).27. Ifchordsofthehyperbolaareataconstantdistancekfromthecentre,findthelocusoftheirpoles.

Ans.:

28. Obtainthelocusofthepointofintersectionoftangentstothehyperbola which

includesanangleβ.

Ans.:4(a2y2−b2x2+a2b2)=(x2+y2−a2+b2)tan2β

29. Ifavariablechordofthehyperbola isatangenttothecirclex2+y2=c2thenprove

thatthelocusofitsmiddlepointis

30. Showthattheconditionforthelinexcosα+ysinα=βtouchesthehyperbola isa2

cos2α−b2sin2α=p2.31. Provethatthetangentatanypointbisectstheanglebetweenfocaldistancesofthepoint.

32. Provethatthemidpointsofthechordsofthehyperbola paralleltothediametery=

mxbeonthediametera2my=b2x.33. IfthepolarofthepointAwithrespecttoahyperbolapassesthroughanotherpointB,thenshow

thatthepolarBpassesthroughA.

34. Ifthepolarsof(x1,y1)and(x2,y2)withrespecttothehyperbola areatrightangles,

thenprovethat.

35. Provethatthepolarofanypointon withrespecttothehyperbola touches

36. Obtaintheequationofthechordjoiningthepointsθandøonthehyperbolaintheform

.Ifθ−øisaconstantandequalto2α,showthatPQ

touchesthehyperbola

37. Ifacirclewithcentre(3α,3β)andofvariableradiuscutsthehyperbolax2−y2=9a2atthepoints

P,Q,RandSthenprovethatthelocusofthecentroidofthetrianglePQRis(x−2α)2−(y−2β)2=

a2.38. IfthenormalatPmeetsthetransverseaxisinrandtheconjugateaxisingandCFbe

perpendiculartothenormalfromthecentrethenprovethatPF·Pr=CB2andPF·Pg=CF2.39. Showthatthelocusofthepointsofintersectionoftangentsattheextremitiesofnormalchordsof

thehyperbola

40. Findtheequationandlengthofthecommontangentstohyperbolas

Ans.:

41. Tangentsaredrawnfromanypointonhyperbolax2−y2=a2+b2tothehyperbola .

Provethattheymeettheaxesinconjugatepoints.42. Provethatthepartofthetangentatanypointofahyperbolainterceptedbetweenthepointof

contactandthetransverseaxisisaharmonicmeanbetweenthelengthsoftheperpendicularsdrawnfromthefocionthenormalatthesamepoint.

43. Ifthechordjoiningthepointsαandβonthehyperbola isafocalchordthenprove

that wherek≠1.

44. LetthetangentandnormalatapointPonthehyperbolameetthetransverseaxisinTandG

respectively,provethatCT·CG=a2+b2.45. Ifthetangentatthepoint(h,k)tothehyperbolacutstheauxiliarycircleinpointswhoseordinates

arey1andy2thenshowthat

46. IfalineisdrawnparalleltotheconjugateaxisofahyperbolatomeetitandtheconjugatehyperbolainthepointsPandQthenshowthatthetangentsatPandQmeetonthecurve

47. Ifanellipseandahyperbolahavethesameprincipalaxesthenshowthatthepolarofanypointoneithercurvewithrespecttotheothertouchesthefirstcurve.

48. IfthetangentatanypointPonthehyperbola whosecentreisC,meetsthetransverse

andconjugateaxesinT1andT2,thenprovethat(i)CN·CT1=a2and(ii)CM·CT2=−b

2wherePMandPNareperpendicularsinthetransverseandconjugateaxes,respectively.

49. IfPisthelengthoftheperpendicularfromC,thecentreofthehyperbola onthe

tangentatapointPonitandCP=r,provethat

8.7ASYMPTOTES

Definition8.7.1Anasymptoteofahyperbolaisastraightlinethattouchesthehyperbolaatinfinitybutdoesnotliealtogetheratinfinity.

8.7.1EquationsofAsymptotesoftheHyperbola

Lettheequationofthehyperbolabe .Lety=mx+cbeanasymptote

ofthehyperbola.Solvingthesetwoequations,wegettheirpointsofintersection.Thexcoordinatesofthepointsofintersectionaregivenby

Ify=mx+cisanasymptote,thentherootsoftheaboveequationareinfinite.Theconditionsforthesearethecoefficientofx2=0andthecoefficientofx=0,b2−a2m2=0andmca2=0.

Theequationsoftheasymptotesare

Thecombinedequationoftheasymptotesis

Note8.7.1.1:

1. Theasymptotesoftheconjugatehyperbola arealsogivenby Therefore,

thehyperbolaandtheconjugatehyperbolahavethesameasymptotes.

2. Theequationofthehyperbolais

Theequationoftheasymptotesis

Theequationoftheconjugatehyperbolais

3. Theequationoftheasymptotesdiffersfromthatofthehyperbolabyaconstantandtheequationoftheconjugatehyperboladiffersfromthatoftheasymptotesbythesameconstantterm.Thisresultholdsgoodevenwhentheequationsofthehyperbolaanditsasymptotesareinthemostgeneralform.

4. Theasymptotespassthroughthecentre(0,0)ofthehyperbola.

5. Theslopesoftheasymptotesare and

Hence,theyareequallyinclinedtothecoordinateaxes,whicharethetransverseandconjugateaxes.

8.7.2AnglebetweentheAsymptotes

Let2θbetheanglebetweentheasymptotes.Then,

Hence,theanglebetweentheasymptotesis2sec−1(e).

Example8.7.1

Findtheequationoftheasymptotesofthehyperbola3x2−5xy−2y2+17x+y+14=0.

Solution

Thecombinedequationoftheasymptotesshoulddifferfromthatofthehyperbolaonlybyaconstantterm.∴Thecombinedequationoftheasymptotesis

Hence,theasymptotesare3x+y+l=0andx−2y+m=0.

Equatingthecoefficientsofthetermsxandyandtheconstantterms,weget

Solvingthesetwoequations,wegetl=2andm=5.

lm=k

∴k=10.Thecombinedequationoftheasymptotesis(3x+y+2)(x−2y+5)=0.

Example8.7.2

Findtheequationoftheasymptotesofthehyperbolaxy=xh+yk.

Solution

Thecombinedequationoftheasymptotesisxy=xh+yk+norxy−xh−yk−n=0.Theasymptotesarex+l=0andy+m=0.

(x+l)(y+m)=xy−xh−yk−n

Equatingthecoefficientsofthetermsxandyandtheconstantterms,weget

Hence,theequationoftheasymptotesis(x−h)(y−k)=0.

Example8.7.3

Findtheequationtothehyperbolathatpassesthrough(2,3)andhasforitsasymptotesthelines4x+3y−7=0andx−2y=1.

Solution

Thecombinedequationoftheasymptotesis(4x+3y−7)(x−2y−1)=0.Hence,theequationofthehyperbolais(4x+3y−7)(x−2y−1)+k=0.Thispassthrough(2,3).


Example8.7.4

Findtheequationofthehyperbolathathas3x−4y+7=0and4x+3y+1=0asasymptotesandpassesthroughtheorigin.

Solution

Thecombinedequationoftheasymptotesis

Hence,theequationofthehyperbolais(3x−4y+7)(4x+3y+1)+k=0.Thispassesthroughtheorigin(0,0).∴7+k=0ork=−7Hence,theequationofthehyperbolais

Example8.7.5

Findtheequationsoftheasymptotesandtheconjugatehyperbolagiventhatthe

hyperbolahaseccentricity ,focusattheoriginandthedirectrixalongx+y+

1=0.

Solution

Fromthefocusdirectrixproperty,theequationofthehyperbolais

Thecombinedequationoftheasymptotesis2xy+2x+2y+k=0,wherekisaconstant.Lettheasymptotesbe2x+l=0andy+m=0.Then,

Equatingliketerms,weget2m=2.∴m=1.Similarly,l=2.Aslm=k,wegetk=2.Therefore,theasymphtesofthecombinedequationoftheasymptotesis2xy+

2x+2y+2=0.Theequationoftheasymptotesoftheconjugatehyperbolashoulddifferby

thesameconstant.Theequationoftheasymptotesoftheconjugatehyperbolais2xy+2x+2y+1=0.

Example8.7.6

Derivetheequationsofasymptotes.

Solution

Theequationofthehyperbolais (i.e.)f(x1,y1)=b2x2−a2y2−a2b2=

0.Thisbeingasecond-degreeequation,itcanhavemaximumtwoasymptotes.Asthecoefficientsofthehighestdegreetermsinxandyareconstants,thereisnoasymptoteparalleltotheaxesofcoordinates.Takex=1andy=minthehighestdegreetermsϕ(m)=b2−a2m2.Similarlyϕ(m)=0.Theslopesofthe

obliqueasymptotesaregivenbyϕ2(m)=0.(i.e.)

Also,

Theequationsoftheasymptotesaregivenby

Therefore,thecombinedequationis

Exercises

1. Provethatthetangenttothehyperbolax2−3y2=3at whenassociatedwiththetwo

asymptotesformanequilateraltrianglewhoseareais squareunits.

2. Provethatthepolarofanypointonanyasymptoteofahyperbolawithrespecttothehyperbolaisparalleltotheasymptote.

3. Provethattherectanglecontainedbytheinterceptsmadebyanytangenttoahyperbolaonitsasymptotesisconstant.

4. Fromanypointofthehyperbolatangentsaredrawntoanotherwhichhasthesameasymptotes.Showthatthechordofcontactcutsoffaconstantareafromtheasymptotes.

5. Findtheequationofthehyperbolawhoseasymptotesarex+2y+3=0and3x+4y+5=0andwhichpassesthroughthepoint(1,−1)

Ans.:(x+2y−13)(3x+4y+3)−8=0

6. Findtheasymptotesofthehyperbola3x2−5xy−2y2+5x+11y−8=0.Ans.:x−2y+3=03x+y−4=0

7. Provethatthelocusofthecentreofthecirclecircumscribingthetriangleformedbythe

asymptotesofthehyperbola andavariabletangentis

8. Findtheequationoftheasymptotesofthehyperbola9y2−4x2=36andobtaintheproductofthe

perpendiculardistanceofanypointonthehyperbolafromtheasymptotes.

9. Showthatthelocusofthepointofintersectionoftheasymptoteswiththedirectricesofthe

hyperbola isthecirclex2+y2=a2.

10. LetCbethecentreofahyperbola.ThetangentatPmeetstheaxesinQandRandtheasymptotesinLandM.ThenormalatPmeetstheaxesinAandB.ProvethatLandMlieonthecircleOABandQandRareconjugatewithrespecttothecircle.

11. IfalinethroughthefocusSdrawnparalleltotheasymptotes ofthehyperbola

meetsthehyperbolaandthecorrespondingdirectrixatPandQthenshowthatSQ=2·

SP.

12. Findtheasymptotesofthehyperbola andshowthatthestraightlineparalleltoan

asymptotewillmeetthecurveinonepointatinfinity.13. Provethattheproductoftheinterceptsmadebyanytangenttoahyperbolaonitsasymptotesisa

constant.14. Ifaseriesofhyperbolasisdrawnhavingacommontransverseaxisoflength2athenprovethatthe

locusofapointPoneachhyperbola,suchthatitsdistancefromoneasymptoteisthecurve(x2−

b2)2=4x2(x2−a2).

8.8CONJUGATEDIAMETERS

Locusofmidpointsofparallelchordsofthehyperbolais

Let(x1,y1)bethemidpointofachordofthehyperbola

Thenitsequationis

Theslopeofthischordis

Letthischordbeparalleltoy=mx.

Then

Thelocusof(x1,y1)is ,whichisastraightlinepassingthroughthe

origin.

Ify=m′xbisectsallchordsparalleltoy=mxthen By

symmetry,wenotethaty=mxwillbisectallchordsparalleltoy=m′x.

Definition8.8.1Twodiametersaresaidtobeconjugateifeachbisectschordsparalleltotheother.Theconditionofthediametersy=mxandy=m′xtobe

conjugatediametersis

Note8.8.2Thesediametersarealsoconjugatediametersoftheconjugate

hyperbola since

Property8.8.1

Ifadiametermeetsahyperbolainrealpoints,itwillmeettheconjugatehyperbolainimaginarypointsanditsconjugatediameterwillmeetthehyperbolainimaginarypointsandtheconjugatehyperbolainrealpointsandviceversa.

Proof

Lettheequationofthehyperbolabe

Thentheequationoftheconjugatehyperbolais

Lety=mxandy=m′xbeapairofconjugatediametersofthehyperbola(8.5).Then

Thepointsofintersectionofy=mxandthehyperbola(8.5)aregivenby

Sincethehyperbolameetsy=mxinrealpointsfrom(8.8)b2−a2m2>0.Thepointsofintersectionof(8.6)withy=mxaregivenby

Therefore,y=mxmeetstheconjugatehyperbolainimaginarypoints.Thepointsofintersectionofy=m′xwiththehyperbola(8.5)aregivenby

Theconjugatediametermeetsthehyperbolainimaginarypoints.Alsoitsintersectionwiththeconjugatehyperbolaisgivenby

y=m′xmeetstheconjugatehyperbolainrealpoints.

Property8.8.2

IfapairofconjugatediametersmeetthehyperbolaanditsconjugatehyperbolainPandD,respectivelythenCP2−CD2=a2−b2.

Proof

LetPbethepoint(asecθ,btanθ)ThenDwillhavecoordinates(−atanθ,−bsecθ).ThenCP2=a2sec2θ+b2tan2θCD2=a2tan2θ+b2sec2θCP2−CD2=a2(sec2θ−tan2θ)−b2(sec2θ−tan2θ)=a2−b2

Property8.8.3

Theparallelogramformedbythetangentsattheextremitiesofconjugatediametersofhyperbolahasitsverticeslyingontheasymptotesandisofconstantarea.

Proof

LetPandDbepoints(asecθ,btanθ)and(atanθ,bsecθ)onthehyperbolaanditsconjugate.

ThenD′andP′are(−atanθ,−bsecθ)and(−asecθ,−btanθ),respectively.Theequationsoftheasymptotesare

TheequationsofthetangentsatP,P′,D,D′are

respectively.ClearlythetangentsatPandP′areparallelandalsothetangentsatDandD′areparallel.Solving(8.9)and(8.11)wegetthecoordinatesofDare[a(secθ+tanθ),b[secθ+tanθ)].

Thisliesontheasymptote .

Similarlytheotherpointsofintersectionalsolieontheasymptotes.

TheequationsofPCP′andDCD′are

Lines(8.11),(8.12)and(8.13)areparallelandalsothelines(8.9),(8.10)and(8.11)areparallel.Therefore,areaofparallelogramABCD=4areaofparallelogramCPAD.

Example8.8.1

Ifapairofconjugatediametersmeethyperbolaanditsconjugate,respectivelyinPandDthenprovethatPDisparalleltooneoftheasymptotesandisbisectedbytheotherasymptote.

Solution

Lettheequationofthehyperbolabe

Theequationoftheconjugatehyperbolais

Theasymptotesofthehyperbola(1)are

LetPbethepoint(asecθ,btanθ).ThenDisthepoint(atanθ,bsecθ).

TheslopeofthechordPDis Theslopeofthe

asymptote(8.18)PDisparalleltotheasymptote(8.18).

ThemidpointofPDisa .Thispointliesonthe

asymptotesgivenby(8.17).Therefore,PDisbisectedbytheotherasymptote.

Example8.8.2

Inthehyperbola16x2−9y2=144findtheequationofthediameterconjugatetothediameterx=2y.

Solution

Theequationofthehyperbolais16x2−9y2=144

Theslopeofthelinex=2yis

Ifmandm′aretheslopesoftheconjugatediametersthen

Therefore,theequationoftheconjugatediameteris or32x−9y=0.

Example8.8.3

FindtheconditionthatthepairoflinesAx2+2Hxy+By2=0tobeconjugate

diametersofthehyperbola

Solution

LetthetwostraightlinesrepresentedbyAx2+2Hxy+By2=0bey=m1xandy=m2x.Then

Iftheselinesaretheconjugatediametersofthehyperbolathen

From(8.19)and(8.20)

Property8.8.4

Anytwoconjugatediametersofarectangularhyperbolaareequallyinclinedtotheasymptotes.

Proof

Lettheequationoftherectangularhyperbolabex2−y2=a2.Theequationoftheasymptotesisx2−y2=0.

Lety=mxand beapairofconjugatediametersoftherectangular

hyperbolabex2−y2=a3.Thenthecombinedequationoftheconjugatediametersis

Thecombinedequationofthebisectorsoftheanglesbetweenthesetwolinesis

Thisisthecombinedequationoftheasymptotes.Therefore,theasymptotesbisecttheanglebetweentheconjugatediameter.

8.9RECTANGULARHYPERBOLA

Definition8.9.1Ifinahyperbolathelengthofthesemi-transverseaxisisequaltothelengthofthesemi-conjugateaxis,thenthehyperbolaissaidtobearectanglehyperbola.

8.9.1EquationofRectangularHyperbolawithReferencetoAsymptotesasAxes

Inarectangularhyperbola,theasymptotesareperpendiculartoeachother.Sincetheaxesofcoordinatesarealsoperpendiculartoeachother,wecantaketheasymptotesasthex-andy-axes.Thentheequationsoftheasymptotesarex=0andy=0.Thecombinedequationoftheasymptotesisxy=0.Theequationofthehyperbolawilldifferfromthatofasymptotesonlybya

constant.Hence,theequationoftherectangularhyperbolaisxy=kwherekisaconstanttobedetermined.LetAA′bethetransverseaxisanditslengthbe2a.Then,AC=CA′=a.DrawALperpendiculartox-axis.Sincetheasymptotes

bisecttheanglebetweentheaxes,

ThecoordinatesofAare Sinceitliesontherectangularhyperbolaxy=

k,weget Hence,theequationoftherectangularhyperbolais orxy=

c2where

Note8.9.1.1:Theparametricequationsoftherectangularhyperbolaxy=c2arex

=ctand

8.9.2EquationsofTangentandNormalat(x1,y1)ontheRectangularHyperbolaxy=c2

Theequationofrectangularhyperbolaisxy=c2.Differentiatingwithrespecttox,weget


since

x1y1=c2.

Theslopeofthenormalat(x1,y1)is


8.9.3EquationofTangentandNormalat ontheRectangularHyperbola

xy=c2

Theequationoftherectangularhyperbolaisxy=c2.Differentiatingwithrespecttox,weget

slopeofthetangentat

Theequationofthetangentatis is

Theslopeofthenormalat‘t’is−t2.Theequationofthenormalat‘t’is

Dividingbyt,weget

8.9.4EquationoftheChordJoiningthePoints‘t1’and‘t2’ontheRectangularHyperbolaxy=c2andtheEquationoftheTangentatt

Thetwopointsare Theequationsofthechordjoiningthetwo

pointsare

Crossmultiplying,weget

Thischordbecomesthetangentat‘t’ift1=t2=t.Hence,theequationofthetangentat‘t’isx+yt2=2ct.

8.9.5Properties

Anytwoconjugatediametersofarectangularhyperbolaareequallyinclinedtotheasymptotes.

ProofLettheequationoftherectangularhyperbolabex2−y2=a2.The

equationoftheasymptotesisx2−y2=0.Lety=mxand beapairof

conjugatediametersoftherectangularhyperbolax2−y2=a2.Then,thecombinedequationoftheconjugatediametersis

Thecombinedequationofthebisectorsoftheanglesbetweenthesetwolinesis

whichisthecombinedequationoftheasymptotes.

Therefore,theasymptotesbisecttheanglebetweentheconjugatediameter.

8.9.6ResultsConcerningtheRectangularHyperbola

1. Theequationofthetangentat(x1,y1)ontherectangularhyperbolaxy=c2is

2. Theequationofthenormalat(x1,y1)is

3. Theequationofthepairoftangentsfrom(x1,y1)is(xy1+yx1−2c2)2=4(xy−c2)(x1y1−c

2).4. Theequationsofthechordhaving(x1,y1)asitsmidpointisxy1+yx1=2x1y1.

5. Theequationofthechordofcontactfrom(x1,y1)isxy1+yx1=2c2.

8.9.7ConormalPoints—FourNormalfromaPointtoaRectangularHyperbola

Let(x1,y1)beagivenpointandtbethefootofthenormalfrom(x1,y1)ontherectangularhyperbolaxy=c2.


Sincethisnormalpassesthrough(x1,y1),

Thisisafourth-degreeequationintandtherearefourvaluesoft(realorimaginary).Correspondingtoeachvalueoftthereisanormal,andhencetherearefournormalsfromagivenpointtotherectangularhyperbola.

Note8.9.7.1:Ift1,t2,t3andt4arethefourpointsofintersection,then

8.9.8ConcyclicPointsontheRectangularHyperbola

Lettheequationoftherectangularhyperbolabexy=c2.Lettheequationofthecirclebex2+y2+2gx+2fy+k=0.

Let beapointofintersectionofrectangularhyperbolaandthecircle.

Then,thepoint alsoliesonthecircle.Substituting inthe

equationofthecircleweget

Thisisafourthdegreeequationint.Foreachvalueoft,thereisapointofintersection(realorimaginary).Hence,therearefourpointsofintersectionforarectangularhyperbolawiththecircle.

Note8.9.8.1:Ift1,t2,t3andt4arethefourpointsofintersection,then

Example8.9.1

Showthatthelocusofpoleswithrespecttotheparabolay2=4axoftangentstothehyperbolax2−y2=a2istheellipse4x2+y2=4a2.

Solution

Let(x1,y1)bethepolewithrespecttotheparabolay2=4ax.Then,thepolarof

(x1,y1)is

Thisisatangenttotherectangularhypherbolax2−y2=a2.Theconditionfortangencyis

Theconditionfortangencyis

Thelocusof(x1,y1)is4x2+y2=4a2whichisanellipse.

Example8.9.2

Pisapointonthecirclex2+y2=a2andPQandPRaretangentstothehyperbolax2−y2=a2.ProvethatthelocusofthemiddlepointofQRisthecurve(x2−y2)2=a2(x2+y2).

Solution

LetP(x1,y1)beapointonthecirclex2+y2=a2

SincePQandPRaretangentsfromPtotherectangularhyperbolax2–y2=a2,QRisthechordofcontactsoftangentsfromP(x1,y1).Therefore,itsequationisxx1+yy1=a2.Let(h,k)bethemidpointofQR.Itsequationisxh–yk=h2–k2.Thesetwoequationsrepresentthesameline.Therefore,identifyingthem,weget

Since

Thelocusof(h,k)is(x2–y2)2=a2(x2+y2).

Example8.9.3

Provethatthelocusofpolesofallnormalchordsoftherectangularhyperbolaxy=c2isthecurve(x2–y2)+4c2xy=0.

Solution

Let(x1,y1)bethepoleofthenormalchordofrectangularhyperbolaxy=c2.Thepolesof(x1,y1)isxy1+yx1=2c2.Letthechordbenormalatt.The

equationofthenormalattis Thesetwoequationsrepresentthe

samestraightline.Identifyingthem,weget

Also

Thelocusof(x1,y1)is(x2–y2)2+4c2xy=0.

Example8.9.4

IfPisanypointontheparabolax2+16ay=0,provethatthepolesofPwithrespecttorectangularhyperbolaxy=2a2willtouchtheparabolay2=ax.

Solution

Let(x1,y1)beanypoint.ThepolarofPwithrespecttothehyperbolaisxy1+x1y

=4a2(i.e.) Thisisatangenttotheparabolay2=ax.Thecondition

is

Thelocusof(x1,y1)isx2+16ay=0.

Example8.9.5

Atangenttotheparabolax2=4aymeetsthehyperbolaxy=c2atPandQ.ProvethatthemiddlepointofPQliesonafixedparabola.

Solution

Let(x1,y1)bethemidpointofthechordPQoftherectangularhyperbolaxy=c2.TheequationofchordPQis

Thisisatangenttotheparabolax2=4ay.Therefore,theconditionis

(i.e.) Thelocusof(x1,y1)is2x2+ay=0,whichisafixed

parabola.

Example8.9.6

Findthelocusofmidpointsofchordsofconstantlength2loftherectangularhyperbolaxy=c2.

SolutionLetR(x1,y1)bethemidpointofthechordPQ.Lettheequationofthechord

Anypointonthislineisx=x1+rcosθ,y=y1+rsinθ.Ifthispointliesontherectangularhyperbolaxy=c2,weget(x1+rcosθ)(y1+rsinθ)=c2.

Thisisaquadraticequationinr.ThetwovaluesofrarethedistancesRPandRQwhichareequalinmagnitudebutoppositeinsign.Theconditionforthisisthecoefficientofrisequaltozero.

Then,equation(8.21)becomes

Fromequation(8.22),

Substitutingtheseinequation(8.23),weget

butr=l.

Therefore,thelocusof(x1,y1)is(x2+y2)(xy–c2)–l2xy=0.

Example8.9.7

IfPP'isadiameteroftherectangularhyperbolaxy=c2showthatthelocusoftheintersectionoftangentsatPwiththestraightlinethroughP′paralleltoeitherasymptoteisxy+3c2=0.

Solution

LetPbethepoint ThenP′isthepoint

TheequationofthetangentatPisx+yt2=2ct.

TheequationofthestraightlineP'Rparalleltox-axisis

Let(x1,y1)bethepointofintersectionofthesetwolines.Then

Substitutinginequation(8.24),

Thelocusof(x1,y1)isxy+3c2=0.

Example8.9.8

Thetangentstotherectangularhyperbolaxy=c2andtheparabolay2=4axattheirpointofintersectionsareinclinedatanglesαandβ,respectively,tothex-axis.Showthattanα+2tanβ=0.

Solution

Let(x1,y1)bethepointofintersectionoftherectangularhyperbolaxy=c2andtheparabolay2=4ax.Theequationoftangentat(x1,y1)totheparabolaisyy1=2a(x+x1).Theequationoftangenttotherectangularhyperbolaisxy1+yx1=2c2.

Theslopeofthetangenttotheparabolais

Theslopeofthetangenttothetangenttotherectangularhyperbolais

since(x1,y1)liesontheparabolay2=4ax.∴tanα+2tanβ=0

Example8.9.9

Ifthenormaltotherectangularhyperbolaxy=c2atthepointtasitintersecttherectangularhyperbolaatt1thenshowthatt3t1=–1.

Solution


Theequationofthechordjoiningthepointstandt1isx+ytt1=c(t+t1).Thesetwoequationsrepresentthesamestraightline.Identifyingthem,weget

Example8.9.10

Showthattheareaofthetriangleformedbythetwoasymptotesoftherectangularhyperbolaxy=c2andthenormalat(x1,y1)onthehyperbolais

Solution


Whenthenormalmeetsthex-axis,y=0.

Whenthenormalmeetsy-axis,x=0

Theareaofthetriangle

(i.e.) sincex1y1=c2andignoringthenegativesign.

Example8.9.11

Iffourpointsbetakenonarectangularhyperbolasuchthatthechordjoininganytwoisperpendiculartothechordjoiningtheothertwoandα,β,γ,δaretheinclinationsofthestraightlinesjoiningthesepointstothecentre.provethattanαtanβtanγtanδ=1.

Solution

Lett1,t2,t3,andt4befourpointsP,Q,R,andSontherectangularhyperbolaxy=c2.Theequationofthechordjoiningt1andt2isx+yt1t2=c(t1+t2).

Theslopeofthischordis

Similarly,theslopeofthechordjoiningt3andt4is

Sincethesetwochordsareperpendicular,

TheslopeofthelineCPis

Similarly,

fromequation(8.26)

Example8.9.12

IfthenormalsatthreepartsP,QandRonarectangularhyperbolaintersectatapointSonthecurvethenprovethatthecentreofthehyperbolaisthecentroidofthetrianglePQR.

Solution

Ifthenormalattmeetsthecurveatt'thent2t'=–1.

Thisisacubicequationint.Ift1,t2andt3aretherootsofthisequationtheycanberegardedastheparametersofthepointsP,QandR,thenormalsatthesepointsmeetatt'whichisS.Fromequation(8.27),wegett1+t2+t3=0andt1t2+t2t3+t3t1=0.Let(h,k)bethecentroidofΔPQR.

Then

Thecentroidisthecentreoftherectangularhyperbola.

Example8.9.13

Showthatfournormalscanbedrawnfromapoint(h,k)totherectangularhyperbolaxy=c2andthatitsfeetformatriangleanditsorthocentre.

Solution


(i.e.)ct4–xt3+yt–c=0

Sincethispassesthrough(h,k),ct4–ht3+kt–c=0.Thisisafourthdegreeequationint.Itsrootsaret1,t2,t3andt4whicharethe

feetofthefournormalsfrom(h,k).

Ift1,t2,t3andt4arethepointsP,Q,RandSontherectangularhyperbolaxy=

c2,itcanbeshownthattheorthocentreofthetriangleis

Thispointis ist1t2t3t4=–1.

∴ThefourpointsP,Q,RandSformatriangleanditsorthocentre.

Example8.9.14

Provethatfromanypoint(h,k)fournormalscanbedrawntotherectangularhyperbolaxy=c2andthatifthecoordinatesofthefourfeetofthenormalsP,Q,RandSbe(xr,yr),r=1,2,3,4.Then(i)x1+x2+x3+x4=h,y1+y2+y3+y4=kand(ii)x1x2x3x4=y1y2y3y4=–c4.

Solution


Sincethispassesthrough(h,k)

Theformvaluesoftcorrespondtothefeetofthefournormalsfromthepoint(h,k).Ift1,t2,t3andt4arethefourfeetofthenormalsthentheyaretherootsoftheaboveequation.

Fromequation(8.28),c(t1+t2+t3+t4)=h

(i.e.)x1+x2+x3+x4=h

Dividingequation(8.30)byequation(8.31),weget

Example8.9.15

Provethatthefeetoftheconcurrentnormalsontherectangularhyperbolaxy=c2whichmeetsat(h,k)lieonanotherrectangularhyperbolawhichpasses

through(0,0)and(h,k).

Solution


Sincethispassesthrough(h,k),

Thelocusof(x1,y1)isx2–y2–hx+ky=0.Clearlythisisarectangularhyperbolapassingthrough(0,0)and(h,k).

Example8.9.16

IfarectangularhyperbolawhosecentreisciscutbyanycircleofradiusrinfourpointsP,Q,R,SthenprovethatCP2+CQ2+CR2+CS2=4r2.

Solution

Lettheequationoftherectangularhyperbolabe


Solvingthesetwoequations,wegettheirpointsofintersections

Substitutinginequation(8.33),

Ifx1,x2,x3,x4aretheabscissaeofthefourpointsofintersectionx1+x2+x3+x4=–2g.

Example8.9.17

A,B,CandDarefourpointsofintersectionofacircleandarectangularhyperbola.IfABpassesthroughthecentreofthehyperbola,showthatCDpassesthroughthecentreofthecircle.

Solution

Lettheequationoftherectangularhyperbolabexy=c2.Lettheequationofthecirclebex2+y2+2gx+2fy+k=0.LetA,B,CandDbethepointst1,t2,t3andt4,respectively.Whenthecircleandrectangularhyperbolaintersectweknowthat

TheequationofthechordABisx+yt1t2=c(t1+t2).SinceABpassesthrough(0,0),weget

t1+t2=0(8.38)

∴Fromequation(8.34),

usingequation(8.38)

[Fromequations(8.39)and(8.40)]

TheequationofthechordCDisx+yt3t4=c(t3+t4).

Thisstraightlinepassesthroughthepoint(–g,–f).Therefore,CDpassesthroughthecentreofthecircle.

Example8.9.18

ShowthatthroughanygivenpointPintheplaneofxy=c2,fournormalscanbedrawntoit.IfP1,P2,P3andP4arefeetofthesenormalsandCiscentrethen

showthat

Solution


Letpbethepoint(h,k).Sincethenormalpassesthrough(h,k),

Thisbringsafourthdegreeequation,therearefournormalsfromP.

Ift1,t2,t3andt4arethefeetofthenormalsthen

Example8.9.19

TheslopesofthesidesoftriangleABCinscribedinarectangularhyperbolaxy=c2aretanα,tanβandtanγ.IfthenormalsatA,BandCareconcurrentshowthatcot2α+cot2β+cot2γ=0.

Solution

LetA,B,Cbethepointst1,t2andt3respectively.

TheslopeofABis

Example8.9.20

Showthataninfinitenumberoftrianglescanbeinscribedinarectangularhyperbolaxy=c2whosesidestouchtheparabolay2=4ax.

Solution

LetABCbeatriangleinscribedintherectangularhyperbolaxy=c2.LetA,BandCbethepointst1,t2,andt3,respectively.SupposethesidesABandACtouchtheparabolay2=4ax.TheequationofthechordABisx+yt1t2=c(t1+t2).

Thistouchestheparabolay2=4ax.(i.e.)

(i.e.)c(t1+t2)+a(t1t2)2=0

(i.e.) SinceACalsotouchestheparabola,

Fromtheseequations,wenotethatt2,t3aretherootsoftheequation

TheequationofthechordBCisx+yt2t3=c(t2+t3).

ThisequationshowsthatBCtouchestheparabolay2=4ax.SinceABCisanarbitrarytriangleinscribedintherectangularhyperbolaxy=c2thereareinfinitenumberofsuchtrianglestouchingtheparabolay2=4ax.

Exercises

1. Provethattheportionofthetangentinterceptedbetweenbyitsasymptotesisbisectedatthepointofcontactandformatriangleofcontactarea.

2. Ifthetangentandnormaltoarectangularhyperbolamakeinterceptsa1anda2ononeasymptoteandb1andb2ontheotherthenshowthata1a2+b1b2=0.

3. PandQarevariablepointsontherectangularhyperbolaxy=c2suchthatthetangentatQpassesthroughthefootoftheordinateofP.ShowthatthelocusoftheintersectionofthetangentsatPandQisahyperbolawiththesameasymptotesasthegivenhyperbola.

4. Ifthelinesx–α=0andy−β=0areconjugatelineswithrespecttothehyperbolaxy=c2then

provethatthepoint(α,β)isonthehyperbolaxy–2c2=0.

5. Ifthechordsofthehyperbolax2−y2=a2touchtheparabolay2=4axthenprovethatthelocusof

theirmiddlepointsisthecurvey2(x–a)=x3.

6. IfPQandPRaretwoperpendicularchordsoftherectangularhyperbolaxy=c2thenshowthatQRisparalleltothenormalatP.

7. Ifthepolarofapointwithrespecttotheparabolay2=4axtouchestheparabolax2=4by,showthatthepointshouldlieonarectangularhyperbola.

8. Showthatthenormalattherectangularhyperbolaxy=c2atthepoint meetsthecurve

againatthepoint .ShowthatPQvariesasCP2whereCisthecentre.

9. IfPQisachordoftherectangularhyperbolaxy=c2whichisthenormalatPshowthat3CP2+

CQ2=PQ2whereCisthecentreoftheconic.10. Tworectangularhyperbolasaresuchthattheaxesofonearealongtheasymptotesoftheother.

Findthedistancebetweenthepointofcontactofacommontangenttothem.11. Provethatanylineparalleltoeitheroftheasymptotesofahyperbolashouldmeetitinonepointat

infinity.12. ThetangentatanypointofthehyperbolameetstheasymptotesatQandR.ShowthatCQ·CRisa

constant.13. Provethatthelocusofthecentreofthecirclecircumscribingthetriangleformedbythe

asymptotesofthehyperbola andavariabletangentis4(a2x2–b2y2)=a2+b2.

14. Showthatthecoordinatesofthepointofintersectionoftwotangentstoarectangularhyperbolaareharmonicmeansbetweenthecoordinatesofthepointofcontact.

15. IfthenormalsatA,B,CandDtotherectangularhyperbolaxy=c2meetinP(h,k)thenprovethat

PA2+PB2+PC2+PD2=3(h2+k2).

16. If(ctanϕ,ccotϕ)beapointontherectangularhyperbolaxy=c2thenshowthatthechordsthroughthepointsϕandϕ'whereϕ+ϕ'isaconstantpassesthroughafixedpointontheconjugateaxisofthehyperbola.

17. Provethatthepoleswithrespecttothecirclex2+y2=a2ofanytangenttotherectangular

hyperbolaxy=c2liesonrectangularhyperbola4c2xy=c2.18. Ifanormaltoarectangularhyperbolamakesanacuteangleθwithitstransverseaxisthenprove

thattheacuteangleatwhichitcutsthecurveagainiscot–1(2tan2θ).

19. Ifacirclecutstherectangularhyperbolaxy=c2infourpointsthenprovethattheproductofthe

abscissaeofthepointsisc4.

20. Lettherectangularhyperbolaxy=c2iscutbyacirclepassingthroughitscentreCinfourpoints

P,Q,RandS.Ifp,qbetheperpendicularsfromconPQ,RSthenshowthatpq=c2.

21. Ifatriangleisinscribedinarectangularhyperbolaxy=c2andtwoofitssidesareparalleltoy=

m1xandy=m2xthenprovethatthethirdsidetouchesthehyperbola4m1m2xy=c2(m1+m2)

2.

22. Ifacirclecutstherectangularhyperbolaxy=c2inP,Q,RandSandtheparametersofthesefourpointsbet1,t2,t3andt4,respectivelythenprovethatthecentreofthemeanpositionofthese

pointsbisectthedistancebetweenthecentresofthetwocurves.

23. Ifthreetangentsaredrawntotherectangularhyperbolaxy=c2atthepoints(xi,yi),i=1,2,3andformatrianglewhosecircumcirclepassesthroughthecentreofthehyperbolathenshowthat

andthatthecentreofthecircleliesonthehyperbola.

24. Ifacirclewithfixedcentre(3p,3q)andofvariableradiuscutstherectangularhyperbolax2–y2=

9c2atthepointsP,Q,RandSthenshowthatthelocusofthecentroidofthetrianglePQRisgiven

by(x–2p)2–(y–2q)2=a2.25. Showthatthesumoftheeccentricanglesofthefourpointsofintersectionofanellipseanda

rectangularhyperbolawhoseasymptotesareparalleltotheaxesoftheellipseisanoddmultipleofπ.

26. Iffromanypointonthelinelx+my+1=0tangentsPQ,PRaredrawntotherectangular

hyperbola2xy=c2andthecirclePQRcutsthehyperbolaagaininTandT'thenprovethatTT'

touchestheparabola(l2+m2)(x2+y2)=(lx+my+1)2.27. Ifacirclecutstwofixedperpendicularlinessothateachinterceptisofgivenlengththenprove

thatthelocusofthecentreofthecircleisarectangularhyperbola.28. IfAandBarepointsontheoppositebranchesofarectangularhyperbola.ThecircleonABas

diametercutsthehyperbolaagainatCandDthenprovethatCDisadiameterofthehyperbola.

29. IfA,BandCarethreepointsontherectangularhyperbolaxy=d2whoseabscissaearea,bandc

respectivelythenprovethattheareais andtheareaofthetriangle

enclosedbytangentsatthesepointsis

30. Iffourpointsonarectangularhyperbolaxy=c2lieonacircle,thenprovethattheproductoftheir

abscissaeisc4.31. Ifx1,x2,x3andx4betheabscissaeoftheangularpointsandtheorthocentreofatriangle

inscribedinxy=c2thenprovethatx1x2x3x4=–c4.

32. Showthatthelengthofthechordoftherectangularhyperbolaxy=c2whichisbisectedatthe

point(h,k)is

33. ProvethatthepointofintersectionoftheasymptotesofarectangularhyperbolawiththetangentatanypointPandoftheaxeswiththenormalatPareequidistantfromP.

34. IfPisanypointonarectangularhyperbolawhoseverticesareAandA'thenprovethatthebisectorsofangleAPA'areparalleltotheasymptotesofthecurve.

35. LetQCQ'isadiameterofarectangularhyperbolaandPisanypointonthecurve.ProvethatPQ,

PQ'areequallyinclinedtotheasymptotesofthehyperbola.36. ThroughthepointP(0,b)alineisdrawncuttingthesamebranchoftherectangularhyperbolaxy=

c2inQandRsuchthatPQ=QR.Showthatitsequationis9c2y+2b2x=9bc2.

37. Ifarectangularhyperbolaxy=c2iscutbyacirclepassingthroughitscentreOinpointsA,B,CandDwhoseparametersaret1,t2,t3andt4thenshowthat(t1+t2)(t3+t4)+t1t2+t3t4=0and

deducethattheproductoftheperpendicularfromOonABandCDisc2.

Chapter9

PolarCoordinates

9.1INTRODUCTION

Acoordinatesystemrepresentsapointinaplanebyanorderedpairofnumberscalledcoordinates.EarlierweusedCartesiancoordinateswhicharedirecteddistancesfromtwoperpendicularaxes.NowwedescribeanothercoordinatesystemintroducedbyNewtoncalledpolarcoordinateswhichismoreconvenientforsomespecialpurposes.

9.2DEFINITIONOFPOLARCOORDINATES

Wechooseapointintheplaneanditiscalledthepole(ororigin)andisdenotedbyO.Thenwedrawaray(halfline)startingatOcalledpolaraxis.Thisisusuallydrawnhorizontallytotherightandcorrespondstopositivex-axisinCartesiancoordinates.

LetPbeanypointintheplaneandrbethedistancefromOtoP.Letθbetheangle(usuallymeasuredinradians)betweenthepolaraxisandthelineOP.ThenthepointPisrepresentedbytheorderedpair(r,θ)and(r,θ)arecalledthepolarcoordinatesofthepointP.Weusetheconventionthatanangleispositiveifmeasuredintheanti-clockwisedirectionfromthepolaraxisandnegativeintheclockwisedirection.

IfPcoincideswithOthenr=θ.Then(r,θ)representthecoordinatesofthepoleforanyvalueofθ.Letusnowextendthemeaningofpolarcoordinates(r,θ)whenrisnegative,agreeingthatthepoints(−r,θ)and(r,θ)lieonthesamelinethroughOandatthesamedistance|r|fromObutonoppositesidesofO.Ifr>0,thepoint(r,θ)liesonthesamequadrantasθ.Ifr<0,thenitliesinthequadrantoftheoppositesideofthepole.Wenotethatthepoint(r,θ)representsthesamepointas(r,θ+π)

Example9.2.1

Representthefollowingpolarcoordinatesinthepolarplane:

Solution

Thecoordinates, and arerepresentedbypointsin

thefollowingdiagram:

InCartesiansystemofcoordinates,everypointhasonlyonerepresentation.Butinpolarcoordinatessystemeachpointhasmanyrepresentations,for

example,point isalsorepresentedby ,etc.

Ingeneral,thepoint(r,θ)isalsorepresentedby(r,θ+2nπ)or(−r,θ+2n+1π)wherenisanyinteger.

9.3RELATIONBETWEENCARTESIANCOORDINATESANDPOLARCOORDINATES

If(x,y)istheCartesiancoordinatesand(r,θ)arethepolarcoordinatesofthe

pointP,then and

Therefore,thetransformationsfromonesystemtoanotheraregivenbyx=rcosθ,y=rsinθ.Tofindrfromxandy,weusetherelationr2=x2+y2andθisgivenby

Wehavealreadystudiedthedistancebetweentwopoints,areaofatriangle,equationsofastraightline,equationstoacircleandequationofconicsinCartesiancoordinatessystem.Letusnowderivetheresultsinpolarcoordinatesystem.

9.4POLAREQUATIONOFASTRAIGHTLINE

ThegeneralequationofastraightlineinCartesiancoordinatesisAx+By+C=0,whereA,BandCareconstants.Let(r,θ)bepolarcoordinatesofapointandthex-axisbetheinitialline.Thenforanypoint(x,y)onthestraightlinex=rcosθ,y=rsinθ.Substitutingtheseintheequationofstraightline,weget

Thiscanbewrittenintheform

whereA,Bandlareconstants.Therefore,equation(9.1)isthegeneralequationofastraightlineinpolarcoordinates.

9.5POLAREQUATIONOFASTRAIGHTLINEINNORMALFORM

Lettheoriginbethepoleandthex-axisbetheinitialline.DrawONperpendiculartothestraightline.LetON=pand∠XON=α.

Thisisthepolarequationoftherequiredstraightline.

Note9.5.1:Polarequationofthestraightlineperpendicularto

isoftheform

or ,wherekisaconstant.

Note9.5.2:Thepolarequationofthestraightlineparallelto

is ,wherekisaconstant.

Note9.5.3:Theconditionforthestraightlines and

tobeperpendiculartoeachotherisAA1+BB1=0.Thisresult

canbeeasilyseenfromtheircartesianequations.

Note9.5.4:Ifthelineisperpendiculartotheinitiallinethenα=0orπ.Therefore,theequationofthestraightlineisrcosθ=porrcosθ=−p.

Note9.5.5:Ifthelineisparalleltotheinitiallinethen or .Inthiscase

theequationofthelineis

Example9.5.6

FindtheequationofthestraightlinejoiningthetwopointsP(r1,θ1)andQ(r2,θ2).

Solution

LetR(r,θ)beanypointonthelinejoiningthepointsPandQ.TheareaofthetriangleformedbythepointsP(r1,θ1),Q(r2,θ2)and(r3,θ3)is

Takingr3=randθ3=θ,weget

SincethepointsP,QandRarecollinear,Δ=0.

Dividingbyrr1r2,weget

Thisistheequationoftherequiredstraightline.

Example9.5.7

Findtheslopeofthestraightline

Solution

Theequationofthestraightlineis

Therefore,theslopeofthestraightlineis

Example9.5.8

Findthepointofintersectionofthestraightlines and

.

Solution

Theequationsofthestraightlinesare


Therefore,theonlypossibilityis .

Thenfromtheequationofthefirststraightline,weget

Hence,thepointofintersectionofthetwogivenlinesis

Example9.5.9

Findtheequationofthelinejoiningthepoints and anddeducethat

thislinealsopassesthroughthepoint .

Solution

Theequationofthelinejoiningthepoints(r1,θ1)and(r2,θ2)is

Therefore,theequationofthelinejoiningthepoints and is

Hence,thepoint liesonthestraightline.

Example9.5.10

Showthatthestraightlinesr(cosθ+sinθ)=±1andr(cosθ−sinθ)=±1encloseasquareandcalculatethelengthofthesidesofthissquare.

Solution

ConvertingintoCartesianformthefourlinesare

Thesefourlinesformaparallelogramandinx+y=±1,x−y=±1theadjacentlinesareperpendicularandhenceABCDisarectangle.

AlsothedistancebetweenABandCD= .

ThedistancebetweenADandBC= .

Therefore,thesefourlinesformasquare.

Example9.5.11

Findtheanglebetweenthelines and

Solution

Exercises

1. Findtheanglebetweenthelines

i. rcosθ=p,rsinθ=p1

ii.

Ans.:

2. Showthatthepoints and arecollinear.

3. Showthattheequationofanylineparallelto throughthepoleis

4. Findtheequationofthelineperpendicularto andpassingthroughthepoint

(r1,θ1).

Ans.:

9.6CIRCLE

9.6.1PolarEquationofaCircle

LetObethepoleandOXbetheinitialline.LetC(c,α)bethepolarcoordinatesofthecentreofthecircle.LetP(r,θ)beanypointonthecircle.Then∠COP=θ−α.Letabetheradiusofthecircle.

Thisisthepolarequationoftherequiredcircle.

Note9.6.1.1:Ifthepoleliesonthecircumferenceofthecirclethenc=a.Thentheequationofthecirclebecomes,

Note9.6.1.2:Theequationofthecircler=2acos(θ−α)canbewrittenintheformr=Acosθ+BsinθwhereAandBareconstants.

Note9.6.1.3:Ifthepoleliesonthecircumferenceofthecircleandtheinitiallinepassesthroughthecentreofthecirclethentheequationofthecirclebecomes,r=2acosθsinceα=0.

Note9.6.1.4:Supposetheinitiallineisatangenttothecircle.Thenc=acosecα.Therefore,fromequation(9.4)theequationofthecirclebecomes,a2=a2

cosec2α+r2–2arcosecαcos(θ–α)

(i.e.)r2–2racosecαcos(θ–α)+a2cot2α=0

Note9.6.1.5:Supposetheinitiallineisatangentandthepoleisatthepointofcontact.Inthiscaseα=90°.Theequationofthecirclebecomes,r2−2rasinθ=0(or)r=2asinθ.

9.6.2EquationoftheChordoftheCircler=2acosθontheLineJoiningthePoints(r1,θ1)and(r2,θ2).

LetPQbethechordofthecircler=2acosθ.

LetPandQbethepoints(r1,θ1)and(r2,θ2).SincethepointsPandQlieonthecircle

LettheequationofthelinePQbe

SincethepointsPandQlieonthisline


Hence,fromequation(9.6),wegetp=2acosθ1cosθ2.Hence,fromequation(9.5)theequationofthechordis2acosθ1cosθ2=

.

Note9.6.2.1:Thischordbecomesthetangentatαifθ1=θ2=α.Therefore,theequationofthetangentatαis2acos2α=rcos(θ–2α).

9.6.3EquationoftheNormalatαontheCircler=2αcosθ

SinceONisperpendiculartoPN,

Theequationofthenormalisp=rcos(θ–α).

9.6.4EquationoftheCircleontheLineJoiningthePoints(a,α)and(b,β)astheendsofaDiameter

Since∠APB=90°

Example9.6.1

Showthatthelocusofthefootoftheperpendiculardrawnfromthepoletothetangenttothecircler=2acosθisr=a(l+cosθ).

Solution

LetPbethepoint(r,α).DrawONperpendiculartothetangentatP.


rcos(θ−2α)=2acos2α

SinceONistheperpendiculardistancefromOonthelinePN,fromthenormalformofthestraightline,weget

ON=p=2acos2αLetthecoordinatesofNbe(r1,θ1),then

Example9.6.2

Showthatthefeetoftheperpendicularsfromtheoriginonthesidesofthetriangleformedbythepointswithvectorialanglesα,β,γandwhichlieonthecircler=2acosθlieonthestraightline2acosαcosβcosγ=rcos(π–α–β–γ).

Solution

Theequationofthecircleisr=2acosθ.LetthevectorialanglesofP,Q,Rbeα,β,γrespectively.TheequationsofthechordPQ,QRandRPare

LetL,MandNbethefeetoftheperpendicularsfromOonthelinesPQ,QRandRPThenfromtheaboveequations,weinferthatthecoordinatesofL,MandN

are

Thesethreepointssatisfytheequation

2acosαcosβcosγ=rcos(θ−α−β−γ)HenceL,MandNliesontheaboveline.

Example9.6.3

Showthatthestraightline touchesthecircler=2acosθifa2

B2+2alA=l2.

Solution


Theequationofthestraightlineis

Solvingthesetwoequationswegettheirpointofintersection.

Dividingbycosθ,weget

Iftheline(9.9)isatangentto(9.8)thenthetwovaluesoftanθoftheequation(9.10)areequal.Theconditionforthatisthediscriminantisequaltozero.

Exercises

1. Showthatr=Acosθ+Bsinθrepresentsacircleandfindthepolarcoordinatesofthecentre.

2. Showthattheequationofthecircleofradiusawhichtouchesthelinesθ=0, isr2–2ar(cos

θ+sinθ)+a2=0.Showthatlocusoftheequationr2−2racos2θsecθ−2a2=0consistsofastraightlineandacircle.

3. Findthepolarequationsofcirclespassingthroughthepointswhosepolarcoordinatesare

andtouchingthestraightlineθ=0.

Ans.:r2−r[(a+b)Sinθ±2bcosθ]+c2=0

4. Acirclepassesthroughthepoint(r,θ)andtouchestheinitiallineatadistancecfromthepole.

Showthatitspolarequationis

5. Showthatr2−krcos(θ−α)+kd=0representsasystemofgeneralcirclesfordifferentvaluesofk.Findthecoordinatesofthelimitingpointsandtheequationofthecommonradicalaxis.

6. Findtheequationofthecirclewholecentreis andradiusis2.

Ans.:

7. Findthecentreandradiusofthecircler2–10rcosθ+9=0.Ans.:(5,0);4

8. Provethattheequationtothecircledescribedonthelinejoiningthepoints and as

diameteris

9. Findtheconditionthattheline maybea

i. tangentii. anormaltothecircler=2cosθ.

10. Findtheequationofthecirclewhichtouchestheinitialline,thevectorialangleofthecentrebeingαandtheradiusofthecirclea.

11. Acirclepassesthroughthepoint(r1,θ1)andtouchestheinitiallineatadistancecfromthepole.Showthatitspolarequationis

9.7POLAREQUATIONOFACONIC

Earlierwedefinedparabola,ellipseandhyperbolaintermsoffocusdirectrix.Nowletusshowthatitispossibletogiveamoreunifiedtreatmentofallthesethreetypesofconicusingpolarcoordinates.Furthermore,ifweplacethefocusattheoriginthenaconicsectionhassimplepolarequation.LetSbeafixedpoint(calledthefocus)andXM,afixedstraightline(called

thedirectrix)inaplane.Letebeafixedpositivenumber(calledthe

eccentricity).ThenthesetofallpointsPintheplanesuchthat iscalleda

conicsection.Theconicis

i. anellipseife<1.ii. aparabolaife=1.iii. ahyperbolaife>1.

9.7.1PolarEquationofaConic

LetSbefocusandXMbethedirectrix.DrawSXperpendiculartothedirectrix.LetSbethepoleandSXbetheinitialline.LetP(r,θ)beanypointontheconic;thenSP=r,∠XSP=θ.DrawPMperpendiculartothedirectrixandPNperpendiculartotheinitialline.

LetLSL′bethedoubleordinatethroughthefocus(latusrectum).Thefocusdirectrixpropertyis

Thisistherequiredpolarequationoftheconic.

Note9.7.1.1:Iftheaxisoftheconicisinclinedatanangleαtotheinitialline

thenthepolarequationofconicis

Totracetheconic,

cosθisaperiodicfunctionofperiod2π.Therefore,totracetheconicitisenoughifweconsiderthevariationofθfrom

–πtoπ.Sincecos(–θ)=cosθthecurveissymmetricalabouttheinitialline.Henceitisenoughifwestudythevariationofθfrom0toπ.Letusdiscussthevariouscasesfordifferentvaluesofθ.Case1:Lete=0.Inthiscase,theconicbecomesr=lwhichisacircleofradiuslwithitscentreatthepole.

Case2:Lete=1.Inthiscase,theequationoftheconicbecomes, Whenθ

variesfrom0toπ,1+cosθvariesfrom2to0.

and variesfrom to∞

Theconicinthiscaseisaparabolaandisshownbelow.

Case3:Lete<1.

Asθvariesfrom0toπ,1+ecosθdecreasesfrom1+eto1–e.rincreasesfrom to

Thecurveisclearlyclosedandissymmetricalabouttheinitialline.Theconicisanellipse.

Case4:Lete>1.

Asθvariesfrom0to ,1+ecosθdecreasesfrom(1+e)to1andhencerincreasesfrom tol.

Asθvariesfrom toπ,1+ecosθdecreasesfrom1to(1–e).

Therefore,thereexistsanangleαsuchthat <α<πatwhich1+ecosθ>0.(i.e.)

Hence,asθvariesfrom toα,rincreasesfrom1to∞.Asθvariesfromαtoπ,1+ecosθremains

negativeandvariesfrom0to(1−e).

rvariesfromto−∞to .

Theconicisshownaboveandisahyperbola.

9.7.2EquationtotheDirectrixCorrespondingtothePole

LetQbeanypointonthedirectrix.Letitscoordinatesbe(r,θ).ThenSX=rcosθ

or Sincethisistrueforallpoints(r,θ)onthedirectrix,the

polarequationofthedirectrixis

Note9.7.2.1:Theequationofthedirectrixoftheconic is

.

Thepolarequationoftheconicfordifferentformofdirectrixisgivenbelow.

Note9.7.2.2:Theaboveconicisanellipseife<1,parabolaife=1andhyperbolaife>1.

9.7.3EquationtotheDirectrixCorrespondingtoFocusOtherthanthePole

Let(r,θ)bethecoordinatesofapointonthedirectrixX′M′.

Then

But

Thisistherequiredequationoftheotherdirectrix.

9.7.4EquationofChordJoiningthePointswhoseVectorialAnglesareα−β

andα+βontheConic

Lettheequationoftheconicbe .

LettheequationofthechordPQbe .

Thischordpassesthroughthepoint(SP,α−β)and(SQ,α+β).

Alsothesetwopointslieontheconic ,


Fromequation(9.12)and(9.14),weget

Subtracting,weget


TheequationofthechordPQis

9.7.5TangentatthePointwhoseVectorialAngleisαontheConic

Theequationofthechordjoiningthepointswithvectorialanglesα−βandα+

βis .

Thischordbecomesthetangentatαifβ=0.

Theequationoftangentatαis .

9.7.6EquationofNormalatthePointwhoseVectorialAngleisαontheConic

Theequationoftheconicis

Theequationoftangentatαontheconic is

Theequationofthelineperpendiculartothistangentis

.

IfthisperpendicularlineisnormalatP,thenitpassesthroughthepoint(SP,α).

Sincethepoint(SP,α)alsoliesontheconic wehave

Fromequation(9.17),weget .

Hence,theequationofthenormalatαis

9.7.7AsymptotesoftheConicis


Theequationofthetangentatαis

Thistangentbecomesanasymptoteifthepointofcontactisatinfinity,thatis,thepolarcoordinatesofthepointofcontactare(∞,α).Sincethispointhastosatisfytheequationoftheconic(9.18)wehavefromequation(9.18),

Theequation(9.19)canbewrittenas

Substituting and wegettheequationofthe

asymptotesas

9.7.8EquationofChordofContactofTangentsfrom(r1,θ1)totheConic

LetQRbethechordofcontactoftangentsfromP(r1,θ1).LetvectorialanglesofQandRbeα−βandα+β.TheequationofthechordQRis

TheequationsoftangentsatQandRare

Thesetwotangentsintersectat(r1,θ1).

Fromtheabovetwoequations,weget

Substitutingthisinequation(9.24),weget

Substitutingequation(9.26)in(9.21),weget

Thisistheequationofthechordofcontact.

9.7.9EquationofthePolarofanyPoint(r1,θ1)withRespecttotheconic

ThepolarofapointwithrespecttoaconicisdefinedasthelocusofthepointofintersectionoftangentsattheextremitiesofavariablechordpassingthroughthepointP(r1,θ1).

LetthetangentsatQandRintersectT.SinceQRisthechordofcontactoftangentsfromT(R,ϕ),itsequationis

SincethispassesthroughthepointP(r1,θ1)wehave

NowthelocusofthepointT(R,ϕ)ispolarofthe(r1,θ1).

Thepolarof(r1,θ1)fromequation(9.28)is

Example9.7.1

Findtheconditionthatthestraightline maybeatangenttothe

conic

Solution

Lettheline touchestheconicatthepoint(r,α).

Thentheequationoftangentat(r,α)is

However,theequationoftangentisgivenas

Equations(9.30)and(9.31)representthesameline.

Identifyingequations(9.30)and(9.31),weget

Squaringandadding,weget(A−e)2+B2=1Thisistherequiredcondition.

Example9.7.2

Showthatinaconic,semilatusrectumistheharmonicmeanbetweenthesegmentsofafocalchord.

Solution

LetPQbeafocalchordoftheconic LetPandQhavethepolar

coordinates(SP,α)and(SQ,α+π).

SincePandQlieontheconic .

Wehave

Addingequations(9.32)and(9.33)

SP,l,SQareinHP(i.e.)listheHMbetweenSPandSQ.

Example9.7.3

Showthatinanyconicthesumofthereciprocalsoftwoperpendicularfocalchordsisaconstant.

Solution

LetPP′andQQ′beperpendicularfocalchordsoftheconic

LetPbethepoint(SP,α).ThevectorialanglesofQ,P′,Q′are

.

SincethepointsP,P′,Q,Q′lieontheconic,

wehave

Example9.7.4

IfachordPQofaconicwhoseeccentricityeandthesemilatusrectuml

subtendsarightangleatthefocusSPthenprovethat

Solution

Lettheequationoftheconicbe .LetthevectorialangleofPbeα.

ThevectorialangleofQis .

SincePandQlieontheconic,

Similarly,


Example9.7.5

LetPSQandPS′RbetwochordsofanellipsethroughthefociSandS′.Show

that isaconstant.

Solution

LetthevectorialangleofPbeα.ThenthevectorialangleofQisα+π.SinceP

andQlieontheconic

Similarly,consideringtheotherfocalchordPS′R

Multiplyequation(9.37)by ,weget

Similarlyfromequation(9.38),weget


Example9.7.6

Provethattheperpendicularfocalchordsofarectangularhyperbolaareequal.

Solution

LetPSP′andQSQ′andbetwoperpendicularfocalchordsofarectangularhyperbola.ThenthevectorialanglesofPandP′areα.

SinceP′liesontheotherbranchofthehyperbola,thepolarequationoftheconic

is

Similarly,

Fromequations(9.41)and(9.42),wegetPP′=QQ′.Thatis,inaRH,perpendicularfocalchordsareofequallength.

Example9.7.7

ThetangentstoaconicatPandQmeetatT.ShowthatifSisafocusthenSTbisects∠PSQ.

Solution

Lettheequationoftheconicbe TheequationofthetangentatP

withvectorialangleαis

TheequationofthetangentatQwithvectorialangle

Atthepointofintersectionofthesetwotangents,

Example9.7.8

IfthetangentsattheextremitiesofafocalchordthroughthefocusSoftheconic

meettheaxisthroughSinTandT′showthat

Solution

LetPSQbeafocalchord.LetthevectorialanglesofPandQbeαandα+π.ThentheequationsoftangentsatPandQare

Whenthetangentsmeettheaxis,atthosepointsθ=0.

Example9.7.9

Ifachordofaconic subtendsanangle2αatthefocusthenshowthat

thelocusofthepointwhereitmeetstheinternalbisectoroftheangleis

Solution

LetPQbeachordoftheconic subtendinganangle2αatthefocus.

LettheinternalbisectorofPSQmeetsPQatT.LetthevectorialanglesofPandQbeβ−αandβ+α.LetthepolarcoordinatesofTbe(r1,β).


ThispassesthroughthepointT(r1,β).

Thelocusof(r1,β)is

Example9.7.10

ThetangentsattwopointPandQoftheconicmeetinTandPQsubtendsa

constantangle2αatthefocus.Showthat isaconstant.

Solution

Lettheequationoftheconicbe LetthevectorialanglesofPandQ

beβ−αandβ+α.

SincethepointsPandQlieontheconic,

AlsotheequationofchordPQis

PQisalsothepolarofthepointTandsoitsequationis

Identifyingequations(9.48)and(9.49),weget

Fromequations(9.46)and(9.47)and(9.50),weget

Example9.7.11

Ifafocalchordofanellipsemakesanangleαwiththemajoraxisthenshowthat

theanglebetweenthetangentsatitsextremitiesis

Solution

Lettheequationoftheconicbe


TheequationofthetangentatQis

Transformingintocartesiancoordinatesbytakingx=rcosθ,y=sinθEquations(9.53)and(9.54)becomes,

Theslopesofthetangentsare

Ifθistheanglebetweenthetangentsthen

Theacuteanglebetweenthetangentsisgivenby

Example9.7.12

AfocalchordSPofanellipseisinclinedatanangleαtothemajoraxis.ProvethattheperpendicularfromthefocusonthetangentatPmakeswiththeaxisan

angle

Solution


TheequationoftangentatPis

TheequationoftheperpendicularlinetothetangentatPis

Iftheperpendicularpassesthroughthefocusthenk=0

Example9.7.13

i. IfAcirclepassingthroughthefocusofaconicwhoselatusrectumis2lmeetstheconicinfour

pointswhosedistancesfromthefocusare,r1,r2,r3,r4thenprovethat

ii. AcircleofgivenradiuspassingthroughthefocusSofagivenconicintersectsinA,b,CandD.ShowthatSA·SB·SC·SDisaconstant.

Solution

Lettheequationsoftheconicbe

Letabetheradiusofthecircleandαbetheanglethediametermakeswiththeinitialline.Thentheequationofthecircleis

Eliminatingθbetweenequations(9.59)and(9.60)wegetanequationwhoserootsarethedistancesofthepointofintersectionfromthefocus.Formequation(9.60),wegetr=2α(cosθcosα+Sinθsinα).


Estimatingθ,weget

Dividingbyr4andrewritingtheequationinpowerof weget

Ifr1,r2,r3,r4arethedistancesofthepointsofintersectionfromthefocusthen

aretherootsoftheaboveequation.

Formequation(9.61),weget

(i.e.)SA·SB·SC·SDisaconstant.

Example9.7.14

Ifachordoftheconic subtendsaconstantangle2βatthepolethen

showthatthelocusofthefootoftheperpendicularfromthepoletothechord(e2

−sec2β)r2−elrcosθ+l2=0.

Solution

LetthevectorialanglesofPandQbeα−βandα+β.


Theequationofthelineperpendiculartothischordis

ThislinepassesthroughthefocusSandsok=0.



Squaringandadding(9.66)and(9.67),weget

Example9.7.15

Avariablechordofconicsubtendsaconstantangle2βatthefocusoftheconic

Showthatthechordtouchesanotherconichavingthesamefocus

anddirectrix.Showalsothatthelocusofpolesofsuchchordsoftheconicisalsoasimilarconic.

Solution

LetPQbeachordoftheconic subtendingaconstantangle2αatthe

focus.

LetT(r1,θ)bethepointofintersectionoftangentsatPandQ.ThenPQisthepolarofTandTisthepoleofPQ.LetthevectorialanglesofPandQbeα−βandα+β.ThentheequationofchordPQis

whereL=ecosβandE=esecθ.ThislineisatangenttotheconicC′.

Thisequationhasthesamefocusas Hence,theconicC′hasthesame

focusandthesameinitiallineasC.Forthegivenconic .

FortheconicC′,

∴SX=SX′.HenceX′coincideswithX.Hence,boththeconicshavethesamefocusandthesamedirectrix.Theequation

oftangentsatPandQare and

ThesetwotangentsintersectatT(r1,θ1)


Substitutingθ1=αinequation(9.68),weget

Thelocusof(r1,θ1)is

Thelocusofpolesistheconichavingthesamefocusandsamedirectrixasthegivenconic.

Example9.7.16

Showthatthelocusofthepointofintersectionoftangentsattheextremitiesofavariablefocalchordisthecorrespondingdirectrix.

Solution


Theequationoftangentatαis

Theequationoftangentatα+πis

Let(r1,θ1)bethepointofintersectionofthesetwotangents.Then,

Addingthesetwoequations,weget

Therefore,thelocusisthecorrespondingdirectrix

Example9.7.17

Showthatthelocusofthepointofintersectionofperpendiculartangentstoaconicisacircleorastraightline.

Solution


LetPandQbethepointsontheconicwhosevectorialanglesareαandβ.TheequationsoftangentsatPandQare

Let(r1,θ1)bethepointofintersectionoftangentsatPandQ.Then


Butα=βisnotpossible.

Formequation(9.73),weget

Expandingequations(9.70)and(9.71),weget

Sincethesetwolinesareperpendicular,wehave

Substitutingfor and ,weget

Therefore,thelocusof(r1,θ1)is(1−e2)r2+2elrcosθ−2l2=0.

Example9.7.18

Provethatpointsontheconic whosevectorialanglesareαandβ,

respectively,willbetheextremitiesofadiameterif

Solution


Letα,βbetheextremitiesofadiameteroftheconic .Thenthe

tangentsatαandβareparallelandhencetheirslopesareequal.Theequationoftangentatαis

Theslopeofthistangentis−

Sincetangentsatαandβareparallel,

Example9.7.19

Ifanormalisdrawnatoneextremityofthelatusrectum,provethatthedistance

fromthefocusoftheotherpointinwhichitmeetstheconicis

Solution


Theequationofthenormalat is

Solvingequations(9.75)and(9.76),wegettheirpointofintersection

Ifcosθ=0then .ThiscorrespondstothepointL.

Attheotherendofthenormal

Substitutinginequation(9.75)weget,

Example9.7.20

IfthetangentsatthepointsPandQonaconicintersectinTandthechordPQmeetsthedirectrixatRthenprovethattheangleTSRisarightangle.

Solution

LetthevectorialanglesofPandQbeαandβ,respectively.LetthetangentsatPandQmeetatT(r1,θ1).

TheequationoftangentsatPandQare and

Sincethesetangentsmeetat(r1,θ1),wehave and

θ1–α=±(θ2−β)whichimplies

LetθbethevectorialangleofR.TheequationofchordPQis

Theequationofthedirectrix

Subtracting,wegetsec

Example9.7.21

AchordPQofaconicsubtendsaconstant2γatthefocusSandtangentsatPand

QmeetinT.Provethat

Solution


LetthevectorialanglesofPandQbeαandβ,respectively.

Sincethesepointslieontheconic, and

IfthetangentsatPandQintersectat(r1,θ1)then

SincePQsubtendsanangle2γatS,

Exercises

1. IfPSP′andQSQ′aretwofocalchordsofaconiccuttingeachotheratrightanglesthenprovethat

=aconstant.

2. Iftwoconicshaveacommonfocusthenshowthattwooftheircommonchordspassthroughthepointofintersectionoftheirdirectrices.

3. Showthat and representthesameconic.

4. InaparabolawithfocusS,thetangentsatanypointsPandQonitmeetatT.Provethat

i. SP·SQ=ST2

ii. ThetrianglesSPTandSQTaresimilar.5. IfSbethefocus,PandQbetwopointsonaconicsuchthattheanglePSQisconstant,provethat

thelocusofthepointofintersectionofthetangentsatPandQisaconicsectionwhosefocusisS.

6. Ifthecircler+2acosθ=0cutstheconic infourrealpointsfindtheequation

inrwhichdeterminesthedistancesofthesepointsfromthepole.Also,showthatiftheiralgebraic

sumequals2aandtheeccentricityoftheconicis2cosα.

7. Provethatthetwoconics and toucheachotherif

8. P,QandRarethreepointsontheconic TangentsatQmeetsSPandSRinMandN

sothatSM=AN=lwhereSisthefocus.ProvethatthechordPQtouchestheconic.9. Provethattheportionofthetangentinterceptedbetweenthecurveanddirectrixsubtendsaright

angleatthefocus.10. Provethatthelocusofthemiddlepointsofasystemoffocalchordsofaconicsectionisaconic

sectionwhichisaparabola,ellipseorhyperbolaaccordingastheoriginalconicisaparabolaellipseorhyperbola.

11. Twoequalellipsesofeccentricityehaveonefocuscommonandareplacedwiththeiraxesatright

angles.IfPQbeacommontangentthenprovethat

12. IfthetangentsatPandQofaconicmeetatapointTandSbethefocusthenprovethatST2=SP·SQiftheconicisaparabola.

13. Aconicisdescribedhavingthesamefocusandeccentricityastheconic andthetwo

conicstouchatθ=α.Provethatthelengthofitslatusrectumis

14. Provethatthreenormalscanbedrawnfromagivenpointtoagivenparabola.Ifthenormalatα,β,

γontheconic meetatthepoint(ρ,ϕ)provethat

15. Ifthenormalsatthreepointsoftheparabola whosevectorialanglesareα,β,γmeet

inapointwhosevectorialangleisϕthenprovethat2ϕ=α+β+γ−π.

16. Ifα,β,γbethevectorialanglesofthreepointson ,thenormalatwhichare

concurrent,provethat

17. IfthenormalatPtoaconiccutstheaxisinG,provethatSN=eSP.18. IfSMandSNbeperpendicularsfromthefocusSonthetangentandnormalatanypointonthe

conic and,STtheperpendicularonMNshowthatthelocustoTisr(e2−1)=elcos

θ.

19. Showthatifthenormalsatthepointswhosevectorialanglesareθ1,θ2,θ3andθ4on

meetatthepoint(r′,ɸ)thenθ1+θ2+θ3+θ4−2ɸ=(2n+1)π.20. Provethatthechordsofarectangularhyperbolawhichsubtendarightangleatafocustoucha

fixedparabola.21. Ifthetangentatanypointofanellipsemakeanangleawithitsmajoraxisandanangleβwiththe

focalradiustothepointofcontactthenshowthatecosα=cosβ

Ans.:A2+B2−2e(Acosγ+Bsinγ)+e2−1=0

22. Provethattheexterioranglebetweenanytwotangentstoaparabolaisequaltohalfthedifferenceofthevectorialanglesoftheirpointsofcontact.

23. Findtheconditionthatthestraightline maybeatangenttotheconic

24. Findthelocusofpolesofchordswhichsubtendaconstantangleatthefocus.25. Provethatifthechordsofaconicsubtendaconstantangleatthefocus,thetangentsattheendof

thechordwillmeetonafixedconicandthechordwilltouchanotherfixedconic.

26. Findthelocusofthepointofintersectionofthetangentstotheconic atPandQ,

where ,kbeingaconstant.

27. IfthetangentandnormalatanypointPofaconicmeetthetransverseaxisofTandG,

respectively,andifSbethefocusthenprovethat isaconstant.

Chapter10

TracingofCurves

10.1GENERALEQUATIONOFTHESECONDDEGREEANDTRACINGOFACONIC

Intheearlierchapters,westudiedstandardformsofaconicnamelyaparabola,ellipseandhyperbola.Inthischapter,westudytheconditionsforthegeneralequationoftheseconddegreetorepresentthedifferenttypesofconic.Inordertostudytheseproperties,weintroducethecharacteristicsofchangeoforiginandthecoordinateaxes,rotationofaxeswithoutchangingtheoriginandreducingtheseconddegreeequationwithoutxy-term.

10.2SHIFTOFORIGINWITHOUTCHANGINGTHEDIRECTIONOFAXES

LetOxandOybetwoperpendicularlinesonaplane.LetO′beapointinthexy-plane.ThroughO′,drawO′XandO′YparalleltoOxandOy,respectively.LetthecoordinatesofO′be(h,k)withrespecttotheaxesOxandOy.DrawO′LperpendiculartoOx.ThenOL=handO′L=k.

LetP(x,y)beanypointinthexy-plane.DrawPMperpendiculartoOxmeetingaxisatN.Then

10.3ROTATIONOFAXESWITHOUTCHANGINGTHEORIGIN

LetOxandOybetheoriginalcoordinateaxes.LetOxandOyberotatedthroughanangleθintheanticlockwisedirection.

LetP(x,y)beapointinthexy-plane.DrawPLperpendiculartoOx,PM

perpendiculartoOXandMNperpendiculartoLP.Then

Let(X,Y)bethecoordinatesofthepointPwithrespecttoaxesOXandOY.Then

From(10.1)and(10.2)weseethatX=xcosθ+ysinθ,Y=−xsinθ+ycosθ.

10.4REMOVALOFXY-TERM

Herewewanttotransformtheseconddegreeequationax2+2hxy+by2+22x+2fy+c=0intoaseconddegreecurvewithoutXY-term,wheretheaxesarerotatedthroughanangleθwithoutchangingtheoriginweget

IfXY-termhastobeabsentthen

Hence,byrotatingtheaxesthroughanangleθaboutOthegeneralseconddegreeexpressionwillresultintoaseconddegreeexpressionwithoutXY-terms.

10.5INVARIANTS

Wewillnowprovethattheexpressionax2+2hxy+by2willchangetoAx2+2hXY+By2if(i)a+b=A+Band(ii)ab−h2=AB−H2.

Proof:LetP(x,y)beanypointwithrespecttoaxes(ox,oy)and(X,Y)beitscoordinateswithrespectto(OX,OY).Thenx=Xcosθ−Ysinθ,y=Xsinθ+Ycosθ.Therefore,wegetx2+y2=X2+Y2

Supposeax2+2hxy+by2=AX2+2HXY+BY2.Thenax2+2hxy+by2+λ(x2+y2)=AX2+2HXY+BY2+δ(X2+Y2).IftheLHSisoftheform(px+qy)2thenitwillbechangedintotheform[p(Xcosθ−Ysinθ)+q(Xsinθ+Ycosθ)]2=(p1X+q1Y)2.

Thiswillbeaperfectsquareifh2=(a+λ)(b+λ)

RHSwillbeaperfectsquareif

Comparing(10.3)and(10.4),wegeta+b=A+B,ab−h2=AB−H2.

10.6CONDITIONSFORTHEGENERALEQUATIONOFTHESECONDDEGREETOREPRESENTACONIC

Thegeneralequationoftheseconddegree

Iftheaxesarerotatedthroughanangleθwiththeanticlockwisedirectionthen

Thentheequationtransformedwiththeseconddegreein(X,Y)is

Now,westudyseveralcasesbasedonthevaluesofAandB.Case1:Ifab–h2=0thenA=0orB=0.SupposeA=0thentheequation(10.6)takestheform

IfG=0thenthisequationwillrepresentapairofstraightlines.IfG≠0thenwehavefrom(3).

Shiftingtheorigintothepoint theaboveequationcanbewritten

intheform whichisaparabola.

Case2:Supposeab–h2≠0thenneitherA=0norB=0.Thenequation(10.6)canbewrittenas

Shiftingtheorigintothepoint theaboveequationtakestheform

IfB=0thenequation(10.8)representsaformofstraightlinesrealorimaginary.IfK≠0thenequation(10.8)canbeexpressedintheform

whichisanellipsedependingonAandB.

IfAandBareofoppositesigns,thatis,ab−h2<0thentheequation(10.9)willrepresentahyperbola.IfB=−A,thatis,a+b=0thenequation(10.9)willrepresentarectangularhyperbola.Hencewehavethefollowingconditionforthenatureoftheseconddegreeequation(10.3)torepresentindifferentforms.Theconditionsareasfollows:

1. Itwillrepresentapairofstraightlineifabc+2fgh−af2−bg2−ch2=0.2. Itwillrepresentacircleifa=bandh=0.

3. Itwillrepresentaparabolaifab−h2=0.

4. Itwillrepresentanellipseifab−h2>0.

5. Itwillrepresentahyperbolaifab−h2<0.6. Itwillrepresentarectangularhyperbolaifa+b=0.

10.7CENTREOFTHECONICGIVENBYTHEGENERALEQUATIONOFTHESECONDDEGREE

Thegeneralequationoftheseconddegreeinxandyis

Sincethisequationhasxandyterms,thecentreisnotattheorigin.Letussupposethecentreisat(x1,y1).Letusnowshifttheoriginto(x1,y1)withoutchangingthedirectionofaxes.ThenX=x+x1,Y=y+y1.Thenequation(10.10)takestheform

Sincetheoriginisshiftedtothepoint(x1,y1)withrespecttonewaxes,thecoefficientofXandYin(10.11)shouldbezero.

Solvingthesetwoequations,weget

Thenthecoordinatesofthecentreare

Ifab−h2=0,thentheconicisaparabola.

10.8EQUATIONOFTHECONICREFERREDTOTHECENTREASORIGIN

FromtheresultobtainedinSection10.7,theequationoftheconicreferredtocentreastheoriginisax2+2hxy+by2+C1=0,where,

Hence,theequationoftheconicreferredtocentreasoriginisax2+2hxy+by2+

C1=0where

Note10.8.1:Iff=ax2+2hxy+by2+2gx+2fy+c=0then

Therefore,thecoordinatesofthecentreoftheconicaregivenbysolvingthe

equations and

Example10.8.1

Findthenatureoftheconicandfinditscentre.Alsowritedowntheequationoftheconicreferredtocentreasorigin:

i. 2x2−5xy−3y2−x−4y+5=0

ii. 5x2−6xy+5y2+22x−26y+29=0

Solution

i. Given:2x2−5xy−3y2−x−4y+5=0

Here,

Therefore,theconicisahyperbola.

Thecoordinatesofthecentrearegivenby

Solvingthesetwoequations,weget

Therefore,thecoordinatesofthecentreare Theequationoftheellipsereferredto

centreasoriginis

Therefore,theequationoftheellipsereferredtothecentreis2x2−5xy−3y+7=0.ii.

Therefore,thegivenequationrepresentsanellipse.Thecoordinatesofthecentrearegivenby

Solvingtheseequationswegetthecentreas(1,2).Theequationoftheconicreferredtocentreasoriginis5x2−6xy+5y2+C1=0whereC1=gx1+fy1+c.

C1=11×(−1)−13(2)+29=−11−26+29=−8Therefore,theequationoftheellipseis5x2−6xy+5y−8=0or5x2−6xy+5y=8.

10.9LENGTHANDPOSITIONOFTHEAXESOFTHECENTRALCONICWHOSEEQUATIONIS

ax2+2hxy+by2=1

Given:

Theequationofthecircleconcentricwiththisconicandradiusris

Homogeneousingequation(10.15)withthehelpof(10.16)weget

Thetwolinesgivenbyabovehomogeneousequationwillbeconsideredonlyiftheradiusofthecircleisequaltolengthofsemi-majoraxisorsemiminoraxis.Theconditionforthatis

Thisisaquadraticequationinr2andsoithastworootsnamely For

anellipsethevalues arebothpositiveandthelengthsofthesemi-axes

are2r1and2r2.Forahyperbolaoneofthevaluesispositiveandtheotheris

negative,thatis,say ispositiveand isnegative.Thenthelengthof

transverseaxisis2r1andthelengthofconjugateaxisis

Usingequation(10.18)inequation(10.17),weget

Thentheequationsofaxesare

Theeccentricityoftheconiccanbedeterminedfromthelengthoftheaxes.

10.10AXISANDVERTEXOFTHEPARABOLAWHOSEEQUATIONISax2+2hxy+by2+2gx+2fy+c=0

Thisequationwillrepresentaparabolaifab−h2=0.Given:

Thenequation(10.20)canbeexpressedintheform

Wechooseλsuchthatand

and

areperpendiculartoeachother.

Nowequation(10.21)canbewrittenas

Sincetheaboveequationrepresentsaparabola,theaxisoftheparabolaispx+

qy+λ=0andthetangentatthevertexis andthe

lengthofthelatusrectumis where

Example10.10.1

Tracetheconic36x2+24xy+29y2−72x+126y+81=0.

Solution

Given:

Therefore,thegivenconicrepresentsanellipse.Thecoordinatesofthecentre

aregivenby and

Solvingequation(10.25)and(10.26)wegetx=2,y=−3.Therefore,thecentreoftheellipseis(2,−3).Theequationofthisellipseinstandardformis36x2+24xy+29y2+C1=0

whereC1=gx1+fy1+c.

Thelengthoftheaxesaregivenby

Solvingforr2wegetr2=9or4∴r1=3=Lengthofthesemi-majoraxisr2=2=Lengthofthesemiminoraxis

Theequationofthemajoraxisis

Theequationofminoraxisis

Referringtothecentretheequationofmajorandminoraxesare

Themajoraxis4x+3y+1=0meetstheaxesat and

Theminoraxismeetstheaxesatthepoint and(6,0).

Theconicmeetstheaxisatpointsaregivenby

whichareimaginary.

Therefore,theconicdoesnotmeetthex-axis.Similarly,bysubstitutingy=0inequation(10.24)weget29y2+126y+81=0.

Therefore,theconicintersectsy-axisinrealpoints.

Example10.10.2

Tracetheconicx2+4xy+y2−2x+2y−6=0.

Solution

ab−h2=1×1−4=−3<0.

Therefore,theconicisahyperbola.Thecoordinatesofthecentrearegivenby

(i.e.)2x+4y−2=0orx+2y−1=0

4x+2y+2=0or2x+y+1=0.Solvingthesetwoequationswegetthe

centre.Therefore,thecoordinatesofthecentreare(−1,1).Theequationoftheconicreferredtothecentreasoriginwithoutchangingthe

directionsoftheaxisisx2+4xy+y2+C1=0whereC1=gx+fy+c

Therefore,thelengthsoftheaxesaregivenby

Hence,thesemi-transverseaxisis

Thelengthofsemi-conjugateaxisis

Semi-latusrectum

Theequationofthetransverseaxisis

Theequationofconjugateaxesis

(i.e.)x+y=0

Thepointswherethehyperbolameetsthex-axisaregivenbyx2−2x−6=0.

Whenthecurvemeetsthey-axis,x=0∴y2+2y−6=0∴y=1.7or−3.7nearlyHence,thecurvepassesthroughthepoints(−1.7,0),(3.7,0),(0,1.7)and(0,

−3.7).Thecurveistracedinfollowingfigure:

Example10.10.3

Tracetheconicx2+2xy+y2−2x−1=0.

Solution

a=1,b=1,h=1,g=−1,f=0,c=−1Here,h2=abandabc+2fgh–af2−bg2−ch2≠0.Therefore,theconicisaparabola.Thegivenequationcanbewrittenas(x+y)2=2x+1.Theequationcanbewrittenas

whereλischosensuchthatx+y+λ=0and2(λ+1)x+2λy+(λ2+1)=0areperpendicular.

Nowequation(10.27)canbewrittenas

whichisoftheformy2=4ax.

Therefore,lengthsoflatusrectumoftheparabolais

Theaxisoftheparabolais

or2x+2y−1=0

Theequationofthetangentatthevertexis

Vertexoftheparabolais

Whentheparabolameetsthex-axis,y=0.

x2−2x−1=0∴x=2.4,−0.4Therefore,thepointsonthecurveare(−0.4,0)and(2.4,0).Whenthecurvemeetsthey-axis,x=0.

y2=1ory=±1Hence(0,−1)and(0,1)arepointsonthecurve.Thegraphofthecurveisgivenasfollows:

Exercises

Tracethefollowingconics:

1. 9x2+24xy+16y2−44x+108y−124=0

2. 5x2−6xy+5y2+22x−26y+29=0

3. 32x2+52xy−7y2−64x−52y−148=0

4. x2+24xy+16y2−86x+52y−139=0

5. 43x2+48xy+57y2+10x+180y+25=0

6. x2−4xy+4y2−6x−8y+1=0

7. x2+2xy+y2−4x−y+4=0

8. 5x2−2xy+5y2+2x−10y−7=0

9. 22x2−12xy+17y2−112x+92y+178=0.

Chapter11

ThreeDimension

11.1RECTANGULARCOORDINATEAXES

Tolocateapointinaplane,twonumbersarenecessary.Weknowthatanypointinthexyplanecanberepresentedasanorderedpair(a,b)ofrealnumberswhereaiscalledx-coordinateofthepointandbiscalledthey-coordinateofthepoint.Forthisreason,aplaneiscalledtwodimensional.Tolocateapointinspace,threenumbersarerequired.Anypointinspaceis

representedbyanorderedtriple(a,b,c)ofrealnumbers.Torepresentapointinspacewefirstchooseafixedpoint‘O’(calledtheorigin)andthenthreedirectedlinesthroughOwhichareperpendiculartoeachother(calledcoordinateaxes)andlabelledx-axis,y-axisasbeinghorizontalandz-axisasverticalandwetaketheorientationoftheaxes.Inordertodothis,wefirstchooseafixedpointO.Inlookingatthefigure,youcanthinkofy-andz-axesaslyingintheplaneofthepaperandx-axisascomingoutofthepapertowardsy-axis.Thedirectionofz-axisisdeterminedbytheneighbourhoodrule.Ifyoucurlthefingersofyourright-handaroundthez-axisinthedirectionofa90°counterclockwiserotationfromthepositivex-axistothepositivey-axisthenyourthumbpointsinthepositivedirectionofthez-axis.

Thethreecoordinateaxesaredeterminedbythethreecoordinateplanes.Thexy-planeistheplanethatcontainsxandy-axes,theyz-planecontainsyandz-axesandthexz-planecontainsx-andz-axes.

Thesethreecoordinateplanesdividethespaceintoeightpartscalledoctants.Thefirstoctantisdeterminedbythepositiveaxes.Lookatanybottomcornerofaroomandcallthecornerasorigin.

Thewallonyourleftisinthexz-plane.Thewallonyourrightisintheyz-plane.Thewallonthefloorisinthexy-plane.Thex-axisrunsalongthe

intersectionofthefloorandtheleftwall.They-axisrunsalongtheintersectionofthefloorandtherightwall.Thez-axisrunsupfromthefloortowardstheceilingalongtheintersectionofthetwowallsaresituatedinthefirstoctantandyoucannowimaginesevenotherroomssituatedintheothersevenoctants(threeonthesamefloorandfourthonthefloorbelow),allareconnectedbythecommonpointO.IfPisanypointinspace,andabethedirecteddistanceinthefirstoctant

fromtheyz-planetoP.Letthedirecteddistancefromthexz-planebebandletcbethedistancefromxy-planetoP.

WerepresentthepointPbytheorderedtriple(a,b,c)ofrealnumbersandwecalla,bandcthecoordinatesofP.aisthex-coordinate,bisthey-coordinateandcisthez-coordinate.Thus,tolocatethepoint(a,b,c)wecanstartattheoriginOandmoveaunitsalongx-axisthenbunitsparalleltoy-axisandcunitsparalleltothez-axisasshownintheabovefigure.

ThepointP(a,b,0)determinesarectangularboxasintheabovefigure.IfwedropaperpendicularfromPtothexy-planewegetapointC′withcoordinatesP(a,b,0)calledtheprojectionofPonthexy-plane.Similarly,B′(a,0,c)and

A′(0,b,c)aretheprojectionsofPonxz-planeandyz-plane,respectively.ThesetofallorderedtriplesofrealnumbersistheCartesianproductR×R×R={(x,y,z)|x,y,z∈R},whichisR3.Wehaveaone-to-onecorrespondencebetweenthepointsPinspaceandorderedtriples(a,b,c)inR3.Itiscalledathree-dimensionalrectangularcoordinatesystem.Wenoticethatintermsofcoordinates,thefirstoctantcanbedescribedastheset{(x,y,z)|x≥0,y≥0,z≥0}.

11.2FORMULAFORDISTANCEBETWEENTWOPOINTS

Considerarectangularbox,wherePandQaretheoppositecornersandthefacesoftheboxareparalleltothecoordinateplanes.IfA(x1,y1,z1)andB(x2,y2,z2)aretheverticesoftheboxindicatedintheabovefigure,then

|PA|=|x2–x1|,|AB|=|y2–y1|,BQ=|z2–z1|SincethetrianglesPBQandPABarebothrightangled,byPythagoras

theorem,

Example11.2.1

Findthedistancebetweenthepoints(2,1,–5)and(4,–7,6).

Solution

Thedistancebetweenthepoints(2,1,–5)and(4,–7,6)is

units

Aliter:LetP(x1,y1,z1)and(x2,y2,z2)betwopoints.DrawPA,QBperpendiculartoxy-plane.ThenthecoordinatesofAandBare(x1,y1,0)and(x2,y2,0).

(i.e.)(x1,y1)and(x2,y2)inthexy-plane.

∴AB2=(x2–x1)2+(y2–y1)2

DrawPCperpendiculartoxy-plane.PAandPBbeingperpendiculartoxy-plane,PAandQBarealsoperpendiculartoAB.

∴PABCisarectangleandsoPC=ABandPA=CB.

Fromtriangle

Example11.2.2

IfOistheoriginandPisthepoint(x,y,z)thenOP2=x2+y2+z2or

11.2.1SectionFormula

Thecoordinatesofapointthatdividesthelinejoiningthepoints(x1,y1,z1)and(x2,y2,z2)areintheratiol:m.LetR(x,y,z)dividethelinejoiningthepointsP(x1,y1,z1)andQ(x2,y2,z2)in

theratiol:m.

DrawPL,QCandRBperpendiculartothexy-plane.ThenACBisastraightlinesinceprojectionofastraightlineonaplaneisastraightline.ThroughRdrawastraightlineLRMparalleltoACBtomeetAP(produced)inAandCQinM.ThentrianglesLPRandMRQaresimilar.

However,

Therefore,from(11.1),

Similarly,bydrawingperpendicularsonxzandyzplaneswecanprovethat

Therefore,thecoordinatesofRare

Note11.2.1.1:IfR′dividesPQexternallyintheratiol:mthen,

∴Therefore,changeminto–mtogetthecoordinatesofR′,theexternalpointofdivision.Thecoordinatesofexternalpointofdivisionare

Note11.2.1.2:TofindthemidpointofPQtakel:m=1:1.

Thenthecoordinatesofmidpointare

11.3CENTROIDOFTRIANGLE

LetABCbeatrianglewithverticesA(x1,y1,z1),B(x2,y2,z2)andC(x3,y3,z3).

ThenthemidpointofBCisD

LetGbethecentroidofthetriangleABC.ThenGdividesADintheratio2:1.ThentheCoordinatesofGare

Hence,thecentroidofΔABCis

11.4CENTROIDOFTETRAHEDRON

LetOBCDbeatetrahedronwithvertices(xi,yi,zi),i=1,2,3,4.

ThecentroidofthetetrahedrondividesADintheratio3:1.Therefore,thecoordinatesofGare

11.5DIRECTIONCOSINES

Directioncosinesinthree-dimensionalcoordinategeometryplayarolesimilartoslopeintwo-dimensionalcoordinategeometry.

Definition11.5.1:Ifastraightlinemakesanglesα,βandγwiththepositivedirectionsofx-,y-andz-axesthencosα,cosβandcosγarecalledthedirectioncosinesoftheline.Thedirectionalcosinesaredenotedbyl,mandn.

∴l=cosα,m=cosβ,n=cosγ.Thedirectioncosinesofx-axisare1,0and0.Thedirectioncosinesofy-axisare0,1and0.Thedirectioncosinesofz-axisare0,0and1.

IfOistheoriginandP(x,y,z)beanypointinspaceandOP=r,thenthedirectioncosinesofthelinearelr,mr,nr.LetObetheoriginandP(x,y,z)isanypointinspace.DrawPNperpendiculartoXOYplane.DrawNL,NMparalleltoy-andx-axes.

ThenOL=x1,OM=y1,PN=z1.

Also,

Similarly,x=rcosαandz=rcosγ.ThenthecoordinatesofPare

Note11.5.2:ThedirectioncosinesofthelineOPare

Note11.5.3:IfOP=1unitthenthedirectioncosinesofthelineare(x,y,z).Thatis,thecoordinatesofthepointParethesameasthedirectioncosinesofthelineOP.

Note11.5.4:IfOP=1unitandPisthepoint(x,y,z)thenOP2=x2+y2+z2orx2+y2+z2=1.

∴l2+m2+n2=1Therefore,directioncosinessatisfythepropertycos2α+cos2β+cos2γ=1.

11.5.1DirectionRatios

Supposea,bandcarethreenumbersproportionaltol1,m1andn1(thedirectioncosinesofaline),then

Therefore,thedirectioncosinesofthelineare

wherethesamesignistakenthroughout.Herea,band

carecalledthedirectionratiosoftheline.If2,3and5arethedirectionratiosofalinethenthedirectioncosinesofthelineare

11.5.2ProjectionofaLine

TheprojectionofalinesegmentABonalineCDisthelinejoiningthefeetoftheperpendicularsfromAandBonCD.IfALmakesanangleθwiththelineCD

then whereALisparalleltoCD.

Therefore,theprojectionofABonCDisLM=ABcosθ.

Note11.5.2.1:TheprojectionofbrokenlinesAB,BCandCDonthelineCDisLM,MNandND.

∴Therefore,thesumoftheprojectionAB,BCandCDisLM+MN+ND=LP.

11.5.3DirectionCosinesoftheLineJoiningTwoGivenPoints

LetP(x1,y1,z1)andQ(x2,y2,z2)bethegivenpoints.WeeasilyseethattheprojectionofPQonx-,y-andz-axesarex2–x1,y2–y1andz2–z1.However,theprojectionsofPQonx-,y-andz-axesarealsoPQcosα,PQcosβandPQcosγ.

Inaddition, SincePQisofconstantlength,thedirectionratios

ofPQare

(x2–x1,y2–y1,z2−z1).

11.5.4AnglebetweenTwoGivenLines

Let(l1,m1,n1),(l2,m2,n2)linesnamelyABandCD.ThroughbethedirectioncosinesofthetwogivenOdrawlinesparalleltoABandCDandtakepointsPandQsuchthatOP=OQ=1unit.

SinceOPandOQareparalleltothetwogivenlinesthentheanglebetweenthetwogivenlinesisequaltotheanglebetweenthelinesOPandOQ.SinceOP=OQ=1unit,thecoordinatesofPandQare(l1,m1,n1)and(l2,m2,n2).Let

ThenPQ2=OP2+OQ2–2·OP·OQcosθ=1+1−2·1·1cosθ

Also,

From(11.2)and(11.3),

Note11.5.4.1:Ifthetwolinesareperpendicularthenθ=90°andcos90°=0.

∴from(11.3),l1l2+m1m2+n1n2=0

Note11.5.4.2:

Note11.5.4.3:

Note11.5.4.4:Ifa1,b1,c1anda2,b2,c2arethedirectionratiosofthetwolinesthen

Ifthetwolinesareparallelthensinθ=0.

Also,ifa1,b1,c1anda2,b2,c2arethedirectionratiosoftwoparallellinesthen


Example11.1

Showthatthepoints(–2,5,8),(–6,7,4)and(–3,4,4)formaright-angledtriangle.

Solution

ThegivenpointsareA(–2,5,8),B(–6,7,4)andC(−3,4,4).

SinceBC2+AC2=AB2,thetriangleisrightangled.SinceBC=AC,thetriangleisalsoisosceles.

Example11.2

Showthatthepoints(3,2,5),(2,1,3),(–1,2,1)and(0,3,3)takeninorderformaparallelogram.

Solution

LetthefourpointsbeA(3,2,5),B(2,1,3),C(–1,2,1)andD(0,3,3).Then,

SinceAB=CDandBC=AD,thefourpointsformaparallelogram.

Aliter:ThemidpointofACis(1,2,3).ThemidpointofBDis(1,2,3).Therefore,inthefigureABCD,thediagonalsbisecteachother.HenceABCD

isaparallelogram.

Example11.3

Showthatthepoints(–1,2,5),(1,2,3)and(3,2,1)arecollinear.

Solution

ThethreegivenpointsareA(–1,2,5),B(1,2,3)andC(3,2,1).

Hence,thethreegivenpointsarecollinear.

Example11.4

Showthatthepoints(3,2,2),(–1,1,3),(0,5,6)and(2,1,2)lieonaspherewhosecentreis(1,3,4).Alsofinditsradius.

Solution

LetthegivenpointsbeS(2,1,2),P(3,2,2),Q(–1,1,3),R(0,5,6)andC(1,3,4).

Therefore,thepointsP,Q,RandSlieonaspherewhosecentreisC(1,3,4)andwhoseradiusis3units.

Example11.5

Findtheratioinwhichthestraightlinejoiningthepoints(1,–3,5)and(7,2,3)isdividedbythecoordinateplanes.

Solution

LetthelinejoiningthepointsP(1,–3,5)andQ(7,2,3)bedividedbyXY,YZandZXplanesintheratiol:1,m:1andn:1,respectively.WhenthelinePQmeetstheXYplanes,theZ-coordinatesofthepointofmeetis0.

(i.e.)TheratioinwhichPQdividestheplaneYZ-planeis1:7externally.

Similarly,

SinceXZ-planedividesPQintheratio3:2internally.

Also or Therefore,XY-planedividesPQintheratio5:3

externally.

Example11.6

PandQarethepoints(3,4,12)and(1,2,2).FindthecoordinatesofthepointsinwhichthebisectoroftheanglePOQmeetsPQ.

Solution

Weknowthat

RdividesPQinternallyintheratio13:3andSdividesPQexternallyintheratio13:3.

Therefore,thecoordinatesofRare

(i.e.)

(i.e.)

SdividesPQexternallyintheratio13:3.

Therefore,thecoordinatesofSare

(i.e.)

Example11.7

Provethatthethreelineswhichjointhemidpointsofoppositeedgesofatetrahedronpassthroughthesamepointandarebisectedatthatpoint.

Solution

LetABCDbeatetrahedronwithvertices(xi,yi,zi),i=1,2,3,4.Thethreepairsofoppositeedgesare(AD,BC),(BD,AC)and(CD,AB).Let(L,N),(P,Q)and(R,S)bethemidpointsofthethreepairsofoppositeedges.ThenListhepoint

Misthepoint

ThemidpointofLMis

Bysymmetry,thisisalsothemidpointofthelinesPQandRS.Therefore,thelinesLM,PQandRSareconcernedandarebisectedatthat

point.

Example11.8

Aplanetriangleofsidesa,bandcisplacedsothatthemidpointsofthesidesareontheaxes.Showthatthelengthsl,mandninterceptedontheaxesaregivenby8l2=b2+c2–a2,8m2=c2+a2–b2and8n2=a2+b2–c2andthatthecoordinatesoftheverticesofthetriangleare(–l,m,n),(l,–m,n)and(l,m,–n).

Solution

LetD,EandFbethemidpointsofthesidesBC,CAandAB,respectively.D,EandFarethepoints(l,0,0),(0,m,0),(0,0,n),respectively.LetA,BandCbethepoints(x1,y1,z1),(x2,y2,z2)and(x3,y3,z3),respectively.Then,

Similarly,

Therefore,theverticesare(–l,m,n),(l,–m,n)and(l,m,–n).

Example11.9

Adirectedlinemakesangles60°and60°withx-andy-axes,respectively.Findtheangleitmakeswithz-axis.

Solution

Ifalinemakesanglesα,βandγwithx-,y-andz-axes,respectivelythencos2α+cos2β+cos2γ=1.

Here,α=60°,β=60°

Example11.10

Findtheacuteanglebetweenthelineswhosedirectionratiosare2,1,–2and1,1,0.

Solution

Thedirectioncosinesofthetwolinesare

Ifθistheanglebetweenthelinesthen

Example11.11

Findtheanglebetweenanytwodiagonalsofaunitcube.

Solution

ThefourdiagonalsofthecubeareOO′,AA′,BB′andCC′.ThenthedirectionratiosofOO′andAA′are(1,1,1)and(−1,1,1).ThedirectioncosinesofOO′andAA′

are and Ifθistheanglebetweenthesetwodiagonals

then

Similarly,theanglebetweenanytwodiagonalsis

Example11.12

Ifα,β,γandδaretheanglesmadebyalinewiththefourdiagonalsofacube,

provethatcos2α+cos2β+cos2γ+cos2

Solution

ThefourdiagonalsareOO′,AA′,BB′andCC′(referfiguregiveninExample11.11).Letl,m,nbethedirectioncosinesofthelinemakinganglesα,β,γandδwiththefourdiagonals.Then,

Squaringandaddingthesefourresults,weget

Example11.13

Ifl1,m1,n1andl2,m2,n2arethedirectioncosinesoftwomutuallyperpendicularlines,showthatthedirectioncosinesofthelinesperpendiculartotheabovetwolinesare

m1n2–m2n1,n1l2−l1n2andl1m2–l2m1.

Solution

Letl,mandnbethedirectioncosinesofthelineperpendiculartothetwogivenlines.Thenll1+mm1+nn1=0;ll2,+mm2+nn2=0

But andsincethetwolinesareperpendicular

Therefore,thedirectioncosinesofthelineperpendiculartothegiventwolinesarem1n2–m2n1,n1l2–n2l1,l1m2–l2m1.

Example11.14

Showthatthreeconcurrentstraightlineswithdirectioncosinesl1,m1,n1;l2,m2,n2andl3,m3,n3arecoplanarif

Solution

Letl,mandnbethedirectioncosinesofthelinewhichisperpendiculartothegiventhreelines.Ifthelinesarecoplanarthenthelinewithdirectioncosinesl,mandnisnormaltothegivencoplanarline.

Eliminatingl,mandnweget

Example11.15

Provethatthestraightlineswhosedirectioncosinesaregivenbytheequations

al+bm+cn=0andfmn+gnl+hlm=0areperpendicularif

Solution

Thedirectioncosinesoftwolinesaregivenby

From(11.5),

Substitutingin(11.6),weget

Dividingbym2,weget|

Ifl1,m1,n1andl2,m2,n2arethedirectioncosinesofthetwogivenlinesthen

and aretherootsoftheequation(11.7),then

Similarly,

Butl1l2+m1m2+n1n2=0

Dividing

Example11.16

Iftwopairofoppositeedgesofatetrahedronareatrightanglesthenshowthatthethirdpairisalsoatrightangles.

Solution

Let(OA,BC),(OB,CA)and(OC,AB)bethreepairofoppositeedges.LetObetheorigin.LetthecoordinatesofA,BandCbe(x1,y1,z1),(x2,y2,z2)and(x3,y3,z3),respectively.ThenthedirectionratiosofOAandBCarex1,y1,z1andx2–x3,y2–y3,z2–z3.SinceOAisperpendiculartoBC,weget

SinceOBisperpendiculartoACweget

Adding(11.8)and(11.9),weget

ThisshowsthatOCisperpendiculartoAB.

Example11.17

Ifl1,m1,n1;l2,m2,n2andl3,m3,n3bethedirectioncosinesofthemutuallyperpendicularlinesthenshowthatthelinewhosedirectionratiosl1+l2+l3,m1+m2+m3andn1+n2+n3makeequalangleswiththem.

Solution

Ifl1,m1,n1;l2,m2,n2andl3,m3,n3arethedirectioncosinesofthreemutually

perpendicularlines also

Letθbetheanglebetweenthelineswiththedirectioncosinesl1,m1,n1anddirectionratiosl1+l2+l3,m1+m2+m3andn1+n2+n3.Then,

Similarly,theothertwoanglesareequaltothesamevalueofθ.Therefore,thelineswiththedirectionratiosl1+l2+l3,m1+m2+m3,n1+n2+n3areequallyinclinedtothelinewithdirectioncosinesl1,m1,n1;l2,m2,n2and

l3,m3,n3.

Example11.18

Showthatthestraightlineswhosedirectioncosinesaregivenbya2l+b2m+c2n=0,mn+nl+lm=0willbeparallelifa±b±c=0.

Solution

Giventhedirectioncosinesoftwogivenlinessatisfytheequations

From(11.11),

Substitutingthisvalueofnin(11.12),weget

Dividingby

Ifl1,m1,n1andl2,m2,n2arethedirectioncosinesofthetwogivenlinesthen

aretherootsoftheequation(11.13).

Also,ifthelinesareparallelthen thentherootsoftheequation(11.13)

areequal.Theconditionforthatisthediscriminantisequaltozero.

Example11.19

Theprojectionsofalineontheaxesare3,4,12.Findthelengthanddirectioncosinesoftheline.

Solution

Let(l,m,n)bethedirectioncosinesofthelineand(x1,y1,z1)and(x2,y2,z2)betheextremitiesoftheline.Thedirectioncosinesofx-,yandz-axesare(1,0,0),(0,1,0)and(0,0,1),respectively.Theprojectionofthelineontheaxisis3.

∴x2–x1=3.Similarly,y2–y1=4,z2–z1=12

Thendirectionratiosofthelineare3,4,12.

Therefore,thedirectioncosinesofthelineare

Exercises

1. Showthatthepoints(10,7,0),(6,6,–1)and(6,9,–4)formanisoscelesright-angledtriangle.2. Showthatthepoints(2,3,5),(7,5,–1)and(4,–3,2)formanisoscelestriangle.3. Showthatthepoints(1,2,3),(2,3,1)and(3,1,2)formanequilateraltriangle.

4. Showthatthepoints(4,0,5),(2,1,3)and(1,3,2)arecollinear.5. Showthatthepoints(1,–1,1),(5,–5,4),(5,0,8)and(1,4,5)formarhombus.6. Provethatthepoints(2,–1,0),(0,–1,–1),(1,1,–3)and(3,1,–2)formtheverticesofarectangle.7. Showthatthepoints(1,2,3),(–1,2,–1),(2,3,2)and(4,7,6)formaparallelogram.8. Showthatthepoints(–4,3,6),(–5,2,2),(–8,5,2),(–7,6,6)formarhombus.9. Showthatthepoints(4,–1,2),(0,–2,3),(1,–5,–1)and(2,0,1)lieonaspherewhosecentreis

(2,–3,1)andfinditsradius.10. Findtheratioinwhichthelinejoiningpoints(2,4,5)and(3,5,–4)isdividedbythexy-plane.

Ans.:(5,4)

11. ThelinejoiningthepointsA(–2,6,4)andB(1,3,7)meetstheYOZ-planeatC.FindthecoordinatesofC.

Ans.:(0,4,6)

12. ThreeverticesofaparallelogramABCDareA(3,–4,7),B(–5,3,–2)andC(1,2,–3).FindthecoordinatesofD.

Ans.:(9,–5,6)

13. Showthatthepoints(–5,6,8),(1,8,11),(4,2,9)and(–2,0,6)aretheverticesofasquare.14. ShowthatthepointsP(3,2,–4),Q(9,8,–10)andR(5,4,–6)arecollinear.Findtheratioinwhich

RdividesPQ.Ans.:(1,2)

15. Findtheratioinwhichthecoordinateplanesdividethelinejoiningthepoints(–2,4,7)and(3,–5,8).

Ans.:7:9;4:5;–7:–8

16. Provethatthelinedrawnfromtheverticesofatetrahedrontothecentroidsoftheoppositefacesmeetinapointwhichdividesthemintheratio3:1.

17. Findthecoordinateofthecircumcentreofthetriangleformedbythepointswithvertices(1,2,1),(–2,2,–1)and(1,1,0).

Ans.:

18. AandBarethepoints(2,3,5)and(7,2,4).FindthecoordinatesofthepointswhichthebisectorsoftheanglesAOBmeetAB.

19. FindthelengthofthemedianthroughAofthetriangleA(2,–1,4),B(3,7,–6)andC(–5,0,2).Ans.:7

20. Provethatthelocusofapoint,thesumofwhosedistancesfromthepoints(a,0,0)and(–a,0,0)

isaconstant2k,isthecurve

21. Whatarethedirectioncosinesofthelinewhichisequallyinclinedtotheaxes?

Ans.:

22. Findtheanglebetweenthelineswhosedirectionratiosare(2,3,4)and(1,–2,1).

Ans.:

23. Avariablelineintwoadjacentpositionshasdirectioncosines(l,m,n),(l+δl,m+δm,n+δn).

Provethatthesmallangleδθbetweentwopositivesisgivenbyδ2θ=(δl)2+(δm)2+(δn)2.24. FindtheanglebetweenthelinesABandCD,whereA,B,CandDarethepoints(3,4,5),(4,6,3),

(–1,2,4)and(1,0,5),respectively.

Ans.:

25. Provebydirectioncosinesthepoints(1,–2,3),(2,3,–4)and(0,–7,10)arecollinear.26. Findtheanglebetweenthelineswhosedirectionratiosare(2,1,–2)and(1,–1,0).

Ans.:

27. Showthatthelinejoiningthepoints(1,2,3)and(1,5,7)isparalleltothelinejoiningthepoints(–4,3,–6)and(2,9,2).

28. P,Q,RandSarethepoints(2,3,–1),(3,5,3),(1,2,3)and(2,5,7).ShowthatPQisperpendiculartoRS.

29. Provethatthethreelineswithdirectionratios(1,–1,1),(1,–3,0)and(1,0,3)lieinaplane.

30. Showthatthelineswhosedirectioncosinesaregivenbyal+bm+cn=0andal2+bm2+cn2=0

areparallelif

31. Showthatthelineswhosedirectioncosinesaregivenbytheequationsal+vm+wn=0andal2+

bm2+cn2=0areparallelifu2(b+c)+v2(c+a)+w2(a+b)=0andperpendicularif

32. Iftheedgesofarectangularparallelepipedarea,bandc,showthattheanglebetweenthefour

diagonalsaregivenbycos-1

33. Ifinatetrahedronthesumofthesquaresofoppositeedgesisequal,showthatitspairsofoppositesidesareatrightangles.

34. Findtheanglebetweenthelineswhosedirectioncosinesaregivenbytheequations:

i. l+m+n=0andl2+m2–n2=0ii. l+m+n=0and2lm–2nl–mn=0

Ans.:(i)

(ii)

35. If(l1,m1,n1)and(l2,m2,n2)arethedirectioncosinesoftwolinesinclinedatanangleq,showthattheactualdirectioncosinesofthedirectionbetweenthelinesare

36. AB,BCarethediagonalsofadjacentfacesofarectangularboxwithcentreattheoriginOitsedgesbeingparalleltoaxes.IftheanglesAOB,BOCandCOAareθ,ϕandω,respectivelythenprovethatcosθ+cosϕ+cosω=−1.

37. Iftheprojectionsofalineontheaxesare2,3,6thenfindthelengthoftheline.Ans.:7

38. ThedistancebetweenthepointsPandQandthelengthsoftheprojectionsofPQonthe

coordinateplanesared1,d2andd3,showthat

39. Showthatthethreelinesthroughtheoriginwithdirectionratios(1,–1,7),(1,–1,0)and(1,0,3)lieonaplane.

40. Showthattheanglebetweenthelineswhosedirectioncosinesaregivenbyl+m+n=0andfmn

+gnl+hlm=0is

Chapter12

Plane

12.1INTRODUCTION

Inthree-dimensionalcoordinategeometry,firstwedefineaplaneandfromaplanewedefineastraightline.Inthischapter,wedefineaplaneandobtainitsequationindifferentforms.Wealsoderiveformulatofindtheperpendiculardistancefromagivenpointtoaplane.Also,wefindtheratioinwhichaplanedividesthelinejoiningtwogivenpoints.

Definition12.1.1:Aplaneisdefinedtobeasurfacesuchthatthelinejoininganytwopointswhollyliesonthesurface.

12.2GENERALEQUATIONOFAPLANE

Everyfirstdegreeequationinx,yandzrepresentsaplane.Considerthefirstdegreeequationinx,yandzas

wherea,b,canddareconstants.LetP(x1,y1,z1)andQ(x2,y2,z2)betwopointsonthelocusofequation(12.1).Thenthecoordinatesofthepointsthatdivide

linejoiningthesetwopointsintheratioλ:1are If

thispointliesonthelocusofequation(12.1)then

SinceP(x1,y1,z1)andQ(x2,y2,z2)aretwopointsonthelocusoftheequation(12.1)thesetwopointshavetosatisfythelocusoftheequation(12.1).

Multiplying(12.4)byλandaddingwith(12.3),weget(ax1+by1+cz1+d)+λ(ax2+by2+cz2+d)=0whichistheequation(12.2).

Therefore,thepoint liesonthelocusofequation

(12.1).Hence,thisshowsthatiftwopointsliesonthelocusofequation(12.1)then

everypointonthislineisalsoapointonthelocusofequation(12.1).Hence,theequation(12.1)representsaplaneandthuswehaveshownthateveryfirstdegreeequationinx,yandzrepresentsaplane.

12.3GENERALEQUATIONOFAPLANEPASSINGTHROUGHAGIVENPOINT

Lettheequationoftheplanepassingthroughagivenpoint(x1,y1,z1)be

since(x1,y1,z1)liesontheplane(12.5).

Subtracting(12.6)from(12.5),wegeta(x–x1)+b(y–y1)+c(z–z1)=0.Thisisthegeneralequationoftheplanepassingthroughthegivenpoint(x1,y1,z1).

12.4EQUATIONOFAPLANEININTERCEPTFORM

Lettheequationofaplanebe

Letthisplanemakeinterceptsa,bandcontheaxesofcoordinates.Ifthisplanemeetsthex-,y-andz-axesatA,BandCthentheircoordinatesare(a,0,0),(0,b,0)and(0,0,c),respectively.SincethesepointslieontheplaneAx+By+Cz+D=0,thecoordinatesofthepointshavetosatisfytheequationAx+By+Cz+D=0.

ByreplacingthevaluesofA,BandC,weget

Thisequationiscalledtheinterceptformofaplane.

12.5EQUATIONOFAPLANEINNORMALFORM

LetaplanemeetthecoordinateaxesatA,BandC.DrawONperpendiculartotheplaneABCandletON=p.LetthedirectioncosinesofONbe(cosα,cosβ,cosγ).SinceON=p,thecoordinatesofNare(pcosα,pcosβ,pcosγ).Letp(x1,y1,z1)beanypointintheplaneABC.Ifalineisperpendiculartoaplanethenitisperpendiculartoeverylinetotheplane.Therefore,ONisperpendiculartoOP.SincethecoordinatesofPandNare(x1,y1,z1)and(pcosα,pcosβ,pcosγ)thedirectionratiosofNare(x1–pcosα,y1–pcosβ,z1–pcosγ)sinceNisperpendiculartoON.

Therefore,thelocusof(x1,y1,z1)isxcosα+ycosβ+zcosγ=p.Thisequationiscalledthenormalformofaplane.

Note12.5.1:Here,thecoefficientsofx,yandzarethedirectioncosinesofnormaltotheplaneandpistheperpendiculardistancefromtheoriginontheplane.

Note12.5.2:Reductionofaplanetonormalform:theequationofplaneingeneralformis

Itsequationinnormalformis

Identifying(12.8)and(12.9),weget

SincephastobepositivewhenDispositive,wehave

12.6ANGLEBETWEENTWOPLANES

Lettheequationoftwoplanesbe

Thedirectionratiosofthenormalstotheaboveplanesare

Theanglebetweentwoplanesisdefinedtobetheanglebetweenthenormalstothetwoplanes.Letθbetheanglebetweentheplanes.

Note12.6.1:Thepositivesignofcosθgivestheacuteanglebetweentheplanesandnegativesigngivestheobtuseanglebetweentheplanes.

Note12.6.2:Iftheplanesareperpendicularthenθ=90°.

∴a1a2+b1b2+c1c2=0

Note12.6.3:Iftheplanesareparallelthendirectioncosinesofthenormalsareproportional.

Note12.6.4:Theequationofplaneparalleltoax+by+cz+d=0canbeexpressedintheformax+by+cz+k=0.

12.7PERPENDICULARDISTANCEFROMAPOINTONAPLANE

Lettheequationoftheplanebe

andP(x1,y1,z1)bethegivenpoint.WehavetofindtheperpendiculardistancefromPtotheplane.Thenormalformoftheplane(12.12)is

DrawPMperpendicularfromPtotheplane(12.12).DrawtheplanethroughPtothegivenplane(12.12).Theequationofthisplaneis

wherep1istheperpendiculardistancefromtheorigintotheplane(12.13).Thisplanepassesthrough(x1,y1,z1).

∴lx1+my1+nz1=p1DrawBNperpendiculartotheplane(12.3)meetingtheplane(12.12)atM.

ThenON=p1andOM=p.

MN=OM–ON=p–p1=p–(lx1+my1+nz1).Comparingequations(12.12)and(12.13),

Therefore,theperpendiculardistancefrom(x1,y1,z1)is= .

Aliter:

LetPMbetheperpendicularfromPontheplaneax+by+cz+d=0.LetP(x1,y1,z1)andM(x2,y2,z2)beapointontheplane(12.12).ThenQMandPMare

perpendicular.Letθbethe ThedirectionratiosofPMandPNare(a,b,c)

and(x1–x2,y1–y2,z1–z2).

Note12.7.1:Theperpendiculardistancefromtheorigintotheplaneax+by+cz

+d=0is .

12.8PLANEPASSINGTHROUGHTHREEGIVENPOINTS

Let(x1,y1,z1),(x2,y2,z2)and(x3,y3,z3)bethreegivenpointsonaplane.Lettheequationoftheplanebe

Anyplanethrough(x1,y1,z1)is

Thisplanealsopassesthrough(x2,y2,z2)and(x3,y3,z3).

Eliminatinga,bandcfrom(12.6),(12.7)and(12.18),weget

Thisistheequationoftherequiredplane.

Aliter:Lettheequationoftheplanebe

Thisplanepassesthroughthepoints(x1,y1,z1),(x2,y2,z2)and(x3,y3,z3).

Eliminatinga,b,andcfrom(12.20),(12.21)and(12.22),weget

Thisistheequationoftherequiredplane.

12.9TOFINDTHERATIOINWHICHTHEPLANEJOININGTHEPOINTS(x1,y1,z1)AND(x2,y2,z2)ISDIVIDEDBYTHEPLANEax+by+cz+d=0.

Theequationoftheplaneis

LetthelinejoiningthepointsP(x1,y1,z1)andQ(x2,y2,z2)meettheplaneatR.LetRdividedPQintheratioλ:1.ThenthecoordinatesofRare

.Thispointliesontheplanegivenby(1).

Note12.8.1:If(ax1+by1+cz1+d)and(ax2+by2+cz2+d)areofthesamesignthenλisnegative.ThenthepointRdividesPQexternallyandsothepointsPandQlieonthesamesideoftheplane.

Note12.8.2:IfP(x1,y1,z1)andtheoriginlieonthesamesideoftheplaneax+by+cz+d=0ifax1+by1+cz1+danddofthesamesign.

12.10PLANEPASSINGTHROUGHTHEINTERSECTIONOFTWOGIVENPLANES

Letthetwogivenplanesbe

Thenconsidertheequation(ax1+by1+cz1+d1)+λ(ax2+by2+cz2+d2)=0.Thisbeingthefirstdegreeequationinx,yandz,representsaplane.Let(x1,y1,z1)bethepointonthelineoftheintersectionofplanesgivenbyequations(12.24)and(12.25).Then(x1,y1,z1)liesonthetwogivenplanes.

Then,clearly(a1x1+b1y1+c1z1+d1)+λ(a2x1+b2y1+c2z2+d2)=0.Fromthisequation,weinferthatthepoint(x1,y1,z1)liesontheplanegiven

by(12.26).Similarly,everypointinthelineofintersectionoftheplanes(12.24)and(12.25)lieontheplanes(12.24)and(12.25).

Hence,equation(12.26)istheplanepassingthroughtheintersectionofthetwogivenplanes.

12.11EQUATIONOFTHEPLANESWHICHBISECTTHEANGLEBETWEENTWOGIVENPLANES

Findtheequationoftheplaneswhichbisecttheanglebetweentwogivenplanes.Letthetwogivenplanesbe

LetP(x1,y1,z1)beapointoneitherofthebisectorsoftheanglebetweenthetwogivenplanes.ThentheperpendiculardistancefromPtothetwogivenplanesareequalinmagnitude.

Bytakingthepositivesign,wegettheequationofoneofthebisectorsandbytakingthenegativesign,wegettheequationtotheotherbisector.

Note12.11.1:Wecandeterminewhichofthetwoplanesbisectstheacuteanglebetweentheplanes.Forthis,wehavetofindtheangleθbetweenthebisectorplanesandoneofthetwogivenplanes.Iftanθ<1(θ<45°),thenthebisectorplanetakenistheinternalbisectorandtheotherbisectorplaneistheexternalbisector.Iftanθ>1thenthebisectorplanetakenistheexternalbisectorandtheotherbisectorplaneistheinternalbisector.

Note12.11.2:Wecanalsodeterminetheequationoftheplanebisectingtheanglebetweentheplanesthatcontaintheorigin.Supposetheequationofthetwoplanesarea1x+b1y+c1z+d1=0anda2x+b2y+c2z+d2=0whered1andd2

arepositive.LetP(x1,y1,z1)beapointonthebisectorbetweentheanglesoftheplanescontainingtheorigin.Thend1anda1x+b1y+c1z+dareofthesign.Sinced1ispositive,a1x+b1y+c1zisalsopositive.Similarly,a2x+b2y+c2z+d2isalsopositive.Therefore,theequationoftheplanebisectingtheangle

containingtheoriginis

Theequationofbisectorplanenotcontainingtheoriginis

12.12CONDITIONFORTHEHOMOGENOUSEQUATIONOFTHESECONDDEGREETOREPRESENTAPAIROFPLANES

Theequationthatrepresentsapairofplanesbeax2+by2+cz2+2fyz+2gzx+2hxy=0.Letthetwoplanesrepresentedbytheabovehomogenousequationofthe

seconddegreeinx,yandzbelx+my+nz=0andl1x+m1y+n1z=0.Then

Comparingtheliketermsonbothsides,weget


Note12.12.1:Tofindtheanglebetweenthetwoplanes:

Letθbetheanglebetweenthetwoplanes.Then

Note12.12.2:Iftheplanesareperpendicularthenθ=90°andtheconditionforthatisa+b+c=0.


Example12.1

Thefootoftheperpendicularfromtheorigintoaplaneis(13,–4,–3).Findtheequationoftheplane.

Solution

Thelinejoiningthepoints(0,0,0)and(13,–4,–3)isnormaltotheplane.Therefore,thedirectionratiosofthenormaltotheplaneare(13,–4,–3).Theequationoftheplaneisa(x–x1)+b(y–y1)+c(z–z1)=0

Example12.2

AplanemeetsthecoordinateaxesatA,BandCsuchthatthecentroidofthetriangleisthepoint(a,b,c).Findtheequationoftheplane.

Solution

Lettheequationoftheplanebe ThenthecoordinatesofA,BandC

are(α,0,0),(0,β,0)and(0,0,γ).ThecentroidofthetriangleABCis .

Butthecentroidisgivenas(a,b,c).

Therefore,theequationoftheplaneis

Example12.3

Findtheequationoftheplanepassingthroughthepoints(2,2,1),(2,3,2)and(–1,3,1).

Solution

Theequationoftheplanepassingthroughthepoint(2,2,1)isoftheforma(x–2)+b(y–2)+c(z–1)=0.Thisplanepassesthroughthepoints(2,3,2)and(–1,3,1).

∴0a+b+c=0and–3a+b+c=0

Solvingweget

Therefore,theequationoftheplaneis1(x–2)+3(y–2)–3(z–1)=0.

∴x+3y–3z–5=0

Example12.4

Findtheequationoftheplanepassingthroughthepoint(2,–3,4)andparalleltotheplane2x–5y–7z+15=0.

Solution

Theequationoftheplaneparallelto2x–5y–7z+15=0is2x–5y–7z+k=0.Thisplanepassesthroughthepoint(2,–3,4).

∴4+15–28+k=0ork=9Hence,theequationoftherequiredplaneis2x–5y–7z+9=0.

Example12.5

Findtheequationoftheplanepassingthroughthepoint(2,2,4)andperpendiculartotheplanes2x–2y–4z–3=0and3x+y+6z–4=0.

Solution

Anyplanepassingthrough(2,2,4)isa(x–2)+b(y–2)+c(z–4)=0.Thisplaneisperpendiculartotheplanes2x–2y–4z–3=0and3x+y+6z–

4=0.

Therefore,thedirectionratiosofthenormaltotherequiredplaneare1,3,–1.Therefore,theequationoftheplaneis(x–2)+0+(z–4)=0(i.e.)(x–2)+

3(y–2)–(z–4)=0.

x+3y–z–4=0

Example12.6

Determinetheconstantsksothattheplanesx–2y+kz=0and2x+5y–z=0areatrightanglesandinthatcasefindtheplanethroughthepoint(1,–1,–1)andperpendiculartoboththegivenplanes.

Solution

Theplanesx–2y+kz=0and2x+5y–z=0areperpendicular.Therefore,2–10–k=0∴k=–8.Anyplanepassingthrough(1,–1,–1)isa(x–1)+b(y+1)+C(x+1)=0.

Thisplaneisperpendiculartotheplanesx–2y–8z=0and2x+5y–z=0.

Therefore,theequationoftherequiredplaneis14(x–1)–5(y+1)+3(z+1)=0or14x–5y+3z–16=0.

Example12.7

AvariableplaneisataconstantdistancepfromtheoriginandmeetstheaxesinA,BandC.ShowthatthelocusofthecentroidofthetetrahedronOABCisx–2+y–2+z–2=16p–2.

Solution

LettheequationoftheplaneABCbe ThenthecoordinatesofO,A,B

andCare(0,0,0),(a,0,0),(0,b,0)and(0,0,c).LetthecentroidofthetetrahedronOABCbe(x1,y1,z1).Butthecentroidof

thetetrahedronis

TheperpendiculardistancefromOandtheplaneABCisp.

Thelocusof(x1,y1,z1)isx–2+y–2+z–2=16p–2.

Example12.8

Twosystemsofrectangularaxeshavethesameorigin.Ifaplanecutsthematdistances(a,b,c)and(a1,b1,c1)respectively,fromtheorigin,provethat

Solution

Let(o,x,y,z)and(O,X,Y,Z)bethetwosystemofcoordinateaxes.Theequationoftheplanewithrespectivefirstsystemofcoordinateaxisis

Theequationofthesameplanewithrespecttothesecondsystemofcoordinateaxesis

Theperpendiculardistancefromtheorigintotheplanegivenbytheequation(12.30)is

Theperpendiculardistancefromtheorigintotheplaneisgivenbytheequation

(12.31)is

Sincetheequations(12.30)and(12.31)representthesameplanethesetwoperpendiculardistancesareequal.

Example12.9

Avariableplanepassesthroughafixedpoint(a,b,c)andmeetsthecoordinateaxesinA,BandC.Provethatthelocusofthepointofintersectionofplanes

throughA,BandCparalleltothecoordinateplanesis

Solution


Thisplanepassesthroughthepoint(a,b,c).

ThentheequationoftheplanesthroughA,BandCparalleltothecoordinateplanesarex=α,y=βandz=γ.Let(x1,y1,z1)bethepointofintersectionoftheseplanes.Thenx1=α,y1=βandz1=γ

Therefore,fromequation(12.33),weget Thelocusof(x1,y1,z1)

is

Example12.10

Avariableplanemakesinterceptsonthecoordinateaxes,thesumofwhosesquaresisconstantandisequaltok2.Provethatthelocusofthefootoftheperpendicularfromtheorigintotheplaneis(x2+y2+z2)(x–2+y–2+z–2)=k2.

Solution


wherea,bandcaretheinterceptsonthecoordinateaxes.Giventhat

LetP(x1,y1,z1)bethefootoftheperpendicularfromOonthisplane.The

directionratiosofthenormalOPare Therefore,theequationofthe

normalOPareax=by=cz.Since(x1,y1,z1)liesonthenormal,ax1=by1+cz1=t(say).


Thepoint(x1,y,z1)alsoliesontheplane.

Eliminatinga,b,cfrom(12.36)and(12.38)

Eliminatingtfrom(12.38)and(12.39)

Therefore,thelocusof(x1,y1,z1)is

Example12.11

FindtheequationoftheplanewhichcutsthecoordinateaxesatA,B,andCsuchthatthecentroidofΔABCisatthepoint(–1,–2,–4).

Solution

Lettheequationoftheplanebe ThenthecoordinatesofA,BandC

are(a,0,0),(0,b,0),(0,0,c).ThecentroidofΔABCis .Butthecentroid

isgivenas(–1,–2,–4).

∴a=–3,b=–6,c=–12

HencetheequationoftheplaneABCis

(i.e.)4x+2y+z+12=0

Example12.12

Findtheequationoftheplanepassingthroughthepoint(–1,3,2)andperpendiculartotheplanesx+2y+2z=5and3x+3y+2z=8.

Solution

Theequationoftheplanepassingthroughthepoint(–1,3,2)isA(x+1)+B(y–3)+C(z–2)=0.Thisplaneisperpendiculartotheplanesx+2y+2z=5and3x+3y+2z=8.Iftwoplanesareperpendicularthentheirnormalsareperpendicular.ThedirectionratiosofthenormaltotherequiredplaneareA,BandC.Thedirectionratiosofthenormalstothegivenplanesare1,2,2and3,3,2.

Therefore,thedirectionratiosofthenormaltotherequiredplaneare2,–4,3.Theequationoftherequiredplaneis2(x+1)–4(y–3)+3(z–2)=0(i.e.)2x–4y+3z+8=0.

Example12.13

Findtheequationoftheplanepassingthroughthepoints(9,3,6)and(2,2,1)andperpendiculartotheplane2x+6y+6z–9+0.

Solution

Anyplanepassingthroughthepoint(9,3,6)is

Theplanealsopassesthroughthepoint(2,2,1).

Theplane(12.40)isperpendiculartotheplane.

Therefore,theequationoftherequiredplaneis3(x–9)+4(y–3)–5(z–6)=0.

∴3x+4y–5z=9

Example12.14

Showthatthefollowingpoints(0,–1,0),(2,1,–1),(1,1,1)and(3,3,0)arecoplanar.

Solution

Theequationoftheplanepassingthroughthepoint(0,–1,0)isAx+B(y+1)+Cz=0.Thisplanepassesthroughthepoints(2,1,–1)and(1,1,1).

Therefore,theequationoftheplaneis4x–3(y+1)+2z=0.

∴4x–3y+2z–3=0.Substitutingx=3,y=3,z=0weget12–9–3=0whichistrue.Therefore,theplanepassesthroughthepoints(3,3,0)andhencethefour

givenpointsarecoplanar.

Example12.15

Findforwhatvaluesofλ,thepoints(0,–1,λ),(4,5,1),(3,9,4)and(–4,4,4)arecoplanar.

Solution

Theequationoftheplanepassingthroughthepoint(4,5,1)isA(x–4)+B(y–5)+C(z–1)=0.Thisplanepassesthroughthepoints(3,9,4)and(−4,4,4).

Therefore,theequationofplaneis5(x–4)–7(y–5)+11(z–1)=0.

Ifthisplanepassesthroughthepoint(0,–1,λ)then0+7+11λ+4=0.∴λ=–1

Example12.16

Avariableplanemovesinsuchawaythatthesumofthereciprocalsoftheinterceptsonthecoordinateaxesisaconstant.Showthattheplanepassesthroughafixedpoint.

Solution


Giventhatthesumofthereciprocalsoftheinterceptsisaconstant.

(12.44)–(12.45)gives

Thisplanespassesthroughthefixedpoint

Example12.17

ApointPmovesonfixedplane andtheplanethroughP

perpendiculartoOPmeetstheaxesinA,BandC.IftheplanesthroughA,BandCareparalleltothecoordinatesplanesmeetinapointthenshowthatthelocus

ofQis

Solution

Thegivenplaneis

LetPbethepoint(α,β,γ).TheplanepassesthroughP.

TheequationoftheplanenormaltoOPis

Theinterceptsmadebythisplaneonthecoordinateaxesare

Iftheseplanesmeetat(x1,y1,z1)then

Nowwehavetoeliminateα,β,γusing(12.47)and(12.49).From(12.49),

From(12.47)and(12.49),

Therefore,thelocusof(x1,y1,z1)is from(12.50)and

(12.51).

Example12.18

IffromthepointP(a,b,c)perpendicularsPL,PMbedrawntoYZ-andZX-planes,findtheequationoftheplaneOLM.

Solution

Pisthepoint(a,b,c).PLisdrawnperpendiculartoYZ-plane.Therefore,thecoordinatesofLare(0,b,0).PMisdrawnperpendiculartoZX-plane.Therefore,thecoordinatesofMare(0,0,c).WehavetofindtheequationoftheplaneOLM.Theequationoftheplanepassingthrough(0,0,0)isAx+By+Cz=0.Thisplanealsopassesthrough(0,b,c),(a,0,c).

TheequationoftheplaneOLMisbcx+cay–abz=0.

Example12.19

Showthat isthecircumcentreofthetriangleformedbythepoints(1,1,

0),(1,2,1)and(–2,2,–1).

Solution

A,BandCarethepoints(1,1,0),(1,2,1)and(–2,2,–1)andPisthepoint

ToprovethatPisthecircumcentreofthetriangleABC,wehaveto

showthat:

i. thepointsP,A,BandCarecoplanarandii. PA=PB=PC.

Theequationoftheplanethroughthepoint(1,1,0)isA(x–1)+b(y–1)+C(z–0)=0.Thisplanealsopassesthrough(1,2,1)and(–2,2,–1).

Therefore,theequationoftheplaneABCis–2(x–1)–3(y–1)+3z=0.

Substitutingthecoordinatesof weget–1+6–0–5=0whichis

true.Therefore,thepointsP,A,BandCarecoplanar.Now

Therefore,PisthecircumcentreofthetriangleABC.

Example12.20

Findtheratioinwhichthelinejoiningthepoints(2,–1,4)and(6,2,4)isdividedbytheplanex+2y+3z+5=0.

Solution

Lettheplanex+2y+3z+5=0dividethelinejoiningthepointsP(2,–1,4)andQ(6,2,4)intheratioλ:1.

Thenthepointofdivisionis

Thispointliesontheplanex+2y+3z+5=0.

Therefore,theplanedividesthelineexternallyintheratio17:27.

Example12.21

Aplanetrianglewhosesidesareoflengtha,b,andcisplacedsothatthemiddlepointsofthesidesareontheaxes.Ifα,βandγareinterceptsontheaxesthen

showthattheequationoftheplaneis where

Solution

Theequationoftheplaneis

LettheplanemeettheaxesatL,M,Nrespectively.L(α,0,0),M(0,β,0),N(0,0,γ)

(12.52)+(12.53)–(12.54)gives,

Therefore,theequationoftheplane whereα,β,γaregivenby

(12.55),(12.56)and(12.57).Let(x1,y1,z1),(x2,y2,z2)and(x3,y3,z3)betheverticesoftheΔABC.Then

Addingweget2(x1+x2+x3)=2αorx1+x2+x3=α

Similarly,

Therefore,theverticesofthetriangleare(–α,β,γ)(α,–β,γ)and(α,β,–γ).

Example12.22

Findtheanglebetweentheplanes2x–y+z=6,x+y+2z=3.

Solution

Thedirectionratiosofthenormaltotheplanesare2,–1,1and1,1,2.The

directioncosinesofthenormalare .Ifθistheangle

betweentheplanes,then

Example12.23

Provethattheplanex+2y+2z=0,2x+y–2z=0areatrightangles.

Solution

Thedirectionratiosofthenormalstotheplanesare1,2,2and2,1,–2.Ifthelinesaretobeperpendicularthena1a2+b1b2+c1c2=0.Hence,a1a2+b1b2+c1c2=2+2–4=0.Therefore,thenormalsareperpendicularandhencetheplanesare

perpendicular.

Example12.24

Findtheequationoftheplanecontainingthelineofintersectionoftheplanesx+y+z–6=0,2x+3y+4z+5=0andpassingthroughthepoint(1,1,1).

Solution

Theequationofanyplanecontainingthelineisx+y+z–6=λ(2x+3y+4z+5)=0.Ifthislinepassesthroughthepoint(1,1,1)then,1+1+1–6+λ(2+3+4+5)=0.

Therefore,theequationoftherequiredplaneis

Example12.25

Findtheequationoftheplanewhichpassesthroughtheintersectionoftheplanes2x+3y+10z–8=0,2x–3y+7z–2=0andisperpendiculartotheplane3x–2y+4z–5=0.

Solution

Theequationofanyplanepassingthroughtheintersectionoftheplanes2x+3y+10z–8=0and2x–3y+7z–2=0is2x+3y+10z–8+λ(2x–3y+7z–2)=0.Thedirectionratiosofthenormaltothisplaneare2+2λ,3–3λ,10+7λ.The

directionratiosoftheplane3x–2y+4z–5=0are3,–2,4.Sincethesetwoplanesareperpendicular,3(2+2λ)–2(3–3λ)+4(10+7λ)=0.

Therefore,therequiredplaneis2x+3y+10z–8–(2x–3y+7z–λ)=0.

Example12.26

Theplanex–2y+3z=0isrotatedthrougharightangleaboutitslineofintersectionwiththeplane2x+3y–4z+2=0.Findtheequationoftheplaneinitsnewposition.

Solution

Theplanex–2y+3z=0isrotatedaboutthelineofintersectionoftheplanes

Thenewpositionoftheplane(12.58)passesthroughthelineofintersectionofthetwogivenplanes.Therefore,itsequationis

Theplane(12.60)isperpendiculartotheplane(12.58).

Therefore,theequationoftheplane(12.58)initsnewpositionis

Example12.27

Thelinelx+my=0isrotatedaboutitslineofintersectionwiththeplanez=0throughanangleα.Provethattheequationoftheplaneis

Solution

Anyplanepassingthroughtheintersectionoflx+my=0andz=0is

Theplanelx+my+λz=0isrotatedthroughanangleαalongtheplane(12.61).

Therefore,theequationoftheplaneinitsnewpositionisgivenby

Example12.28

Findtheequationoftheplanepassingthroughthelineofintersectionoftheplanes2x−y+5z−3=0and4x+2y−z+7=0andparalleltoz-axis.

Solution

Theequationoftheplanepassingthroughthelineofintersectionofthegivenplanesis2x−y+5z−3+λ(4x+2y−z+7)=0.Iftheplaneisparalleltoz-axis,itsnormalisperpendiculartoz-axis.Thedirectionsofthenormaltotheplaneare2+4λ,−1+2λ,5−λ.Thedirectionratiosofthez-axisare0,0,1.

Hence,theequationoftherequiredplaneis(2x−y+5z−3)+5(4x+2y−z+7)=0.

Example12.29

Findthedistanceofthepoints(2,3,−5),(3,4,7)fromtheplanex+2y−2z=9andprovethatthesepointslieontheoppositesidesoftheplane.

Solution

LetthelinejoiningthepointsP(2,3,−5)andQ(3,4,7)bedividedbytheplaneintheratioλ:1.

Therefore,thepointsPandQlieontheoppositesideoftheplane.Theperpendiculardistancefrom(2,3,−5)totheplanex+2y−2z−9=0is

Theperpendiculardistancefrom(3,4,7)totheplaneis

Note12.29.1:Sincep1andp2areofoppositesignsthepointsareontheoppositesidesoftheplane.

Example12.30

Provethatthepoints(2,3,−5)and(3,4,7)lieontheoppositesidesoftheplanewhichmeetstheaxesinA,BandCsuchthatthecentroidofthetriangleA,BandCisthepoints(1,2,4).

Solution

Lettheequationoftheplanebe .ThenthecoordinatesofA,BandC

are(a,0,0),(0,b,0)and(0,0,c).Thecentroidis Butthecentroidis

givenas(1,2,4).

Therefore,theequationoftheplaneABCis

LetthelinejoiningthepointsP(2,3,−5)andQ(3,4,7)bedividedbytheplaneintheratioλ:1.

Then

Therefore,thepointslieontheoppositesidesoftheplane.

Example12.31

Findthedistancebetweentheparallelplanes2x−2y+z+3=0,4x−4y+2z+5=0.

Solution

Let(x1,y1,z1)beapointontheplane2x−2y+z+3=0

Thenthedistancebetweentheparallelplanesisequaltothedistancefrom(x1,y1,z1)totheotherplane.

Note12.31.1:Thedistancebetweentheparallelplanesax+by+cz+d=0and

ax+by+cz+d1=0is

Ondividingtheequation4x−4y+2z+5=0by2,weget

Distancebetweentheplanes=

Example12.32

Aplaneisdrawnthroughthelineofx+y=1,z=0tomakeanangle

withtheplanex+y+z=0.Provethattwosuchplanescanbedrawn.Findtheir

equation.Showthattheanglebetweentheplanesis

Solution

Theequationoftheplanethroughtheline

Thedirectionratiosofthisplaneis1,1,λ.Alsothedirectionratiosoftheplanex+y+z=0are1,1,1.Ifθistheanglebetweenthesetwoplanesthen

From(12.63),theequationsoftherequiredplanesarex+y+2z=1and5x+

5y+2z−5=0.Ifθistheanglebetweenthesetwoplanesthen

Example12.33

Findthebisectorsoftheanglesbetweentheplanes2x−y+2z+3=0,3x−2y+6z+8=0;alsofindoutwhichplanebisectstheacuteangle.

Solution

Thetwogivenplanesare

Theequationsofthebisectorsare

Letθbetheanglebetweentheplanes(12.64)and(12.66)then

Henceθ>45°.Theplane5x−y−4z−3=0bisectstheobtuseanglebetweentheplanes(12.64)and(12.65).Therefore,23x−13y+32z+45=0bisectstheacuteanglebetweenthe

planes(12.64)and(12.65).

Example12.34

Provethattheequation2x2−2y2+4z2+2yz+6zx+3xy=0representsapairof

planesandanglebetweenthemis

Solution

Herea=2,b=−2,c=4,f=1,g=3,

Now,abc+2fgh−af2−bg2−ch2=0

⇒–16+9−2+18−9=0Hence,thegivenequationrepresentsapairofplanes.Letθbetheangle

betweentheplanes.Then

Exercises

SectionA

1. IfPisthepoint(2,3,−1),findtheequationoftheplanepassingthroughPandperpendiculartoOP.

Ans.:2x+3y−z−14=0

2. Thefootoftheperpendicularfromtheorigintoaplaneis(12,−4,−3).Finditsequation.Ans.:12x−4y−3z+69=0

3. Findtheinterceptsmadebytheplane4x−3y+2z−7=0onthecoordinateaxes.

Ans.:

4. AplanemeetsthecoordinateaxesatA,BandCsuchthatthecentroidofthetriangleisthepoint

(a,b,c).Showthattheequationoftheplaneis

5. Findtheequationoftheplanethatpassesthroughthepoint(2,−3,1)andisperpendiculartothelinejoiningthepoints(3,4,−1)and(2,−1,5).

Ans.:x+5y−6z+19=0

6. OistheorginandAisthepoint(a,b,c).FindtheequationoftheplaneperpendiculartoA.

Ans.:ax+by+cz−(a2+b2+c2)=0

7. Findtheequationoftheplanepassingthroughthepoints:i. (8,−2,2),(2,1,−4),(2,4,−6)ii. (2,2,1),(2,3,2),(−1,3,0)iii. (2,3,4),(−3,5,1),(4,−1,2)

Ans.:(i)2x−2y−2z=14,(ii)2x+3y−3z−7=0,(iii)x+y−z−1=0

8. Showthatthepoints(0,−1,−1),(4,5,1),(3,9,4)and(−4,4,4)lieonaplane.9. Showthatthepoints(0,−1,0),(2,1,−1),(1,1,1)and(−3,3,0)arecoplanar.10. Findtheequationoftheplanethroughthethreepoints(2,3,4),(−3,5,1)and(4,−1,2).Alsofind

theangleswhichthenormaltotheplanemakeswiththeaxesofreference.

Ans.:

11. Findtheequationoftheplanewhichpassesthroughthepoint(2,−3,4)andisparalleltotheplane2x−5y−7z+15=0.

Ans.:2x−5y−7z+9=0

12. Findtheequationoftheplanethrough(1,3,2)andperpendiculartotheplanesx+2y+2z−5=0and3x+3y+3z−8=0.

Ans.:2x−4y+3z+8=0

13. Findtheequationoftheplanewhichpassesthroughthepoint(2,2,4)andperpendiculartotheplanes2x−2y−4z+3=0and3x+y+6z−4=0.

Ans.:x−3y−z−4=0

14. Findtheequationoftheplanewhichpassesthroughthepoints(9,3,6)and(2,2,1)andperpendiculartotheplane2x+4y+6z−9=0.

Ans.:3x+4y−5z−9=0

15. Findtheequationofthestraightlinepassingthroughthepoints(−1,1,1)and(1,−1,1)andperpendiculartotheplanex+2y+2z−5=0.

Ans.:2x+2y−3z+3=0

16. Findtheequationoftheplanewhichpassesthroughthepoints(2,3,1),(4,−5,3)andareparalleltothecoordinateaxes.

Ans.:y+4z−7=0,x−z−1=0,4x+y−11=0

17. Findtheequationoftheplanewhichpassesthepoint(1,2,3)andparallelto3x+4y−5z=0.Ans.:3x+4y−5z+4=0

18. Findtheequationoftheplanebisectingthelinejoiningthepoints(2,3,−1)and(−5,6,3)atrightangles.

Ans.:x−y−z+7=0

19. AvariableplaneisataconstantdistancepfromtheoriginandmeetstheaxesinA,BandC.ShowthatthelocusofcentroidofthetetrahedronOABCis

x−2+y−2+z−2=16p−2.20. OABCisatetrahedroninwhichOA,OBandOCaremutuallyperpendicular.Provethatthe

perpendicularfromOtothebaseABCmeetsitatitsorthocentre.21. ThroughthepointP(a,b,c)aplaneisdrawnatrightanglestoOPtomeettheaxesinA,BandC.

ProvethattheareaofthetriangleABCis wherepisthelengthofOP.

22. AplanecontainsthepointsA(−4,9,−9)andB(5,−9,6)andisperpendiculartothelinewhichjoinsBandC(4,−6,k).Obtainkandtheequationoftheplane.

Ans.:

23. Findthedistancebetweentheparallelplanes2x+y+2z−8=0and4x+2y+4z+5=0.

Ans.:

24. Findthelocusofthepoint,thesumofthesquaresofwhosedistancesfromtheplanesx+y+z=

0,x=z=0,x−2y+z=0is9.

Ans.:x2+y2+z2=9

25. Findtheequationoftheplanewhichisatadistance1unitfromtheoriginandparalleltotheplane3x+2y−z+2=0.

Ans.:

26. Theplane meetsthecoordinateaxesinA,BandC,respectively.

ShowthattheareaofthetriangleABCis

27. Showthattheequationsby+cz+d=0,cz+ax+d=0,ax+by+d=0representplanesparalleltoOX,OYandOZ,respectively.

28. Showthatthepoints(2,3,−5)and(3,4,7)lieontheoppositesidesoftheplanemeetingtheaxesinA,BandCsuchthatthecentroidofthetriangleABCisthepoint(1,2,4).

29. Findthelocusofthepointsuchthatthesumofthesquaresofitsdistancesfromtheplanesx+y+z=0andx−2y+z=0isequaltoitsdistancefromtheplanex−z=0.

Ans.:y2−2xz=0

30. Findthelocusofthepointwhosedistancefromtheoriginis7timesitsdistancefromtheplane2x+3y−6z=2.

Ans.:3x2+8y2+53z2−36yz−24zx+12xy−8x−12y+24z+14=0

31. Provethattheequationoftheplanepassingthroughthepoints(1,1,1),(1,−1,1)and(−7,−3,−5)andisparalleltoaxisofy.

32. Determinetheconstantksothattheplanesx−2y+kz=0and2x+5y−z=0areatrightangles,andinthatcasefindtheplanethroughthepoint(1,−1,−1)andperpendiculartoboththegivenplanes.

Ans.:k=−8,14x−5y+3z−16=0

33. Provethat3x–y–z+11=0istheequationoftheplanethrough(−1,6,2)andperpendiculartothejoinofthepoints(1,2,3)and(−2,3,4).

34. A,BandCarepoints(a,0,0),(0,b,0)and(0,0,c).FindtheequationoftheplanethroughBCwhichbisectsOA.BysymmetrywritedowntheequationsoftheplanethroughCAbisectingOB

andthroughABbisectingOC.Showthattheseplanespassthrough

SectionB

1. Findtheequationoftheplanethroughtheintersectionoftheplanesx+3y+6=0and3x−y−4z=0whoseperpendiculardistancefromtheoriginisunity.

Ans.:2x+y−2z+3=0,x−2y−2z−3=0

2. Findtheequationoftheplanethroughtheintersectionoftheplanesx−2y+3z+4=0and2x−3y+4z−7=0andthepoint(1,−1,1).

Ans.:9x−13y−17z−39=0

3. Findtheequationoftheplanethroughtheintersectionoftheplanesx+2y+3z+4=0and4x+3y+3z+1=0andperpendiculartotheplanex+y+z+9=0andshowthatitisperpendiculartoxz-plane.

Ans.:x−z=2

4. Findtheequationoftheplanethroughthepoint(1,−2,3)andtheintersectionoftheplanes2x−y+4z−7=0andx+2y−3z+8=0.

Ans.:17x+14y+11z+44=0

5. Findtheequationoftheplanepassingthroughtheintersectionoftheplanesx+2y+3z+4=0and4x+3y+2z+1=0andthroughthepoint(1,2,3).

Ans.:11x+4y−3z−10=0

6. Findtheequationoftheplanepassingthroughthelineofintersectionoftheplanesx−2y−z+3=0and3x+5y−2z−1=0whichisperpendiculartotheyz-plane.

Ans.:11y+z−10=0

7. Theplanex+4y−5z+2=0isrotatedthrougharightangleaboutitslineofintersectionwiththeplane3x+2y+z+1=0.Findtheequationoftheplaneinitsnewposition.

Ans.:20x+10y+12z+5=0

8. Aretheplanesgivenbytheequations3x+4y+5z+10=0and9x+12y+15z+20=0parallel?Ifsofindthedistancebetweenthem.

Ans.:

9. Findtheequationoftheplanepassingthroughthelineofintersectionoftheplanes2x−y=0and3x−y=0anperpendiculartotheplane4x+3y−3z=8.

Ans.:24x−17y+15z=0

10. Findtheequationoftheplanepassingthroughthelineofintersectionoftheplanesax+by+cz+d=0anda1x+b1y+c1z+d1=0perpendiculartoxy-plane.

Ans.:(ac1−a1c)x+(bc1−b1c)y+(dc1−d1c)z=0

11. Findtheequationoftheplanepassingthroughthelineofintersectionoftheplanes2x+3y+10z−8=0,2x−3y+7z−2=0andisperpendiculartotheplane3x−2y+4z−5=0.

Ans.:2y+z−2=0

12. Obtaintheequationoftheplanesbisectingtheanglesbetweentheplanesx+2y−2z+1=0and12x−4y+3z+5=0.Alsoshowthatthesetwoplanesareatrightangles.

Ans.:23x−38y+35z+2=049x+14y−17z+28=0

13. Findtheequationoftheplanethatbisectstheanglebetweentheplanes3x−6y−2z+5=0and4x−12y+3z−3=0whichcontaintheorigin.Doesthisplanebisecttheacuteangle?

Ans.:yes,67x+162y+47z+44=0

14. Findtheequationoftheplanethatbisectstheacuteanglebetweentheplanes3x−4y+12z−26=0andx+2y−2z−9=0.

Ans.:22x+14y+10z−195=0

15. Findtheequationoftheplanethatbisectstheobtuseanglebetweentheplanes4x+3y−5z+1=0and12x+5y−13=0.

Ans.:8x−14y−13=0

16. Showthattheoriginliesintheacuteanglebetweentheplanesx+2y−2z−9=0,4x−3y+12z+13=0.Findtheplanesbisectingtheanglebetweenthemandfindtheplanewhichbisectstheacuteangle.

Ans.:x+35y−10z−156=0

17. Findtheequationoftheplanewhichbisectstheacuteanglebetweentheplanesx+2y+2z−3=0and3x+4y+12z+1=0.

Ans.:11x+19y+31z−18=0

18. Provethattheequation2x2−6y2–12z2+18yz+2zx+xy=0representsapairofplanesand

showthattheanglebetweenthemis

19. Provethattheequation representsapairofplanes.

20. Iftheequationɸ(x,y,z)=ax2+by2+cz2+2fyz+2gzx+2hxy=0representsapairofplanes

thenprovethattheproductofthedistancesoftheplanesfrom

Chapter13

StraightLine

13.1INTRODUCTION

TheintersectionoftwoplanesP1andP2isthelocusofallthecommonpointsonboththeplanesP1andP2.Thislocusisastraightline.Anygivenlinecanbeuniquelydeterminedbyanyofthetwoplanescontainingtheline.Thus,alinecanberegardedasthelocusofthecommonpointsoftwointersectingplanes.Letusconsiderthetwoplanes

Anysetofcoordinates(x,y,z)whichsatisfythesetwoequationssimultaneouslywillrepresentapointonthelineofintersectionofthesetwoplanes.Hencethesetwoequationstakentogetherwillrepresentastraightline.Itcanbenotedthattheequationofx–axisarey=0,z=0.Theequationofthey–axisisx=0,z=0andtheequationofthez–axisisx=0,y=0.Therepresentationofthestraightlinebytheequationsax+by+cz+d=0anda1x+b1y+c1z+d1=0iscallednon–symmetricalform.Letusnowderivetheequationsofastraightlineinthesymmetricalform.

13.2EQUATIONOFASTRAIGHTLINEINSYMMETRICALFORM

LetA(x1,y1,z1)beapointonthestraightlineandP(x,y,z)beanypointonthestraightline.Letl,m,nbethedirectioncosinesofthestraightline.LetOP=r.TheprojectionsofAPonthecoordinateaxesarex−x1,y−y1,z−z1.AlsotheprojectionsofAPonthecoordinateaxesaregivenbylr,mr,nr.Thenx−x1=lr,y−y1=mrandz−z1=nr.

Theseequationsarecalledthesymmetricalformofthestraightline.

Aliter:Wenowderivetheequationsinsymmetricalformfromthevectorequationofthestraightlinepassingthroughapointandparalleltoavector.

LetAbeagivenpointonastraightlineandPbeanypointonthestraight

line.Let beavectorparalleltotheline.LetObetheoriginand

Then

But

wheretisascalar.From(13.3)and(13.4),

ThisequationistrueforallpositionsofPonthestraightlineandthereforethisisthevectorequationofthestraightline.Let

Thenfromequation(13.4),wehave

Equatingthecoefficientsof , and ,wehave

Thesearethecartesianequationsofthestraightlineinsymmetricalform.

Note13.2.1:Toexpresstheequationsofastraightlineinsymmetricalformwerequire(i)thecoordinateofapointonthelineand(ii)thedirectioncosinesofthestraightline.

Note13.2.2:Anypointonthislineis(x1+lr,y1+mr,z1+nr).Evenifl,mandnarethedirectionratiosoftheline,(x1+lr,y1+mr,z1+nr)willrepresentapointonthelinebutrwillnotbedistancebetweenthepoints(x,y,z)and(x1,y1,z1).

13.3EQUATIONSOFASTRAIGHTLINEPASSINGTHROUGHTHETWOGIVENPOINTS

LetP(x1,y1,z1)andQ(x2,y2,z2)betwogivenpoints.Thedirectionratiosofthelinearex2−x1,y2−y1,z2−z1.

Therefore,theequationsofthestraightlineare .

Aliter:LetObetheoriginandPandQbethepointsonthestraightlineandRbeanypointonthestraightline.

Then

But

Thisisthevectorequationofthestraightline.

Let

Thenfrom(13.6),weget

Equatingthecoefficientsof , and ,weget

13.4EQUATIONSOFASTRAIGHTLINEDETERMINEDBYAPAIROFPLANESINSYMMETRICALFORM

Wehavealreadyseenthatastraightlineisdeterminedbyapairofplanesax+by+cz+d=0anda1x+b1y+c1z+d1=0wenowexpresstheseequationsinsymmetricalform.Tofinditweneedtofindthedirectioncosinesofthelineandthecoordinates

ofapointontheline.Letl,m,nbethedirectioncosinesoftheline.Thislineisperpendiculartothenormaltothetwogivenplanessincethelineliesontheplane.Thedirectionratiosofthetwonormalsarea1,b1,c1anda2,b2,c2.Thedirectioncosinesofthelinearel,m,n.Sincethenormalsareperpendiculartothelinewehave,

Solvingforl,m,nweget

Therefore,thedirectionratiosofthelineare

Tofindapointontheline,letusfindthepointwherethelinemeetstheplane.z=0anda1x+b1y+d=0anda1x+b1y+d1=0.Solvingthelasttwoequations,weget

Therefore,apointonthelineis

Thentheequationsofthestraightlinesare .

Note13.4.1:Wecanalsofindthepointwherethelinemeetstheyz–planeorzx–plane.

13.5ANGLEBETWEENAPLANEANDALINE

Lettheequationoftheplanebeax+by+cz+d=0.

Lettheequationofthelinebe .

Letθbetheanglebetweentheplaneandtheline.Thedirectionratiosofthenormaltotheplanearea,b,c.Thedirectionratiosofthelinearel,m,n.Sinceθ

istheanglebetweentheplaneandtheline, istheanglebetweenthe

normaltotheplaneandtheline.

Note13.5.1:Ifthelineisparalleltotheplane,θ=0.

∴al+bm+cn=0

13.6CONDITIONFORALINETOBEPARALLELTOAPLANE


Lettheequationofthelinebe

Ifthelineisparalleltotheplanethenthenormaltotheplaneisperpendiculartotheline.Theconditionforthisis

Since(x1,y1,z1)isapointonthelineanddoesnotlieontheplanegivenby(13.10).

∴ax1+by1+cz1+d≠0Hencetheconditionsfortheline(13.11)tobeparalleltotheplane(13.10)are

al+bm+cn=0andax1+by1+cz1+d≠0.

13.7CONDITIONSFORALINETOLIEONAPLANE

Lettheequationofthelinebe


Sincethelineliesontheplane,

Sincethelineliesontheplaneeverypointonthelineisalsoapointontheplane.(x1,y1,z1)isapointonthelineandthereforeitshouldalsolieontheplanegivenby(13.14).Hence,ax1+by1+cz1+d=0.Therefore,theconditionsfortheline(13.13)tobeparalleltotheplane(13.14)areal+bm+cn=0andax1+by1+cz1+d=0.

13.8TOFINDTHELENGTHOFTHEPERPENDICULARFROMAGIVENPOINTONALINE

LetthegivenpointbeP(p,q,r)andthegivenlineQRbe

ThenL(x1,y1,z1)isapointontheline.DrawPMperpendiculartotheline.

AlsoLMistheprojectionofPLonQR.

Thenfrom(13.17),

13.9COPLANARLINES

Findtheconditionforthelines and

tobecoplanarandalsofindtheequationoftheplane

containingthesetwolines.Considerequations,


Sincetheplanescontainslines(13.20),wehave


Sincetheplanealsocontainstheline(13.21)thepoint(x2,y2,z2)liesontheplane(13.22).

Eliminatinga,b,cfromequation(13.24),(13.26)and(13.27),weget

Thisistherequiredconditionforthelines(13.20)and(13.21)tobecoplanar.Eliminatinga,b,cfromtheequation(13.23),(13.24)and(13.25),wegetthe

equationoftheplanecontainingthetwogivenlinesas

Aliter:Iftheplanesarecoplanartheymayintersect.Anypointontheline(13.20)isx1+l1r1,y1+m1r1,z1+n1r1.Anypointontheline(13.21)isx2+l2r2,y2+m2r2,z2+n2r2.Ifthetwolinesintersectthenthetwopointsarethesame.

Eliminatingr1andr2fromtheaboveequations,weget

Thisistherequiredconditionforcoplanarlines.

13.10SKEWLINES

Twonon–intersectingandnon–parallellinesarecalledskewlines.Therealsoexistsashortestdistancebetweentheskewlinesandthelineoftheshortestdistancewhichiscommonperpendiculartobothofthese.

13.10.1LengthandEquationsoftheLineoftheShortestDistance

Lettheequationoftheskewlinesbe

LetPQbethelineoftheshortestdistancebetweenlines(13.28)and(13.29).Letl,m,nbethedirectioncosinesofthelinesoftheshortestdistancePQ.

TheconditionforPQtobeperpendiculartoABandCDare

Solvingthesetwo,weget

Therefore,thedirectionratiosofthelinePQarem1n2−m2n1,n1l2−n2l1,l1m2−l2m1.Therefore,thedirectionratiosofthelineoftheSDare

l(x1,y1,z1)andm(x2,y2,z2)arepointsonthelines(13.28)and(13.29).ThenthelengthoftheSD=PQ=ProjectionofLMonPQ=(x1−x2)l+(y1−

y2)m+(z1−z2)n,wherel,m,narethedirectioncosinesofthelinePQ.

TheequationoftheplanecontainingthelinesABandPQis

TheequationoftheplanecontainingthelinesCDandPQis

Therefore,theequationofthelineoftheSDistheintersectionofthesetwoplanesanditsequationsaregivenby

Note13.10.1:Ifthelines(13.28)and(13.29)arecoplanarthentheSDbetweenthelinesiszero.Hencetheconditionforthelines(13.28)and(13.29)tobe

coplanar,from(13.30)is

Aliter:Letthevectorequationsofthetwolinesbe

wheretandsarescalars.

Ifthelines(13.33)and(13.34)arecoplanarthentheplaneisparalleltothe

vectors and .Thereby isperpendiculartotheplanecontaining and .

Alsoas and arethepointsontheplane, isalineontheplaneandis

perpendicularto .Theconditionforthisis or

Thescalarformoftheequationis .Thevector

equationoftheplanecontainingthetwolinesis or

But .

Therefore,thescalarequationoftheplaneis

Aliter:

Letthevectorequationofthetwolinesbe and

Let and , .

LetDQbetheSDbetweenthelinesABatCD.Then isperpendicularto

both and .Then isparallelto .Let and bethepositionvectorsof

pointsLandMonABandCD,respectively.

13.10.2EquationoftheLineofSD

TheequationofthelineoftheshortestdistanceistheequationofthelineofintersectionoftheplanesthroughthegivenlinesandtheSD.Theequationofthe

planecontainingtheline andtheSDPQisparallelto and

thereforeperpendicularto is

Similarlytheequationoftheplanecontainingtheline andPQis

TheequationofthelineofSDistheequationofthelineofintersectionof(13.35)and(13.36).(i.e.)

Inscalarforms,theequationofthelineare

13.11EQUATIONSOFTWONON-INTERSECTINGLINES

Wewillnowshowthattheequationsofanytwoskewlinescanbepartintotheformy=mx,z=candy=−mx,z=−c.

LetABandCDbetwoskewlines.LetLMbethecommonperpendiculartotheskewlines.LetLM=2candθbe

itsmiddlepoint.ChooseObetheoriginanddrawlinesOPandOQparallelto

ABandCD,respectively.Letthebisectorsof bechosenasaxesofxandy.

LetOEbetakenasz–axis.Let sothat .ThenthelineOP

makesangleθ, and withx−,y−,z−axes.Itsdirectioncosinesarecosα,

sinα,0.ThecoordinatesofLare(0,0,c).ABisastraightlinepassingthroughL

andparalleltoAB.TheequationsofthelineOPare ory=

xtanθ,z=c(i.e.)y=mx,z=cwherem=tanθ

ThelineOQmakesangles−θ, −θand withx-,y-,z-axes.

ThedirectioncosinesofthelineOQarecosθ,−sinθ,θ.ThecoordinatesofMare(0,0,−c).CDisastraightlinepassingthroughF

andparalleltoCD.Itsequationsare .

(i.e.)y=−mx,z=−cwherem=tanθ

Note13.11.1:AnypointonthelineABis(r1,mr1,c)andonaxis(r1,−mr1,−c).

13.12INTERSECTIONOFTHREEPLANES

Threeplanesmayintersectinalineorapoint.Letusfindtheconditionsforthreegivenplanestointersect(i)inalineand(ii)inapoint.

Lettheequationsofthreegivenplanesbe

Theequationofanyplanepassingthroughtheintersectionofplanes(13.37)and(13.38)is

Ifplanes(13.37),(13.38)and(13.39)intersectinalinethenequations(13.39)and(13.40)representthesameplaneforsamevaluesofλ.Identifyingequations(13.40)and(13.39),weget

Eliminatingλandμfromtheequationtakenthreeatatime,weget

Therefore,theconditionsforthethreeplanestointersectinalineareΔ1=0,Δ2=0,Δ3=0andΔ4=0.

Note13.12.1:Ofthesefourconditionsonlytwoareindependentsinceiftwoplaneshavetwopointsincommonthentheyshowthelinejoiningthesetwopointsshouldalsohaveincommon.Itcanbeprovedifanytwooftheseconditionsaresatisfied,thentheothertwowillalsosatisfy.

Aliter:Theequationsofthelineofintersectionof(13.37)and(13.38)aregivenby

Iftheplanes(13.37),(13.38)and(13.39)intersectinaplanethentheconditionsare(i)theline(13.41)mustbeparalleltotheplane(13.39)and(ii)thepoint

mustlieontheplanegivenby(13.39).Theconditionsfor

(13.37)is

Thecondition(ii)isgivenby

Therefore,theconditionsforplanestointersectinalineareΔ3=0andΔ4=0(ii)Conditionfortheplanetointersectatapoint:Solvingequations(13.37),(13.38)and(13.39),weget

IftheplanesintersectatapointthenΔ4≠0.HencetheconditionforaplanetointersectatapointisΔ4≠0.

Aliter:Iftheplanesmeetatapointthenthelineofintersectionofanytwoplanesisnon–paralleltothethirdplane.Letl,m,nbethedirectioncosinesoftheintersectionofplanes(13.37)and(13.38).Then

Solvingthetwoequationsforl,m,nweget,

Therefore,thedirectionratiosofthelinesareb1c2−b2c1,a2c1−a1c2,a1b2−a2b1.Alsothelineofintersectionwillnotbeparalleltothethirdplane.


13.13CONDITIONSFORTHREEGIVENPLANESTOFORMATRIANGULARPRISM

Thelineofintersectionofplanes(13.37)and(13.38)isgivenby

Thethreeplanesformatriangularprismifthelineisparalleltothethirdplane.Theconditionsforthisarethelineisnormaltotheplane(13.39)andthe

point doesnotlieontheplane(13.39).

(i.e.)Δ4=0andΔ3≠0.Thesearetherequiredconditions.


Example13.1

Findtheequationofthelinejoiningthepoints(2,3,5)and(−1,2,−4).

Solution

Thedirectionratiosofthelineare2+1,3−2,5+4(i.e.)3,1,9.Therefore,the

equationsofthelineare .

Example13.2

Findtheequationofthelinepassingthroughthepoint(3,2,−6)andperpendiculartotheplane3x−y−2z+2=0.

Solution

Thedirectionratiosofthelinearethedirectionratiosofthenormaltotheplane.Therefore,thedirectionratiosofthelineare3,−1,−2.Giventhat(4,2,−6)isapointontheplane.

Therefore,theequationsofthelineare .

Example13.3

Findtheequationsofthelinepassingthroughthepoint(1,2,3)andperpendiculartotheplanesx−2y−z+5=0andx+y+3z+6=0.

Solution

Letl,m,nbethedirectionratiosofthelineofintersectionoftheplanesx−2y−z+5=0andx+y+3z+6=0.Thenl−2m−n=0andl+m+3n=0

Sincethelinealsopassesthroughthepoint(−1,2,3),itsequationsis

Example13.4

Expressthesymmetricalformoftheequationsofthelinex+2y+z−3=0,6x+8y+3z−13=0.

Solution

Toexpresstheequationsofalineinsymmetricalformwehavetofind(i)thedirectionratiosofthelineand(ii)apointontheline.Letl,m,nbethedirectionratiosofline.Thenl+2m+n=0and6l+8m+3n

=0.

Letusfindthepointwherethelinemeetsthexy–plane(i.e.)z=0.

Therefore,theequationsofthelineare

Example13.5

Findtheperpendiculardistancefromthepoint(1,3,−1)totheline

Solution

Theequationsofthelineare

Anypointonthislineare(5r+13,−8r−8,r+31).DrawPQperpendiculartotheplane.Thedirectionratiosofthelineare(5r+12,−8r−11,r+32).SincethelinePQisperpendiculartoQR,wehave

Qisthepoint(3,8,29)andPis(1,3,−1)

Example13.6

Findtheequationofplanepassingthroughtheline andparallel

totheline .

Solution

Anyplanecontainingtheline

is

where

Alsothelineisparalleltotheplane

SolvingforA,BandCfrom(13.44)and(13.46),weget or

Therefore,theequationoftherequiredplaneis11(x−1)+2(y+1)−5(z−3)=0.

(i.e.)11x+2y−5z+6=0

Example13.7

Findtheimageofthepoint(2,3,5)ontheplane2x+y−z+2=0.

Solution

LetQbetheimageofthepointP(2,3,5)ontheplane2x+y−z+2=0.The

equationofthelinePQis

Anypointonthislineis(2r+2,r+3,−r+5).Whenthelinemeetstheplane,thispointliesontheplane2x+y−z+2=0.

Example13.8

Findtheimageoftheline intheplane2x−y+z+3=0.

Solution

Theequationsofthelineare

Anypointonthislineis(3r+1,5r+3,2r+4).Asthispointliesontheplane,

HencethecoordinatesofRare(−5,−7,0).TheequationsofthelinePLperpendiculartotheplaneare

Anypointonthislineis(2r1+1,−r1+3,r1+4).Ifthispointliesontheplane(13.48),weget2(2r1+1)−(−r1+3)+(r1+4)+3=0.

(i.e.)6r1+6=0orr1=−1.Therefore,thecoordinatesofLwherethislinemeetstheplane(13.47)are(−1,4,3).IfQ(x1,y1,z1)istheimageofPintheplane

HencetheequationsofthereflectionlineRQare .

Example13.9

Findtheequationofthestraightlinesthroughtheorigineachofwhichintersects

thestraightline andareinclinedatanangleof60°toit.

Solution

TheequationsofthelinePQare

ThepointPonthislineisP(2r+3,r+3,r).

ThedirectionratiosofOPare2r+3,r+3,r.Since ,

orr2+3r+2=0orr=−1,−2.Therefore,thecoordinatesofPandQare(1,−2,−1)and(−1,1,−2).

HencetheequationsofthelinesOPandOQare and .

Example13.10

Findthecoordinatesofthepointwherethelinegivenbyx+3y−z=6,y−z=4cutstheplane2x+2y+z=6.

Solution

Letl,m,nbethedirectioncosinesofthelinex+3y−z=6,y−z=4.Then

Therefore,thedirectionratiosofthelineare2,−1,−1.Whentheline

meetsthexy–planewhoseequationisz=0,wehavex+3y=6,y=4.Therefore,thepointwherethelinemeetsxy–planeis(−6,4,0).

Therefore,theequationsofthelineare

Anypointonthislineis(2r−6,−r+4,−r).Thispointliesontheplane2x+2y+z=0.

Hencetherequiredpointis(2,0,−4).

Example13.11

Findthedistanceofthepoint(1,−2,3)fromtheplanex−y+z=5measured

paralleltotheline

Solution

Theequationsofthelinethrough(1,−2,3)andparalleltotheline are

.Anypointonthislineis(2r+1,3r−2,−6r+3).Ifthispoint

liesontheplanex−y+z=5then(2r+1)−(3r−2)+(−6r+3)=5.

(i.e.)−7r+1=0or

Therefore,thepointPis .

Therefore,thedistancebetweenthepointsA(1,−2,3)and is

.

Example13.12

Provethattheequationofthelinethroughthepoints(a,b,c)and(a′,b′,c′)passesthroughtheoriginifaa′+bb′+cc′=pp′wherepandp′arethedistancesofthepointsfromtheorigin.

Solution

Theequationsofthelinethrough(a,b,c)and(a′,b′,c′)are

Ifthispassesthroughtheoriginthen

Letpandp′bethedistancesofthepoints(a,b,c)and(a′,b′,c′)fromtheorigin.

ByLagrange’sidentity,

Example13.13

IffromthepointP(x,y,z),PMisdrawnperpendiculartotheline andis

producedtoQsuchthatPM=MQthenshowthat

Solution

TheequationofthelineOAis .

Anypointonthislineis(lr,mr,nr).

IfMisthispointthen

ThedirectionratiosofthelineMParex−lr,y−mr,z−nr.SinceMPisperpendiculartoOA,

From(13.50),weget .

Example13.14

Reducetheequationsofthelinesx=ay+b,z=cy+dtosymmetricalformandhencefindtheconditionthatthelinebeperpendiculartothelinewhoseequationsarex=a′y+b′,z=c′y+d′.

Solution

Theline

canbeexpressedinthesymmetricalformas

Theline

insymmetricalformis

Ifthelines(13.51)and(13.52)areperpendicularthenaa′+bb′+cc′=0.Thisistherequiredcondition.

Example13.15

FindtheequationofthelinepassingthroughGperpendiculartotheplaneXYZrepresentedbytheequationlx+my+nz=pwherel2+m2+n2=1andcalculatethedistanceofGfromtheplane.

Solution

TheequationoftheplaneXYZis

wherel2+m2+n2=1.Whenthisplanemeetsthex–axis,y=0andz=0.

HenceXisthepoint .Similarly,Yis andZis .

ThecentroidofΔXYZis .

TheequationofthelinethroughGperpendiculartotheplaneXYZis

WhenthislinemeetstheYOZplane,x=0(13.56)

Then

Here,risthedistanceofGfromtheplane(13.55)sincepandl2arepositive,r=GA

Example13.16

Findtheperpendiculardistanceofangularpointsofacubefromadiagonalwhichdoesnotpassthroughtheangularpoint.

Solution

Letabethesideofthecube.BB′isadiagonalofthecubenotpassingthroughO.ThedirectionratiosofBB′area,−a,a.(i.e.)1,−1,1.Thedirectioncosinesof

BB′are .TheprojectionsofOB′onBB′

Example13.17

Provethattheequationsofthelinethroughthepoint(α,β,γ)andatrightangles

tothelines are .

Solution

Letl,m,nbethedirectioncosinesofthelineperpendiculartothetwogivenlines.Thenwehave

Therefore,thedirectionratiosofthelinearem1n2−m2n1,n1l2−n2l1,l1m2−l2m1.Thelinealsopassesthroughthepoint(x1,y1,z1).

Itsequationsare

Exercises1

1. Showthattheline isparalleltotheplane2x+3y−z+4=0.

2. Findtheequationoftheplanethroughtheline andthepoint(0,7,−7).Show

furthertheplanecontainstheline .

Ans.:x+y+z=0

3. Findtheequationoftheplanewhichpassesthroughtheline3x+5y+7z−5=0=x+y+z−3andparalleltotheline4x+y+z=0=2x−3y−5z.

Ans.:2x+4y+y+6z=2

4. Findtheequationsofthelinethroughthepoint(1,0,7)whichintersecteachofthelines

Ans.:7x−6y−z=0,9x−7y−z−2=0

5. Findtheequationoftheplanewhichpassesthroughthepoint(5,1,2)andisperpendiculartothe

line Findalsothecoordinatesofthepointinwhichthislinecutstheplane.

Ans.:x−2y−2z−1=0;(1,2,3)

6. Findtheequationoftheplanethrough(1,1,2)and(2,10,−1)andperpendiculartothestraight

line

Ans.:3x−y−7z+2=0

7. Findtheprojectionoftheline3x−y+2z=1,x+2y−z=2ontheplane3x+2y+z=0.Ans.:3x+2y+z=0,3x−8y+7z+4=0

8. Findtheprojectionofthelinex=3−6t,y=2t,z=3+2tintheplane3x+4y−5z−26=0.

Ans.:

9. Findtheequationoftheplanewhichcontainsthelineandisperpendiculartotheplanex+2y+z=12.

Ans.:9x−2y−5z+4=0

10. Findtheequationoftheplanewhichpassesthroughthez–axisandisperpendiculartotheline

.

Ans.:xcosα+ysinα=0

11. Findtheequationsoftwoplanesthroughtheoriginwhichareparalleltotheline

anddistant fromit.Showalsothatthetwoplanesareperpendicular.

Ans.:x+2y−2z=0,2x+2y+z=0

12. Findtheequationstothelineofthegreatestslopethroughthepoint(1,2,−1)intheplanex−2y+3z=0assumingthattheaxesaresoplacedthattheplane2x+3y−4z=0ishorizontal.

Ans.:

13. Assumingtheline asvertical,findtheequationofthelineofthegreatestslopeinthe

plane2x+y−5z=12andpassingthroughthepoint(2,3,−1).

Ans.:

14. Withthegivenaxesrectangulartheline isvertical.Findthedirectioncosinesofthe

lineofthegreatestslopeintheplane3x−2y+z=0andtheangleofthislinemakeswiththehorizontalplane.

Ans.:

15. Showthatthelines willbecoplanarif

16. Showthattheequationoftheplanethroughtheline andwhichisperpendiculartothe

planecontainingthelines and is∑(m−n)x=0.

17. Showthattheline and willlieinaplaneifα=βorβ=γ

orγ=α.

18. Findtheequationoftheplanepassingthroughtheline andperpendiculartothe

planex+2y+z=12.Ans.:9x−2y+5z+4=0

19. Findtheequationsofthelinethrough(3,4,0)andperpendiculartotheplane2x+4y+7z=8.

Ans.:

20. Findtheequationoftheplanepassingthroughtheline areparalleltotheline

.

Ans.:4y−3z+1=0,2x−7z+1=0,3x−2y+1=0.

21. Showthattheequationoftheplanesthroughthelinewhichbisecttheanglebetweenthelines

(wherel,m,nandl′,m′,n′aredirectioncosines)andperpendiculartothe

planecontainingthemare(l+l′)x+(m+m′)y+(n+n′)z=0.

22. Findtheequationoftheplanethroughtheline andparalleltothecoordinate

planes.Ans.:xcosθ+ysinθ=0

23. Provethattheplanethroughthepoint(α,β,γ)andthelinex=py+q=zx+risgivenby

.

24. ThelineLisgivenby .FindthedirectioncosinesoftheprojectionsofLonthe

plane2x+y−3z=4andtheequationoftheplanethroughLparalleltotheline2x+5y+3z=4,x−y−5z=6.

Ans.:

Exercises2

1. Findtheequationofthelinejoiningthepointsi. (2,3,5)and(−1,2,−4)ii. (1,−1,3)and(3,3,1)

Ans.:

2. Findtheequationsofthelinepassingthroughthepoint(3,2,−8)andisperpendiculartotheplane3x−y−2z+2=0.

Ans.:

3. Findtheequationsofthelinepassingthroughthepoint(3,1,−6)andparalleltoeachoftheplanesx+y+2z−4=0and2x−3y+z+5=0.

Ans.:

4. Findtheequationsofthelinethroughthepoint(1,2,3)andparalleltothelineofintersectionoftheplanesx−2y−z+5=0,x+y+3z−6=0.

Ans.:

5. Findthepointatwhichtheline meetstheplane2x+4y−z+1=0.

Ans.:

6. Findthecoordinatesofthepointatwhichtheline meetstheplane2x+3y+z=0.

Ans.:

7. Provethattheequationsofthenormaltotheplaneax+by+cz+d=0throughthepoint(α,β,γ)

are

8. Expressinsymmetricalformthefollowinglines:i. x+2y+z=3,6x+8y+3z=13ii. x−2y+3z−4=0,2x−3y+4z−5=0iii. x+3y−z−15=0,5x−2y+4z+8=0

Ans.:

9. Provethatthelines3x+2y+z−5=0,x+y−2z−3=0and8x−4y−4z=0,7x+10y−8z=0areatrightangles.

10. Provethatthelinesx−4y+2z=0,4x−y−3z=0andx+3y−5z+9=0,7x−5y−z+7=0areparallel.

11. Findthepointatwhichtheperpendicularfromtheoriginonthelinejoiningthepoints(−9,4,5)and(11,0,−1)meetsit.

Ans.:(1,2,2).

12. Provethatthelines2x+3y−4z=0,3x−4y+7=0and5x−y−3z+12=0,x−7y+5z−6=0areparallel.

13. Findtheperpendicularfromthepoint(1,3,9)totheline

Ans.:21

14. Findthedistanceofthepoint(−1,−5,−10)fromthepointofintersectionoftheline

andtheplanex−y+z=5.

Ans.:13

15. Findthelengthoftheperpendicularfromthepoint(5,4,−1)totheline .

Ans.:

16. Findthefootoftheperpendicularfromthepoint(−1,11,5)totheline

Ans.:

17. Obtainthecoordinatesofthefootoftheperpendicularfromtheoriginonthelinejoiningthepoints(−9,4,5)and(11,0,−1).

18. Findtheimageofthepoint(4,5,−2)intheplanex−y+3z−4=0.Ans.:(6,3,4)

19. Findtheimageofthepoint(1,3,4)intheplane2x−y+z+3=0.Ans.:(1,0,7)

20. Findtheimageofthepoint(2,3,5)intheplane2x+y−z+2=0.

Ans.:

21. Findtheimageofthepoint(p,q,r)intheplane2x+y+z=6andhencefindtheimageoftheline

.

Ans.:

22. Findthecoordinatesofthefootoftheperpendicularfrom(1,0,2)totheline

Alsofindthelengthoftheperpendicular.

Ans.:

23. Findtheequationinsymmetricalformoftheprojectionoftheline ontheplane

x+2y+z=12.

Ans.:

24. Provethatthepointwhichtheline meetstheplane2x+35y−39z+12=0is

equidistantfromtheplanes12x−15y+16z=28and6x+6y−7z=8.

25. Findtheequationoftheprojectionofthestraightline ontheplanex+y+2z=5

insymmetricalform.

Ans.:

26. Provethattwolinesinwhichtheplanes3x−7y−5z=1and5x−13y+3z+2=0cuttheplane8x−11y+2z=0includearightangle.

27. Reducetosymmetricalformthelinegivenbytheequationsx+y+z+1=0,4x+y−2z+2=0.Hencefindtheequationoftheplanethrough(1,1,1)andperpendiculartothegivenline.

Ans.:

28. Showthatthelinex+2y−z−3=0,x+3y−z−4=0isparalleltothexz–planeandfindthecoordinatesofthepointwhereitmeetsyz–plane.

Ans.:(0,1,−1)

29. Findtheanglebetweenthelinesx−2y+z=0,x+y−z−3=0,andx+2y+z−5=0,8x+12y+5z=0.

Ans.:

30. Findtheequationoftheplanepassingthroughtheline andparalleltotheline

.

Ans.:11x+2y−5z+6=0

31. Theplane meetstheaxesinA,BandC.Findthecoordinatesoftheorthocentreofthe

ΔABC.

Ans.:

32. TheequationtoalineABare .ThroughapointP(1,2,3),PNisdrawnperpendicularto

ABandPQisdrawnparalleltotheplane2x+3y+4z=0tomeetABinQ.FindtheequationsofPNandPQandthecoordinatesofNandQ.

Ans.:

ILLUSTRATIVEEXAMPLES(COPLANARLINESANDSHORTESTDISTANCE)

Example13.18

Provethatthelines and arecoplanarandfind

theequationoftheplanecontinuingthesetwolines.

Solution

(−1,−10,1)isapointonthefirstlineand−3,8,2arethedirectionratiosofthefirstline.(−3,−1,4)isapointonthesecondlineand−4,7,1arethedirection

ratiosofthesecondline.Ifthelinesarecoplanarthen

Therefore,thetwolinesarecoplanar.Theequationoftheplanecontainingthe

linesis

Example13.19

Showthatthelines and intersect.Findthepointof

intersectionandtheequationoftheplanecontainingthesetwolines.

Solution

Thetwogivenlinesare

Anypointonthefirstlineis(−3r−1,2r+3,r−2).Anypointonthesecondlineis(r1,−3r1+7,2r1−7).Ifthetwolinesintersectthenthetwopointsareoneandthesame.

Solving(13.60)and(13.61),wegetr=−1andr1=2.Thesevaluessatisfyequation(13.59).Thepointofintersectionis(2,1,−3).Theequationoftheplanecontainingthetwolinesis

Example13.20

Showthatthelines andx+2y+3z−8=0,2x+3y+4z−11=0

arecoplanar.Findtheequationoftheplanecontainingthesetwolines.

Solution

Thetwolinesare

Anyplanecontainingthesecondlineis

Ifthelinegivenby(13.62)liesonthisplanethenthepoint(−1,−1,−1)alsoliesontheplane.

Theequationoftheplane(13.64)is

Alsothenormaltothisplaneshouldbeperpendiculartotheline(13.62).Thedirectionratiosofthenormaltotheplaneare4,1,−2.Thedirectionratiosoftheline(13.62)are1,2,3.Alsoll1+mm1+nn1=4+2−6=0whichistrue.Hence,theplanecontainingthetwogivenlinesis4x+y−2z+3=0.Anypointonthefirstlineis(r−1,2r−1,3r−1).Ifthetwogivenlinesintersectatthispointthenitshouldlieonthesecondlineandhenceontheplanex+2y+3z−8=0.

Therefore,thepointofintersectionofthetwogivenlinesis(0,1,2).

Example13.21

Showthatthelinesx+2y+3z−4=0,2x+3y+4z−5=0and2x+3y+3z−5=0,3x−2y+4z−6=0arecoplanarandfindtheequationoftheplanecontainingthetwolines.

Solution

Letusexpressthefirstlineinsymmetricalform.Letl,m,nbethedirectioncosinesofthefirstline.Thenthislineisperpendiculartothenormalsoftheplanesx+2y+3z−4=0and2x+3y+4z−5=0.

Solving,weget

Therefore,thedirectionratiosofthefirstlineare1,−2,1.TofindapointonthefirstlineletusfindwherethislinemeetstheXOYplane

(i.e.)z=0.

Solvingthesetwoequationswegetthepointas(−2,3,0).Therefore,theequationsofthefirstlineare

Anyplanecontainingthesecondlineis

Iftheplanecontainsthesecondlinethenthepoint(−2,3,0)shouldlieontheplane(13.67).

Hencetheequationsoftheplane(13.67)becomes

Alsoitshouldsatisfythecondition.Thatthenormaltotheplaneshouldbeperpendiculartotheline(13.66).Thedirectionratiosofthenormaltotheplane(13.68)are1,1,1.Thedirectionratiosofthelineare1,−2,1.Also1−2+1=0whichis

satisfied.Hencetheequationoftherequiredplaneisx+y+z−1=0.

Example13.22

Provethatthelinesx=ay+b=cz+dandx=αy+β=γz+δarecoplanarif(r−c)(αβ−bd)−(α−a)(αδ−δγ)=0.

Solution

Firstletusexpressthegivenlinesinsymmetricalform.Thetwogivenlines

Thentwolinesarecoplanarif

Example13.23

Provethatthelinesa1x+b1y+c1z+d1=0=a2x+b2y+c2z+d2anda3x+b3y

+c3z+d3=0=a4x+b4y+c4z+d4arecoplanarif .

Solution

Letthetwolinesintersectat(x1,y1,z1).Then(x1,y1,z1)shouldlieontheplanescontainingtheselines.

Eliminating(x1,y1,z1)fromtheaboveequationsweget


Example13.24

Findtheshortestdistanceandtheequationtothelineofshortestdistance

betweenthetwolines and .

Solution

Thetwogivenlinesare and .

ThecoordinatesofanypointPonthefirstlineare(3r−7,4r−4,−2r−3).ThecoordinatesofanypointQonthesecondlineare(6r1+21,−4r1−5,−r+

2).ThedirectionratiosofthelinePQare3r−6r1−28,4r+4r1+1,−2r+r1−

5.IfPQisthelineoftheshortestdistancethenthetwolinesareperpendicular.

Thedirectionratiosofthetwolinesare3,4,−2and6,−4,−1.Then3(3r−6r1−28)+4(4r+4r1+1)−2(−2r+r1−5)=0and6(3r−6r1−2r)−4(4r+4r1+1)−1(−2r+r1−8)=0

Solvingforrandr1,weget

ThecoordinatesofPandQaregivenbyP(−1,4,−7)andQ(3,7,5).

Theequationsofthelineoftheshortestdistanceare (i.e.)

Example13.25

Showthattheshortestdistancebetweenz–axisandthelineofintersectionofthe

plane2x+3y+z−1=0with3x+2y+z−2=0is units.

Solution

Theequationsoftheplanecontainingthegivenlineis

Thedirectionratiosofthenormaltothisplaneare2+3λ,3+2λ,4+λ.Thedirectionratiosofthez-axisare0,0,1.Ifz-axisisparalleltothelinethen

0(2+3λ),0(3+2λ)+1(4+λ)=0.

∴λ=−4Therefore,theequationoftheplane(13.69)is2x+3y+4z−1−4(3x+2y+z−2)=0

Example13.26

Findthepointsonthelines and whicharenearest

toeachother.Hencefindtheshortestdistancebetweenthelinesandalsoitsequation.

Solution

Thegivenlinesare

Anypointontheline(13.71)isP(3r+6,−r+7,r+4).Anypointontheline(13.72)isQ(−3r1,2r1−9,4r1+2).ThedirectionratiosofPQare(3r+3r1+6,−r−2r1+16,r−4r1+2).SincePQisperpendiculartothetwogivenlines.

Therefore,thepointsPandQare(3,8,3)and(−3,−7,6).TheSDisthedistancePQ.

ThedirectionratiosofPQare6,15,−3(i.e.)2,5,−1.Pis(3,8,3).

Therefore,theequationsofthelineofSDare .

Example13.27

Findtheshortestdistancebetweenthelines and

.Findalsotheequationofthelineoftheshortestdistance.

Solution

Letl,m,nbethedirectionratiosofthelineoftheSD.Sinceitisperpendiculartoboththelines

Solvingforl,m,n,weget

ThedirectionratiosofthelineofSDare2,3,6.Thedirectioncosinesofthe

lineofSDare

ThelengthofthelineoftheSD=|(x1−x2)l+(y1−y2)m+(z1−z2)n|where(x1,y1,z1)and(x2,y2,z2)arethedirectioncosinesofthelineofSD.

TheequationoftheplanecontainingthefirstlineandthelineofSDis

TheequationoftheplanecontainingthesecondlineandthelineofSDis

Therefore,theequationsofthelineofSDare117x+4y+71z−490=0,63x−28y+7z−238=0.

Example13.28

If2distheshortestdistancebetweenthelinesx=0, andy=0,

thenprovethat .

Solution

Thetwogivenlinesare

Theequationofanyplanecontainingthefirstlineis

Theequationofthesecondlineinsymmetricalformis

Theplanegivenbyequation(13.75)isparalleltotheline(13.76).If

Hencefrom(13.75),theequationoftheplanecontainingline(13.73)and

paralleltotheplane(13.74)is .

ThentheSDbetweenthegivenlines=theperpendiculardistancefromanypointontheline(13.74)totheplane(13.75).(0,0,−c)isapointontheline(13.76).

Example13.29

Showthattheshortestdistancebetweenanytwooppositeedgesofthetetrahedronformedbytheplanesy+z=0,z+x=0,x+y=0andx+y+z=a

is andthethreelinesoftheshortestdistanceintersectatthepointx+y+z=

a.

Solution

Theequationsoftheedgedeterminedbytheplanesy+z=0,z+x=0is

Theequationoftheoppositeedgesarex+y=0,x+y+z=a

Letl,m,nbethedirectioncosinesofthelineoftheSDbetweentwolines.Thenl+m−n=0,l−m+0.n=0.

Solvingforl,m,nweget .

Therefore,thedirectioncosinesofthelineoftheSDare .

Theequationoftheplanecontainingtheedgegivenby(13.77)andthelineof

theSDis .

Therefore,theequationsoftheSDaregivenby

Thislinepassesthroughthepoint(a,a,a).Similarly,bysymmetrywenotethattheothertwolinesofSDalsopassthroughthepoint(a,a,a).

Example13.30

AsquareABCDofdiagonal2aisfoldedalongthediagonalAC,sothattheplanesDAC,BACareatrightangles.FindtheshortestdistancebetweenDCandAB.

Solution

Letabethesideofthesquare.LetustakeOB,OC,ODastheaxesofcoordinates.ThecoordinatesofB,C,DandAare(a,0,0),(0,a,0),(0,0,a),(0,0,−a).

TheequationsofABare

TheequationsofCDare

Theequationsoftheplanepassingthroughthestraightline(13.80)andparallel

to(13.81)is .

Therefore,therequiredshortestdistance=perpendicularfromthepoint(0,a,0)totheplane(13.82).

Example13.31

Provethattheshortestdistancebetweenthediagonalofrectangular

parallelepipedandtheedgenotmeetingitis wherea,b,c

aretheedgesoftheparallelepiped.

Solution

LetOA,OBandOCbethecoterminousedgesofarectangularparallelepiped.ThediagonalsareOO′,AA′,BB′andCC′.ThecoordinatesofO′are(a,b,c).ThecoordinatesofBandC′are(a,0,0)and(a,b,0).

TheequationsofOO′are

TheequationsofBCare .

Letl,m,nbethedirectioncosinesofthelineoftheSD.Then

Hence,l,m,nare−c,o,a.

ThedirectioncosinesofthelineoftheSDare

SimilarlywecanprovethattheothertwoSDare, and .

Exercises3

1. Provethatthelines and arecoplanarandfindtheequationof

theplanecontainingtheline.Ans.:x−y+z=0

2. Provethatthelines and intersect.Findthepointof

intersectionandtheplanecontainingtheline.

Ans.:

3. Showthatthelines and intersectandfindtheequationof

theplanecontainingthelines.Ans.:(5,−7,6)

4. Provethattheline and arecoplanar.Findalsothepointof

intersectionandtheequationoftheplanethroughthem.Ans.:(−1,5,8),4x−11y+7z+3=0

5. Showthatthelines and arecoplanar.Findtheequationof

theplanecontainingtheline.Ans.:x−2y+z=0

6. Showthatthelines and arecoplanar.Findthepointof

intersectionandtheequationoftheplanecontainingthem.Ans.:(1,3,2),17x−47y−24z+172=0

7. Showthatthelines and arecoplanarandfindtheequation

oftheplanecontainingthem.Ans.:x−2y+z=0

8. Showthatthelines and arecoplanarandfindthe

equationoftheplanecontainingthem.Ans.:6x−5y−z=0

9. Showthatthelines and intersectandfindtheequationof

theplanecontainingtheselines.

Ans.:

10. Showthatthelines and3x+2y+z−2=0,x−3y+2z−13=0intersect.

Findalsotheequationoftheplanecontainingthem.

Ans.:(−1,2,3),6x−5y−z=0.

11. Showthatthelinesx−3y+2z+4=0,2x+y+4z+1=0and3x+2y+5z−1=0,2y+z=0arecoplanar.Findtheirpointofintersectionandtheequationoftheplanecontainingtheselines.

Ans.:(3,1,−2),3x+4y+6z−1=0.

12. Showthatthelinesx+y+z−3=0,2x+3y+4z−5=0and4x−y+5z−7=0,2x−5y−z−3=0arecoplanar.Findtheequationoftheplanecontainingtheselines.

Ans.:x+2y+3z−2=0.

13. Showthatthelines7x−4y+7z+16=0,4x+3y−2z+3=0andx−3y+4z+6=0,x−y+z+1=0arecoplanar.

14. Showthatthelines7x−2y−2z+3=0,9x−6y+3=0and5x−4y+z=0,6y−5z=0arecoplanar.Findtheequationoftheplaneinwhichtheylie.

Ans.:x−2y+z=0

15. Showthatthelines andx+2y+z+2=0,4x+5y+3z+6=0arecoplanar.

Findthepointofintersectionofthesetwolines.

Ans.:

16. Showthatthelines andx+2y+3z−14=0,3x+4y+5z−26=0are

coplanar.Findtheirpointofintersectionandtheequationoftheplanecontainingthem.Ans.:(1,2,3),11x+2y−7z+6=0.

17. Showthatthelines3x−y−z+2=0,x−2y+3z−6=0and3x−4y+3z−4=0,2x−2y+z−1=0arecoplanar.Findtheirpointofintersectionandtheequationoftheplanecontainingtheselines.

Ans.:(1,2,3),x−z+2=0

18. Showthatthelines2x−y−z−3=0,x−3y+2z−4=0andx−y+z−2=0,4x+y−6z−3=0arecoplanarandfindtheequationoftheplanecontainingthesetwolines.

Ans.:(1,−1,0),x−z−1=0

19. Showthatthelinesx+2y+3z−4=0;2x+3y+4z−5=0and2x−3y+3z−5=0,3x−2y+4z−6=0arecoplanar.Findtheequationoftheplanecontainingthesetwolines.

Ans.:x+y+z−1=0

20. Showthattheequationoftheplanethroughtheline andwhichisperpendiculartothe

planecontainingthelines and is(m−n)x+(n−l)y+(l−m)z=0.

21. Provethatthelines andax+by+cz+d=0,a1x+b1y+c1z+d1=0,are

coplanarif .

22. Showthatthelines and arecoplanarif

.

23. A,A′;B,B′andC,C′arepointsontheaxes,showthatthelinesofintersectionoftheplanes(A′BC,AB′C′),(B′CA,BC′A′)and(C′AB,CA′B′)arecoplanar.

24. Findtheshortestdistancebetweenthelines and andalsothe

equationsofthelineoftheSD.

Ans.: ,4x+y−5z=0,9x+y−8z−31=0

25. Findtheshortestdistancebetweenthelines and andfind

theequationofthelineoftheshortestdistance.

Ans.: ,4x−5y−17z+79=0,22x−5y+19z−83=0

26. Findtheshortestdistancebetweenthelines and Findalso

theequationofthelineofSDandthepointswherethelineofSDintersectthetwogivenlines.

Ans.: ,(3,5,7),(–1,–1,–1)

27. Showthattheshortestdistancebetweenz–axesandthelineofintersectionoftheplane2x+3y+

4z−1=0with3x+2y+z−2=0is .

28. Showthattheshortestdistancebetweenthelines and is

anditsequationare11x+2y−7z+6=0,7x+y−3z+7=0.

29. Findthelengthoftheshortestdistancebetweenthelines and2x+3y−6z−6=0,

3x−2y−z+5=0.

Ans.:

30. Findtheshortestdistancebetweenz–axisandthelineax+by+cz+d=0,a′x+b′y+c′z+d′=0.

Ans.:

31. Findtheshortestdistancebetweenanedgeofacubeandadiagonalwhichdoesnotmeetit.

Ans.:

32. Alinewithdirectioncosinesproportionalto1,7,−5isdrawntointersectthelines

and .Findthecoordinatesofthepointofintersectionand

thelengthinterceptedonit.

Ans.:(2,8,−3),(0,1,2),

33. Alinewithdirectioncosinesproportionalto2,7,−5isdrawntointersectthelines

Findthecoordinatesofthepointsofintersectionandthe

lengthinterceptedonit.

Ans.:(2,8,−3),(0,1,2);

34. Thetwolines and arecutbyathirdlinewhose

directioncosinesareλ,μ,γ.Showthatthelengthinterceptedonthethirdlineisgivenby

÷ andshowthatthelengthoftheshortestdistanceis

35. Thelengthsoftwooppositeedgesofatetrahedronarea,b,c;theshortestdistanceisequaltod

andtheanglebetweenthemisθ.Provethatthevolumeofthetetrahedronis abdsinθ.

36. Showthattheequationoftheplanecontainingthelinex=0, andparalleltotheliney=

0, is .Ifdistheshortestdistancebetweenthelinesthenshowthat

.

37. Showthattheshortestdistancebetweenthelinesy=az+b,z=αx+βandy=a′z+b,z=α′x+

β′yis .

38. Findtheshortestdistancebetweenthelinesx=2z+3,y=3z+4andx=4z+5,y=5z+6.Whatconclusiondoyoudrawfromyouranswer?

Ans.:Zero;Coplanarlines

Chapter14

Sphere

14.1DEFINITIONOFSPHERE

Thelocusofamovingpointinspacesuchthatitsdistancefromafixedpointisconstantiscalledasphere.Thefixedpointiscalledthecentreofthesphere.Theconstantdistanceiscalledtheradius.

14.2THEEQUATIONOFASPHEREWITHCENTREAT(a,b,c)ANDRADIUSr

LetP(x,y,z)beanypointonthesphere.LetC(a,b,c)bethecentre.

Then,

Thisistheequationoftherequiredsphere.

Showthattheequationx2+y2+z2+2ux+2vy+2wz+d=0alwaysrepresentsasphere.Finditscentreandradius.

Addu2+v2+w2tobothsides.

Thisequationshowsthatthisisthelocusofapoint(x,y,z)movingfromthe

fixedpoint(–u,–v,–w)keepingaconstantdistance fromit.

Therefore,thelocusisaspherewhosecentreis(–u,–v,–w)andwhoseradius

is .

Note14.2.1:Ageneralequationofseconddegreeinx,y,zwillrepresentasphereif(i)coefficientsofx2,y2,z2arethesameand(ii)thecoefficientsofxy,yz,zxarezero.

14.3EQUATIONOFTHESPHEREONTHELINEJOININGTHEPOINTS(x1,y1,z1)AND(x2,y2,z2)ASDIAMETER

Findtheequationofthesphereonthelinejoiningthepoints(x1,y1,z1)and(x2,y2,z2)astheextremitiesofadiameter.

A(x1,y1,z1)andB(x2,y2,z2)betheendsofadiameter.Let(x,y,z)beanypointonthesurfaceofthesphere.Then∠APB=90°Therefore,APisperpendiculartoBP.ThedirectionratiosofAParex–x1,y–y1,z–z1.ThedirectionratiosofBParex–x2,y–y2,z–z2.SinceAPisperpendiculartoBP,

Thisistheequationoftherequiredsphere.

14.4LENGTHOFTHETANGENTFROMP(x1,y1,z1)TOTHESPHEREx2+y2+z2+2ux+2vy+

2wz+d=0

FindthelengthofthetangentfromP(x1,y1,z1)tothespherex2+y2+z2+2ux+2vy+2wz+d=0.

Thecentreofthesphereis(–u,–v,–w).

Theradiusofthesphereis .

Note14.4.1:IfPT2>0,thepointPliesoutsidethesphere.IfPT2=0,thenthepointPliesonthesphere.IfPT2<0,thenthepointPliesinsidethesphere.

14.5EQUATIONOFTHETANGENTPLANEAT(x1,y1,z1)TOTHESPHEREx2+y2+z2+2ux+

2vy+2wz+d=0

Findtheequationofthetangentplaneat(x1,y1,z1)tothespherex2+y2+z2

+2ux+2vy+2wz+d=0.

Thecentreofthesphereis(–u,–v,–w).P(x1,y1,z1)isapointonthesphereandtherequiredplaneisatangentplane

tothesphereatP.Therefore,thedirectionratiosofCParex1+u,y1+v,z1+w.

Therefore,theequationofthetangentplaneat(x1,y1,z1)is(x1+u)(x–x1)+(y1+v)(y–y1)+(z1+w)(z–z1)=0.

Addingux1+vy1+wz1+dtobothsides,weget

Therefore,theequationofthetangentplaneat(x1,y1,z1)isxx1+yy1+zz1+u(x+x1)+v(y+y1)+w(z+z1)+d=0.

14.6SECTIONOFASPHEREBYAPLANE

LetCbethecentreofthesphereandPbeanypointonthesectionofthespherebytheplane.DrawCNperpendiculartotheplane.ThenNisthefootoftheperpendicular

fromPontheplanesection.JoinCP.SinceCNisperpendiculartoNP,CNPisarightangledtriangle.

SinceCPandCNareconstants,NP=constantshowsthatthelocusofPisacirclewithcentreatNandradiusequaltoNP.

Note14.6.1:Iftheradiusofthecircleislessthantheradiusofthespherethenthecircleiscalledasmallcircle.Inotherwords,acircleofthespherenotpassingthroughthecentreofthesphereiscalledasmallcircle.

Note14.6.2:Iftheradiusofthecircleisequaltotheradiusofthespherethenthecircleiscalledagreatcircleofthesphere.Inotherwords,acircleofthespherepassingthroughthecentreofthesphereiscalledagreatcircle.

14.7EQUATIONOFACIRCLE

Thesectionofaspherebyaplaneisacircle.Supposetheequationofthesphereis

andtheplanesectionis

Thenanypointonthecirclelieonthesphere(14.1)aswellastheplanesection(14.2).Hence,theequationsofthecircleofthespherearegivenbyx2+y2+z2+2ux+2vy+2wz+d=0andax+by+cz+k=0.

14.8INTERSECTIONOFTWOSPHERES

Thecurveofintersectionoftwospheresisacircle.

Thecurveofintersectionoftwospheresisacircle.

Letthetwospheresbe

Equation(14.3)isalinearequationinx,y,zandthereforerepresentsaplaneandthisplanepassesthroughthepointofintersectionofthegiventwospheres.Inaddition,weknowthatsectionofthespherebyaplaneisacircle.Hence

thecurveofintersectionofthespheresisgivenbyS1−S2=0.

14.9EQUATIONOFASPHEREPASSINGTHROUGHAGIVENCIRCLE

LetthegivencirclebeS=x2+y2+z2+2ux+2vy+2wz+d=0and

ConsidertheequationS+λP=0.

Thisequationrepresentsasphere.Suppose(x1,y1,z1)isapointonthegivencircle.Then

Equations(14.6)and(14.7)showthatthepoint(x1,y1,z1)liesonthespheregivenbyequation(14.5).Since(x1,y1,z1)isanarbitrarypointonthecircle,itfollowsthateverypoint

onthecircleisapointonthespheregivenby(14.5).Henceequation(14.5)representstheequationofaspherepassingthroughthe

circle(14.4).

14.10CONDITIONFORORTHOGONALITYOFTWOSPHERES

Letthetwogivenspheresbe

ThecentresofthespheresareA(–u,–v,–w)andB(–u1,–v1,–w1).Theradius

Twospheresaresaidtobeorthogonal,ifthetangentplanesatthispointofintersectionareatrightangles.

(i.e.)Theradiidrawnthroughthepointofintersectionareatrightangles.


14.11RADICALPLANE

Thelocusofapointwhosepowerswithrespecttotwospheresareequaliscalledtheradicalplaneofthetwospheres.

14.11.1ObtaintheEquationstotheRadicalPlaneofTwoGivenSpheres

Letthetwogivenspheresbe

Let(x1,y1,z1)beapointsuchthatthepowerofthispointwithrespecttospheres(14.10)and(14.11)beequal.Then

Thelocusof(x1,y1,z1)is

Thisisalinearequationinx,yandzandhencethisequationrepresentsaplane.Henceequation(14.12)istheequationtotheradicalplaneofthetwogiven

spheres.

Note14.11.1.1:Whentwospheresintersect,theplaneoftheirintersectionistheradicalplane.

Note14.11.1.2:Whenthetwospherestouch,thecommontangentplanethroughthepointofcontactistheradicalplane.

14.11.2PropertiesofRadicalPlane

1. Theradicalplaneoftwospheresisperpendiculartothelinejoiningtheircentres.Proof:Lettheequationsofthetwospheresbe

Thecentresofthetwospheresare

C1(–u1,–v1,–w1)andC2(–u2,–v2,–w2).Thedirectionratiosofthelineofcentresareu1–u2,v1–v2,w1–w2.

Theradicalplaneofspheres(14.13)and(14.14)is2(u1–u2)x+2(v1–v2)y+2(w1–w2)z+(d1–d2)=0.Thedirectionratiosofthenormaltotheplaneareu1–u2,v1–v2,w1–w2.Therefore,thelineofcentreisparalleltothenormaltotheradicalplane.Hence,theradicalplaneoftwospheresisperpendiculartothelinejoiningthecentres.

2. Theradicalplanesofthreespherestakeninpairspassthroughaline.Proof:LetS1=0,S2=0,S3=0betheequationsofthethreegivenspheresineachofwhichthe

coefficientsofx2,y2andz2areunity.ThentheequationsoftheradicalplanestakeninpairsareS1–S2=0,S2–S3=0,S3–S1=0.TheseequationshowthattheradicalplanesofthethreespherespassthroughthelineS1=S2=

S3.Hencetheresultisproved.

Note14.11.2.1:Thelineofconcurrenceofthethreeradicalplanesiscalledradicallineofthethreespheres.

3. Theradicalplanesoffourspherestakeninpairsmeetinapoint.Proof:LetS1=0,S2=0,S3=0andS4=0betheequationsofthefourgivenspheres,ineachofwhich

thecoefficientsofx2,y2,z2areunity.

Thentheequationsoftheradicalplanestakentwobytwoare

TheseequationsshowthattheradicalplanesofthefourspheresmeetinatapointgivenbyS1=S2=S3=S4.

Note14.11.2.2:Thepointofconcurrenceoftheradicalplanesoffourspheresiscalledtheradicalcentreofthefourspheres.

14.12COAXALSYSTEM

Definition14.12.1:Asystemofspheresissaidtobecoaxalifeverypairofspheresofthesystemhasthesameradicalplane.

14.12.1GeneralEquationtoaSystemofCoaxalSpheres

LetS=x2+y2+z2+2ux+2vy+2wz+d=0andS1=x2+y2+z2+2u′x+2v′y+2w′z+d′=0betheequationofanytwospheres.Nowconsidertheequation

Nowconsidertheequation

whereλisaconstant.Clearlythisequationrepresentsasphere.Considertwodifferentspheresofthissystemfortwodifferentvaluesofλ.

Thecoefficientsofx2,y2,z2termsin(14.15)are1+λ.

representtwospheresofthesystemwithunitcoefficientsforx2,y2,z2terms.Therefore,theequationoftheradicalplaneof(14.18)and(14.19)is

Sinceλ2≠λ1,S–S′=0whichistheequationtotheradicalplaneofspheres(14.16)and(14.17).Sincethisequationisindependentofλ,everypairofthesystemofspheres

(14.15)hasthesameradicalplane.Henceequation(14.15)representsthegeneralequationtothecoaxalsystemofthespheres.

14.12.2EquationtoCoaxalSystemistheSimplestForm

Inacoaxalsystemofspheres,thelineofcentresisnormaltothecommonradicalplane.

Therefore,letuschoosethex-axisasthelineofcentresandthecommonradicalplaneastheyz-plane,thatis,(x=0).Lettheequationtoasphereofthecoaxalsystembex2+y2+z2+2ux+2vy+

2wz+d=0.Sincethelineofcentresisthex-axis,isyandzcoordinatesarezerov=0,w=

0.Thentheequationoftheabovespherereducestotheformx2+y2+z2+2ux+

d=0.Letusnowconsidertwosphereofthissystemsay,x2+y2+z2+2ux+d=0

andx2+y2+z2+2u1x+d1=0.Theradicalplaneofthesetwospheresis

Buttheequationoftheradicalplaneisx=0.Therefore,from(14.20),d–d1=0ord1=d

Hencetheequationtoanysphereofthecoaxalsystemisoftheformx2+y2+z2

+2λx+d=0whereλisavariableanddisaconstant.

14.12.3LimitingPoints

Limitingpointsaredefinedtobethecentresofpointspheresofthecoaxalsystem.Lettheequationtoacoaxalsystembe

Centreis(–λ,0,0)andradiusis

Forpointsphereradiusiszero.

Therefore,thelimitingpointsofthesystemofspheresgivenby(14.21)are(

,0,0)and(– ,0,0).

Note14.12.3.1:Limitingpointsarerealorimaginaryaccordingasdispositiveornegative.

14.12.4IntersectionofSpheresofaCoaxalSystem

Lettheequationtoacoaxalsystemofspherebex2+y2+z2+2λx+d=0.Nowconsidertwospheresofthesystemsay

Nowconsidertwospheresofthesystemsay

TheintersectionofthesetwospheresisS1–S2=0.

(i.e.)2(λ1–λ2)x=0(i.e.)x=0sinceλ1≠λ2substitutingx=0in(14.22)or(14.23)weget,

Therefore,thisequationisacircleintheyz-planeandalsoitisindependentofλ.Henceeverysphereofthesystemmeetstheradicalplanewithsamecircle.

Note14.12.4.1:Thiscircleiscalledthecommoncircleofthecoaxalsystem.

Note14.12.4.2:Ifd<0,thecommoncircleisrealandthesystemofspheresaresaidtobeintersectingtype.Ifd=0,thecommoncircleisapointcircleandinthiscaseanytwospheresofthesystemtoucheachother.

Ifd>0,thecommoncircleisimaginaryandthespheresaresaidtobeofnon-intersectingtype.


Example14.1

Findtheequationofthespherewithcentreat(2,–3,–4)andradius5units.

Solution

Theequationofthespherewhosecentreis(a,b,c)andradiusris(x–a)2+(y–b)2+(z–c)2=r2.Therefore,theequationofthespherewhosecentreis(2,–3,–4)andradius5is(x–2)2+(y+3)2+(z–4)2=52.

Example14.2

Findthecoordinateofthecentreandradiusofthesphere16x2+16y2+16z2–16x–8y–16z–35=0.

Solution

Theequationofthesphereis16x2+16y2+16z2–16x–8y–16z–35=0.

Dividingby16,

Centreofthesphereis .

Example14.3

Findtheequationofthespherewiththecentreat(1,1,2)andtouchingtheplane2x–2y+z=5.

Solution

Theradiusofthesphereisequaltotheperpendiculardistancefromthecentre(1,1,2)ontheplane2x–2y+z–5=0.

Theequationofthespherewithcentreat(1,1,2)andradius1unitis(x–1)2+(y–1)2+(z–2)2=1.

(i.e.)x2+y2+z2–2x–2y–4z+5=0

Example14.4

Findtheequationofthespherepassingthroughthepoints(1,0,0),(0,1,0),(0,0,1)and(0,0,0).

Solution

Lettheequationofthespherebex2+y2+z2+2ux+2vy+2wz+d=0.Thispassesthroughthepoints(1,0,0),(0,1,0),(0,0,1)and(0,0,0).

Theequationofthesphereisx2+y2+z2–x–y–z=0.

Example14.5

Findtheequationofthespherewhichpassesthroughthepoints(1,0,0),(0,1,0)and(0,0,1)andhasitscentreontheplanex+y+z=6.

Solution

Lettheequationofthespherebex2+y2+z2+2ux+2vy+2wz+d=0.Thisspherepassesthroughthepoints(1,0,0),(0,1,0)and(0,0,1).

Thecentreofthesphereis(–u,–v,–w).Thisliesontheplanex+y+z–6=0.

Theequationofthesphereisx2+y2+z2–4x–4y–4z+4=0.

Example14.6

Findtheequationofthespheretouchingtheplane2x+2y–z=1andconcentricwiththesphere2x2+2y2+2z2+x+2y–z=0.

Solution

Centreis .

Thespheretouchestheplane2x+2y–z–1=0.

Theequationofthesphereis

Example14.7

Findtheequationofthespherewhichpassesthroughthepoints(2,7,–4)and(4,5,–1)hasitscentreonthelinejoiningthethesetwopointsasdiameter.

Solution

Aliter:Thetwogivenpointsaretheextremitiesofadiameterofthesphere.

Therefore,theequationofthesphereis

Example14.8

Theplane cutsthecoordinateaxesinA,BandC.Findtheequationof

thespherepassingthroughA,B,CandO.Findalsoitscentreandradius.

Solution

Theplane cutsthecoordinatesofA,BandC.ThecoordinatesofA,B

andCare(a,0,0),(0,b,0)and(0,0,c).

LettheequationofthespherepassingthroughA,BandCbex2+y2+z2+2ux+2vy+2wz+d=0.SincethispassesthroughoriginO,d=0.SincethispassesthroughA,Band

C.

Hencetheequationofthesphereisx2+y2+z2–2ax–2by–2cz=0.

Centreofthesphereis(a,b,c)andradiusofthesphere=

Example14.9

Findtheequationofthespherecircumscribingthetetrahedronwhosefacesare

and

Solution

Thefacesofthetetrahedronare

Noweasilyseenthattheverticesofthetetrahedronare(0,0,0),(a,b,–c),(a,–b,c)and(–a,b,c).Lettheequationofthespherebex2+y2+z2+2ux+2vy+2wz+d=0.Thisspherepassesthroughthepoints(0,0,0),(a,b,–c,),(a,–b,c)and(–a,b,

c).

Adding(14.28)and(14.29),2(a2+b2+c2)+4ua=0

Similarly,

Therefore,theequationofthesphereisx2+y2+z2–(a2+b2+c2)

Example14.10

Asphereisinscribedinatetrahedronwhosefacesarex=0,y=0,z=0and2x+6y+3z=14.Findtheequationofthesphere.Alsofinditscentreandradius.

Solution

Lettheequationofthespherebex2+y2+z2+2ux+2vy+2wz+d=0.Sincethespheretouchestheplanex=0,theperpendiculardistancefromthe

centre(–u,–v,–w)onthisplaneisequaltotheradius.

∴–u=r,–v=r,–w=r.Alsothespheretouchestheplane2x+6y+3z–14=0.

When ,theequationofthesphereis

Forthissphere,centreis andradius= .

When ,

Example14.11

Findtheequationofthespherepassingthroughthepoints(1,0,–1),(2,1,0),(1,1,–1)and(1,1,1).

Solution

Lettheequationofthespherebex2+y2+z2+2ux+2vy+2wz+d=0.Thisspherepassesthrough(1,0,–1),(2,1,0),(1,1,–1)and(1,1,1).

From(14.31),–2+d=–2⇒d=0.Therefore,therequiredequationofthecircleisx2+y2+z2–2x–y=0.

Example14.12

Findtheequationofthespherewhichtouchesthecoordinateaxes,whosecentreliesinthepositiveoctantandhasaradius4.

Solution

Lettheequationofthespherebex2+y2+z2+2xu+2vy+2wz+d=0.

Theequationofthex-axisis .

Anypointonthislineis(t,0,0).Thepointliesonthegivenspheret2+2ut+d=0.

Sincethespheretouchesthex-axisthetworootsofthisequationareequal.

∴4u2–4d=0oru2=d.

Similarly,v2=dandw2=d

Theradiusofthesphereis

Sincethecentreliesonthex-axis,–u=–v=–w=2 .

Therefore,therequiredequationisx2+y2+z2–4 (x+y+z)+8=0.

Example14.13

Findtheradiusandtheequationofthespheretouchingtheplane2x+2y–z=0andconcentricwiththesphere2x2+2y2+2z2+x+2y–z=0.

Solution

Sincetherequiredsphereisconcentricwiththesphere2x2+2y2+2z2+x+2y–

z=0itscentreisthesameasthatofthegivensphere

Centreis .Theradiusoftherequiredsphereisequaltothe

perpendiculardistancefromthispointtotheplane2x+2y–z=0.

Theequationoftherequiredsphereis

Example14.14

Findtheequationofthespherewhichpassesthroughthepoints(1,0,0),(0,2,0),(0,0,3)andhasitsradiusassmallaspossible.

Solution

Lettheequationofthespherebex2+y2+z2+2ux+2vy+2wz+k=0.Thisspherepassesthroughthepoints(1,0,0),(0,2,0)and(0,0,3).

Theradiusofthesphereisgivenbyr2=u2+v2+w2–k.

Therequiredequationofthesphereis

Example14.15

Findtheequationofthespheretangentialtotheplanex–2y–2z=7at(3,–1,–1)andpassingthroughthepoint(1,1,–3).

Solution

TheequationofnormalatAis

Anypointinthislineis(r+3,–2r–1,–2r–1).IfthispointisthecentreofthespherethenCA=CB.

Therefore,centreofthesphereis(0.5,5).

Radius=

Therefore,theequationofthesphereis(x–0)2+(y–5)2+(z–5)2=81.

(i.e.)x2+y2+z2–10y–10z–31=0

Example14.16

Showthattheplane4x–3y+6z–35=0isatangentplanetothespherex2+y2

+z2–y–2z–14=0andfindthepointofcontact.

Solution

Iftheplaneisatangentplanetothespherethentheradiusisequaltotheperpendiculardistancefromthecentreontheplane.

Thecentreofthespherex2+y2+z2–y–2z–14=0is .

Perpendiculardistancefromthecentreontheplaneis

Therefore,theplanetouchesthesphere.Theequationsofthenormaltothe

tangentplaneare

Anypointonthislineis .

Ifthispointliesontheplane4x–3y+6z–35=0then,

Therefore,thepointofcontactis(2,–1,4).

Example14.17

AsphereofconstantradiusrpassesthroughtheoriginOandcutstheaxesinA,BandC.FindthelocusofthefootoftheperpendicularfromOtotheplaneABC.

Solution

Lettheequationofthespherebex2+y2+z2+2ux+2vy+2wz+d=0.Thispassesthroughtheorigin.

∴d=0

ThespherecutstheaxesatA,BandCwhereitmeetsthex-axis.

y=0,z=0∴x2+2ux=0∴x=–2uTherefore,thecoordinatesofAare(–2u,0,0).SimilarlythecoordinatesofBandCareB(0,–2v,0)andC(0,0,–2w).Therefore,theequationsofthesphereisx2+y2+z2–2ux–2vy–2wz=0.Radius=r

TheequationsoftheplaneABCis

Thedirectionratiosofthenormaltothisplaneare .

Theequationsofthenormalare

Let(x,y,z)bethefootoftheperpendicularfromOontheplane.Then(x1,y1,z1)lieson(14.37).

Substitutingin(14.35),

Thepoint(x1,y1,z1)alsoliesontheplane(14.36)

or

Multiplying(14.38)and(14.39),weget

Thelocusof

Example14.18

Asphereofconstantradius2kpassesthroughtheoriginandmeetstheaxesinA,BandC.ShowthatthelocusofthecentroidofthetetrahedronOABCisx2+y2

+z2=k2.

Solution


Sincex≠0,x=–2u.Therefore,thecoordinatesofAare(–2u,0,0).SimilarlythecoordinatesofBandCare(0,–2v,0)and(0,0,–2w).

Let(x1,y1,z1)bethecentroidofthetetrahedronOABC.

Theradiusofthesphereisr.

∴u2+v2+w2=4r2

Using(14.40),

Thelocusof(x1,y1,z1)isthespherex2+y2+z2=r2.

Example14.19

AsphereofconstantradiusrpassesthroughtheoriginandmeetstheaxesinA,BandC.ProvethatthecentroidofthetriangleABCliesonthesphere9(x2+y2

+z2)=4r2.

Solution

Lettheequationofthespherebex2+y2+z2+2ux+2vy+2wz+d=0.Thislinepassingthroughtheorigin.

∴d=0.

Whenthecirclemeetsthex-axis,y=0,z=0

∴x2+2ux=0as

x≠0,x=−2u∴Aisthepoint(–2u,0,0).SimilarlyBandCarethepoints(0,–2v,0)and(0,0,–2w).Alsogiventhe

radiusisr.

Let(x1,y1,z1)bethecentroidofthetriangleABC.

Butthecentroidis

∴from(14.41):

Thelocusof(x1,y1,z1)is9(x2+y2+z2)=4r2.

Example14.20

Aplanepassesthroughthefixedpoint(a,b,c)andmeetstheaxesinA,B,C.

Provethatthelocusofthecentreofthesphereis

Solution


∴d=0Whenthisspheremeetsthex-axis,y=0andz=0.

∴x2+2ux=0.Asx≠0,x=–2u.Therefore,thecoordinatesofAare(–2u,0,0).SimilarlythecoordinatesofBandCare(0,–2v,0)and(0,0,–2w).

TheequationoftheplaneABCis

Thisplanepassesthroughthepoint(a,b,c).

Thelocusofthecentre(–u,–v,–w)is

Example14.21

Findthecentreandradiusofthecirclex2+y2+z2–8x+4y+8z–45=0,x–2y+2z=3.

Solution

Thecentreofthespherex2+y2+z2–8x+4y+8z–45=0is(4,–2,–4).

CNistheperpendicularfromthecentreofthesphereontheplanex–2y+2z=3.

Therefore,theradiusofthecircleis units.TheequationofthelineCNis

Anypointonthislineist+4,–2t–2,2t–4.Thispointisthecentreofthecirclethenthisliesontheplanex–2y+2z–3=0thent+4–2(2t–2)+2(2t–4)–3=0.

Therefore,thecentreofthecircleis

Example14.22

Showthatthecentresofallsectionsofthespherex2+y2+z2=r2byplanesthroughthepoint(α,β,γ)lieonthespherex(x–α)+y(y–β)+z(z–γ)=0.

Solution

Let(x1,y1,z1)beacentreofasectionofthespherex2+y2+z2=r2byaplanethrough(α,β,γ).Thentheequationoftheplaneisx1(x–x1)+y1(y–y1)+z1(z–z1)=0.Thisplanepassesthroughthepoint(α,β,γ).

x1(α–x1)+y1(β–y1)+Z(γ–z1)=0Therefore,thelocusof(x1,y1,z1)isx(α–x)+y(β–y)+z(γ–z)=0.(i.e.)x(x–α)+y(β–y)+z(γ–z)=0whichisasphere.

Example14.23

Findtheequationofthespherehavingthecirclex2+y2+z2=5,x–2y+2z=5foragreatcircle.Finditscentreandradius.

Solution

Anyspherecontainingthegivencircleisx2+y2+z2–5+2λ(x–2y+2z−5)=0.Thecentreofthissphereis(–λ,2λ,–2λ).Sincethegivencircleisagreat

circle,thecentreofthesphereshouldlieontheplanesectionx–2y+2z=5.

Therefore,theequationofthesphereisx2+y2+z2–5– (x–2y+2z–5)=0.

9(x2+y2–z2–5)–10(x–2y+2z–5)=0.

Centreofthesphereis .

Example14.24

Findtheequationsofthesphereswhichpassesthroughthecirclex2+y2+z2=5,x+2y+3z=3andtouchtheplane4x+3y=15.

Solution

Anyspherecontainingthegivencircleisx2+y2+z2–5+λ(x+2y+3z–3)=0.

Centreis .

Ifthespheretouchestheplane4x+3y=15thentheradiusofthesphereisequaltotheperpendiculardistancefromthecentreontheplane.

Therearetwospherestouchingthegivenplanewhoseequationsarex2+y2+z2

–5+2(x+2y+3z–3)=0andx2+y2+z2–5– (x+2y+3z–3)=0

Example14.25

Provethatthecirclesx2+y2+z2–2x+3y+4z–5=0,5y+6z+1=0andx2+y2+z2–3x–4y+5z–6=0,x+2y–7z=0lieonthesamesphereandfinditsequation.

Solution

Theequationofthespherethroughthefirstcircleis

Theequationofthespherethroughthesecondcircleis

Thegivencircleswilllieonthesamesphereifequation(14.42)and(14.43)areidentical.Therefore,comparingequations(14.42)and(14.43)weget,

Thesetwovaluesλandμsatisfy(14.42),theequations(14.45)and(14.46).Hence,thetwogivencircleslieonthesamesphere.Theequationofthe

sphereisx2+y2+z2–3x–4y+5z–6+x+2y–7z=0.

(i.e.)x2+y2+z2–2x–2y–2z–6=0

Example14.26

Theplane meetsthecircleO,A,BandC.Findtheequationsofthe

circumcircleofthetriangleABCandalsofinditscentre.

Solution

TheequationoftheplaneABCis .

Therefore,thecoordinatesofA,BandCare(a,0,0),(0,b,0)and(0,0,c)respectively.AlsoweknowthattheequationofthesphereOABCisx2+y2+z2–ax–by–

cz=0.Therefore,theequationofthecircumcircleofthetriangleABCarex2+y2+z2

–ax–by–cz=0and .

ThecentreofthesphereOABCis

TheequationofthenormalCNis

Anypointonthislineis .

Thus,pointliesontheplane

Hencethecentreofthecircleis

Example14.27

Obtaintheequationstothespherethroughthecommoncircleofthespherex2+y2+z2+2x+2y=0andtheplanex+y+z+4=0whichintersectstheplanex+y=0incircleofradius3units.

Solution

Theequationofthespherecontainingthegivencircleisx2+y2+z2+2x+2y+λ(x+y+z+4)=0.

Centreofthissphereis

CN=PerpendicularfromthecentreContheplanex+y=0.

Therefore,theequationsoftherequiredspheresarex2+y2+z2+2x+2y–2(x+y+z+4)=0andx2+y2+z2+2x+2y+18(x+y+z+4)=0.

Example14.28

Findtheequationofthespherewhichtouchesthespherex2+y2+z2+2x–6y+1=0atthepoint(1,2,–2)andpassesthroughtheorigin.

Solution

Theequationofthetangentplaneat(1,2,–2)isx+2y–2z+(x+1)–3(y+2)+1=0.

Theequationofthespherepassingthroughtheintersectionof(14.48)and(14.49)isx2+y2+z2+2x–6y+1+λ(2x–y–2z–4)=0.Thisspherepassesthroughtheorigin.

Therefore,theequationoftherequiredsphereis4(x2+y2+z2+2x–6y+1)+(2x–y–2z–4)=0.

(i.e.)4(x2+y2+z2)+10x–25y–2z=0

Example14.29

Showthattheconditionforthespherex2+y2+z2+2ux+2vy+2wz+d=0tocutthespherex2+y2+z2+2u1x+2v1y+2w1z+d1=0inagreatcircleis

wherer1istheradiusofthelattersphere.

Solution

TheintersectionofthesetwosphereisS–S1=0.

(i.e.)2(u–u1)x+2(v–v1)y+2(w–w1)z+d–d1=0.

ThecentreofthesphereS1=0is(–u1,–v1,–w1).SinceS1=0cutsS2=0inagreatcircle,thecentreofthesphereliesontheplaneofintersectionS1–S2=0.

Example14.30

Atangentplanetothespherex2+y2+z2=r2makesinterceptsa,bandconthecoordinateaxes.Provethata–2+b–2+c–2=r–2.

Solution

LetP(x1,y1,z1)beapointonthespherex2+y2+z2=r2.

TheequationofthetangentplaneatPisxx1+yy1+zz1=r2.

Therefore,theinterceptsmadebytheplaneonthecoordinateaxesare

Example14.31

Twospheresofradiir1andr2intersectorthogonally.Provethattheradiusofthe

commoncircleis .

Solution

Lettheequationofthecommoncirclebe

Thentheequationofthespherethroughthegivencircleisx2+y2+z2–r2+λz=0whereλisarbitrary.Lettheequationofthetwospheresthroughthegivencirclebex2+y2+z2–r2+λ1z=0andx2+y2+z2–r2+λ2z=0Ifr1andr2aretheradiioftheabovetwospheresthen

Sincethetwospherescutorthogonally.

Eliminatingλ1andλ2from(14.52)and(14.53),weget

Example14.32

Findtheequationofthespherewhichtouchestheplane3x+2y–z+2=0atthepoint(1,–2,1)andcutsorthogonallythespherex2+y2+z2–4x+6y+4=0.

Solution

Lettheequationoftherequiredspherebe

Thisspheretouchestheplane3x+2y–z+2=0at(1,–2,1).Theequationofthetangentplaneat(1,–2,1)isxx1+yy1+zz1+u(x+x1)+v(y+y1)+w(z+z1)+d=0.

Butthetangentplaneisgivenas

Identifyingequations(14.55)and(14.56)weget,

Thesphere(14.54)cutsorthogonallythesphere

Therefore,theequationofthesphereisx2+y2+z2+7x+10y–5z+12=0.

Example14.33

Findtheequationsoftheradicalplanesofthespheresx2+y2+z2+2x+2y+2z+2=0,x2+y2+z2+4y=0andx2+y2+z2+3x–2y+8z+6=0.Alsofindtheradicallineandtheradicalcentre.

Solution

Considertheequations,

Theradicalplaneofthespheres(14.62)and(14.63)isS1–S2=0.

Theradicalplaneofthespheres(14.63)and(14.64)isS2–S3=0.

(i.e.)3x–6y–8z+6=0

Theradicalplaneofthesphere(14.62)andS3isS1–S3=0.

(i.e.)x–4y+6z+4=0Theequationoftheradicallineofthespheresaregivenby

Alsotheradicallineisgivenby

3x–6y+8z+6=0,2x–3y+7z+4=0.Theradicalcentreisthepointofintersectionoftheabovetwolines.Sowehavetosolvetheequations

Solvingtheseequationsweget

Therefore,theradicalcentreis .

Example14.34

Findtheequationofthespherethroughtheoriginandcoaxalwiththespheresx2

+y2+z2=1andx2+y2+z2+x+2y+3z–5=0.

Solution

TheradicalplaneofthetwogivenspheresisS–S1=0.

(i.e.)x+2y+3z–4=0TheequationofanyspherecoaxalwithgivenspheresisS+λP=0.

(i.e.)x2+y2+z2–1+λ(x+2y+3z–4)=0Thisspherepassesthroughtheorigin.

Therefore,theequationoftherequiredsphereisx2+y2+z2–1– (x+2y+3z

–4)=0.

Example14.35

Findthelimitingpointsofthecoaxalsystemofspheresdeterminedbyx2+y2+z2+4x–2y+2z+6=0andx2+y2+z2+2x–4y–2z+6=0.

Solution

Theradicalplaneofthetwogivenspheresis2x+2y+4z=0.Theequationtoanysphereofthecoaxalsystemisx2+y2+z2+4x–2y+2z

+6+λ(x+y+2z)=0.

Thecentreis .

Radiusis

Forlimitingpointofthecoaxalsystemradius=0.

Therefore,thelimitingpointsarethecentresofpointspheresofthecoaxalsystem.Therefore,thelimitingpointsare(–2,1,–1)and(–1,2,1).

Example14.36

Thepoint(–1,2,1)isalimitingpointofacoaxalsystemofspheresofwhichx2

+y2+z2+3x–2y+6=0isamember.Findtheequationoftheradicalplaneofthissystemandthecoordinatesofotherlimitingpoint.

Solution

Thepointgivesbelongingtothecoaxalsystemcorrespondingtothelimitingpoint(–1,2,1)is(x+1)2+(y–2)2+(z–1)2=0.

(i.e.)x2+y2+z2+2x–4y–2z+6=0Twomembersofthesystemofthesystemarex2+y2+3x–3y+6=0andx2+y2+z2+2x–4y–2z+6=0.Theradicalplaneofthecoaxalsystemisx+y+2z=0.Anymemberofthe

systemisx2+y2+2x–4y–2z+6+λ(x+y+2z)=0.

Centreis

Forlimitingpointsradius=0

Whenλ=0centreis(–1,2,1)whichisthegivenlimitingpoint.Whenλ=2,thecentreis(–2,1,–1)whichisotherlimitingpoint.

Example14.37

Showthatthespheresx2+y2+z2=25andx2+y2+z2–24x–40y–18z+225=0touchexternally.Findtheirpointofcontact.

Solution

Thecentreandradiusofsphere(14.65)are

C1(0,0,0),r1=5Centreandradiusofthesphere(14.66)are

Thedistancebetweenthecentres

Hencethetwogivenspherestouchexternally.

Therefore,thepointofcontactdividesthelinesofcentresintheratio4:1Therefore,thecoordinatesofthepointofcontactis

Exercises1

1. Findtheequationofthespherewithi. centreat(1,–2,3)andradius5units.

ii. centreat andradius1unit.

iii. centreat(1,2,3)andradius4units.

2. Findthecoordinatesofthecentreandradiusofthefollowingspheres:

i. x2+y2+z2+2x–4y–6z+15=0

ii. 2x2+2y2+2z2–2x–4y–6z–1=0

iii. ax2+ay2+az2+2ux+2vy+2wz+d=0

3. Findtheequationofthespherewhosecentreisatthepoint(1,2,3)andwhichpassesthroughthepoint(3,1,2).

Ans.:x2+y2+z2–2x–4y–6z+8=0

4. Findtheequationofthespherepassingthroughpoints:i. (0,0,0),(0,1,–1),(–1,2,0)and(1,2,3)ii. (2,0,1),(1,–5,–1),(0,–2,3)and(4,–1,3)iii. (0,–1,2),(0,–2,3),(1,5,–1)and(2,0,1)iv. (–1,1,1),(1,–1,1),(1,1,–1),(0,0,0)

5. Findtheequationofthesphereonthelinejoiningthepoints(2,–3,–1)and(1,–2,–1)attheendsofadiameter.

Ans.:x2+y2+z2–3x+5y+7=0

6. Findtheradiusofthespheretouchingtheplane2x+2y–z–1=0andconcentricwiththesphere

2x2+2y2+2z2+x+2y–z=0.

Ans.: units

7. Findtheequationofthespherepassingthroughthepoints(0,2,3),(1,1,–1),(–5,4,2)andhavingitscentreontheplane3x+4y+2z–6=0.

Ans.:9(x2+y2+z2)+28x+7y–20z–96=0

8. Provethataspherecanbemadetopassthroughthemidpointsoftheedgesofatetrahedronwhose

facesarex=0,y=0,z=0and .Finditsequation.

Ans.:x2+y2+z2–ax–by–cz=0

9. Findtheconditionthattheplanelx+my+nz=pmaytouchthespherex2+y2+z2+2ux+2vy+2wz+d=0.

Ans.:(ul+vm+wn+p)2=(l2+m2+n2)(u2+v2+w2–d)

10. Provethatthespherecircumscribingthetetrahedronwhosefacesarey+z=0,z+x=0,x+y=0

andx+y+z=1isx2+y2+z2–3(x+y+z)=0.11. Apointmovessuchthat,thesumofthesquaresofitsdistancesfromthesixfacesofacubeisa

constant.Provethatitslocusisthespherex2+y2+z2=3(k2–a2).

12. Provethatthespheresx2+y2+z2=100andx2+y2+z2–12x+4y–6z+40=0touchinternallyandfindthepointofcontact.

Ans.:

13. Provethatthespheresx2+y2+z2=25andx2+y2+z2–24x–40y–18z+225=0touchexternally.Findthepointofcontact.

Ans.:

14. Findtheconditionthattheplanelx+my+nz=ptobeatangenttothespherex2+y2+z2=r2.

Ans.:r2(l2+m2+n2)=p2

15. Findtheequationofthespherewhichtouchesthecoordinateplanesandwhosecentreliesinthefirstoctant.

Ans.:x2+y2+z2–2vx–2vy–2vz+2v2=0

16. Findtheequationofthespherewithcentreat(1,–1,2)andtouchingtheplane2x–2y+z=3.

Ans.:x2+y2+z2–2x+2y+z+5=0

17. Findtheequationofthespherewhichhasthepoints(2,7,–4)and(4,5,–1)astheextremitiesofadiameter.

Ans.:x2+y2+z2–6x–12y+5z+47=0

18. Findtheequationofthespherewhichtouchesthethreecoordinateplanesandtheplane2x+y+2z=6andbeinginthefirstoctant.

Ans.:x2+y2+z2–6x–6y–6z+18=0

19. ApointPmovesfromtwopointsA(1,3,4)andB(1,–2,–1)suchthat3.PA=2.PB.Showthatthe

locusofPisthespherex2+y2+z2–2x–4y–16z+42=0.ShowalsothatthisspheredividesAandBinternallyandexternallyintheration2:3.

20. Aplanepassesthroughafixedpoint(a,b,c).Showthatthelocusofthefootoftheperpendicular

toitfromtheoriginisthespherex2+y2+z2–ax–by–cz=0.21. AvariablespherepassesthroughtheoriginOandmeetsthecoordinateaxesinA,BandCsothat

thevolumeofthetetrahedronOABCisaconstant.Findthelocusofthecentreofthesphere.

Ans.:xyz=k2

22. Findtheequationofthesphereonthelinejoiningthepoints:

i. (4,–1,2)and(2,3,6)astheextremitiesofadiameterii. (2,–3,4)and(–5,6,–7)astheextremitiesofadiameter

Ans.:x2+y2+z2–6x–2y–8z+17=0

x2+y2+z2+3x–3y+3z–56=0

23. Aplanepassesthroughafixedpoint(a,b,c)andcutstheaxesinA,BandC.Showthatthelocus

ofcentreofthesphereABCis

24. Asphereofconstantradius2kpassesthroughtheoriginandmeetstheaxesinA,BandC._Prove

thatthelocusofthecentroidofΔABCis9(x2+y2+z2)=a2.

25. Thetangentplaneatanypointofthespherex2+y2+z2=a2meetsthecoordinateaxesatA,BandC.FindthelocusofthepointofintersectionoftheplanesdrawnparalleltothecoordinateplanesthroughA,BandC.

Ans.:x−2+y−2+z–2=a−2

26. OA,OBandOCarethreemutuallyperpendicularlinesthroughtheoriginandtheirdirectioncosinesarel1,m,n;l2,m2,n2andl3,m3,n3.IfOA=a,OB=b,OC=cthenprovethatthe

equationofthesphereOABCisx2+y2+z2–x(al1+bl2+cl3)–y(am1+bm2+cm3)–z(an1+bn2+cn3)=0.

Exercises2

1. Findthecentreandradiusofasectionofthespherex2+y2+z2=1bytheplanelx+my+nz=1.

Ans.:

2. Findtheequationofthespherethroughthecirclex2+y2+z2=5,x+2y+3z=3andthepoint(1,2,3).

Ans.:5(x2+y2+z2)–4x–8y–12z–13=0.

3. Provethattheplanex+2y–z=4cutsthespherex2+y2+z2–x+z–2=0inacircleofradiusunityandfindtheequationofthespherewhichhasthiscircleforoneofitsgreatcircles.

Ans.:x2+y2+z2–2x–2y+2z–2=0

4. Findthecentreandradiusofthecircleinwhichthespherex2+y2+z2=25iscutbytheplane2x+y+2z=9.

Ans.:(2,1,2);4

5. Showthattheintersectionofthespherex2+y2+z2–2x–4y–6z–2=0andtheplanex–2y+

2z–20=0isacircleofradius withitscentreat(2,4,5).

6. Findthecentreandradiusofthecirclex2+y2+z2–2x–4z+1=0,x+2y+2z=11.

Ans.:

7. Provethattheradiusofthecirclex2+y2+z2+x+y+z=4,x+y+z=0is2.

8. Findthecentreandradiusofthecirclex2+y2+z2+2x–2y–4z–19=0,x+2y+2z+7=0.

Ans.:

9. Findtheradiusofthecircle3x2+3y2+3z2+x–5y–2=0,x+y=2.

10. Showthetwocircles2(x2+y2+z2)+8x+13y+17z–17=0,2x+y–3z+1=0andx2+y2+

z2+3x–4y+3z=0,x–y+2z−4=0lieonthesamesphereandfinditsequation.

Ans.:x2+y2+z2+5x–6y+7z–8=0

11. Provethatthetwocirclesx2+y2+z2–2x+3y+4z–5=0,5y+6z+1=0andx2+y2+z2–3x–4y+5z−6=0,x+2y−7=0lieonthesamesphereandfinditsequation.

Ans.:x2+y2+z2–2x–2y–2z–6=0

12. Findtheareaofthesectionofthespherex2+y2+z2+12x–2y–6z+30=0bytheplanex–y+2z+5=0.

Ans.:

13. Findtheequationofthespherewhichhasitscentreontheplane5x+y–4z+3=0andpassing

throughthecirclex2+y2+z2–3x+4y–2z+8=0,4x–5y+3z–3=0.

Ans.:x2+y2+z2+9x–11y+7z–1=0

14. Findtheequationofthespherehavingthecirclex2+y2+z2+10x–4z–8=0,x+y+z–3=0asagreatcircle.

Ans.:x2+y2+z2+6x–4y–3z+4=0

15. Avariableplaneisparalleltoagivenplane andmeetstheaxesatA,BandC.Prove

thatthecircleABCliesonthesurface

16. Findtheequationofthesphereswhichpassthroughthecirclex2+y2+z2–4x–y+6z+12=0,2x+3y–7z=10andtouchtheplanex–2y–2z=1.

Ans.:x2+y2+z2–2x+2y–4z+2=0

x2+y2+z2–6x–4y+10z+22=0

17. Findtheequationofthespherewhichpassthroughthecirclex2+y2+z2=5,x+2y+3z=5andtouchtheplanez=0.

Ans.:x2+y2+z2–2x+y+5z+5=0

5(x2+y2+z2)–(2x–4y+5z+1=0)

18. Findthecentreandradiusofthecirclex2+y2+z2–2x+4y+2z–6=0,x+2y+2z–4=0.

Ans.:(2,0,1),

19. Findtheequationofthespherewhichpassesthroughthepoint(3,1,2)andmeetsXOYplaneinacircleofradius3unitswiththecentreatthepoint(1,–2,0).

Ans.:x2+y2+z2−2x+4y–4z–4=0

20. Findthecentreandradiusofthecirclex2+y2+z2+12x–12y–16z+111=0,2x+2y+z=17.Ans.:(–4,8,9),r=4

21. Findthecentreandradiusofthecirclex2+y2+z2+2x+2y–4z–19=0,x+2y+2z+7=0.

Ans.:

22. Findthecentreandradiusofthecirclex2+y2+z2=9,x+y+z=1.

23. Findtheequationofthespherethroughthecirclex2+y2+z2=9,2x+3y+4z=5andthepoint(1,2,1).

Ans.:3(x2+y2+z2)–2x–2y–4z–22=0

24. Findtheequationofthespherecontainingthecirclex2+y2+z2–2x=9,z=0andthepoint(4,5,6).

Ans.:x2+y2+z2–2x–10z–9=0

25. Findtheequationofthespherepassingthroughthecirclex2+y2=a2,z=0andthepoint(α,β,λ).

Ans.:r(x2+y2+z2–a2)–z(α2+β2+γ2–a2)=0

26. Findtheequationofthespherethroughthecirclex2+y2+z2+2x+3y+6=0,x–2y+4z–9=0andthecentreofthesphere.

Ans.:x2+y2+z2–2x+4y–6z+5=0

27. Findtheequationsofthespherethroughthecirclex2+y2+z2=1,2x+4y+5z=6andtouchingtheplanez=0.

Ans.:x2+y2+z2–2x–4y–5z+5=0

5(x2+y2+z2)–2x–4y–5z+1=0

28. Findtheequationofthespherehavingthecirclex2+y2+z2+10y–4z–8=0,x+y+23=0asagreatcircle.

Ans.:x2+y2+z2–4x+6y–8z+4=0

29. Showthatthetwocirclesx2+y2+z2–y+2z=0,x–y+z–2=0andx2+y2+z2+x–3y+z–5=0,2x−y+4z−1=0lieonthesamesphereandfinditsequation.

Ans.:x2+y2+z2+3x–4y+5z–6=0

30. Provethatthecirclesx2+y2+z2–2x+3y+4z–5=0,5y+6z+1=0andx2+y2+z2–3x+4y+5z–6=0,x+2y–7z=0lieonthesamesphere.Finditsequation.

Ans.:x2+y2+z2–2x–2y–2z–6=0

31. Findtheconditionsthatthecirclesx2+y2+z2+2ux+2vy+2wz+d=0,lx+my+nz=pand

(x2+y2+z2)2u′x+2v′y+2w′z+d′=0,l′x+m′y+n′z=p′tolieonthesamecircle.

32. Findthecentreandradiusofthecircleformedbytheintersectionofthespherex2+y2+z2=2225andtheplane2x–2y+z=27.

Ans.:(6,–6,3),12

33. Findthecentreandradiusofthecirclex2+y2+z2=25,x+2y+2z=9.Ans.:(1,2,2),4

34. Findtheequationofthecirclewhichliesonthespherex2+y2+z2=25andhasthecentreat(1,2,3).

Ans.:x2+y2+z2=25,x+2y+3z=14

35. Aplanepassesthroughapoint(α,β,γ)andintersectsthespherex2+y2+z2=a2.Showthatthelocusofthecentreofthecircleofintersectionisthespherex(x–α)+y(y–β)+z(z–γ)=0.

36. Findtheequationofthespherethroughthecirclex2+y2+z2–4=0andthepoint(2,1,1).

Ans.:x2+y2+z2–2x+y–2z–1=0

37. Findtheequationofthespherethroughthecirclex2+y2+z2=9,2x+3y+4z=5andthroughtheorigin.

Ans.:5(x2+y2+z2)–18x–27y–36z=0

38. Showthatthetwocircles2(x2+y2+z2)+8x–13y+17z–17=0,2x+y–3z+1=0andx2+

y2+z2+3x–4y+3z=0,x–y+2z–4=0lieonthesamesphereandfinditsequation.

Ans.:x2+y2+z2+5x–6y–7z–8=0

39. Findtheequationofthespherewhichhasitscentreontheplane5x+y–4z+3=0andpassing

throughthecirclex2+y2+z2–3x+4y–2z+8=0,4x–5y+3z–3=0.

Ans.:x2+y2+z2+9x–11y+7z–1=0

40. FindtheequationofthespherewhichhasthecircleS=x2+y2+z2+2x+4y+6z–11=0,2x+

y+2z+1=0asgreatcircle.

Ans.:x2+y2+z2–2x+2y+2z–13=0.

41. Findtheequationofthespherewhoseradiusis1andwhichpassesthroughthecircleof

intersectionofthespheresx2+y2+z2+2x+2y+2z–6=0andx2+y2+z2+3x+3y–z–1=0.

Ans.:3x2+3y2+3z2+16x+16y+4z+32=0

42. Ifristheradiusofthecirclex2+y2+z2+2ux+2vy+2wz+d=0,lx+my+nz=0thenprove

that(r2+d2)(l2+m2+n2)=(mw–nv)2+(nv–lw)2+(lv–mu)2.

43. Findtheequationofthespherethroughthecirclex2+y2=4,z=0meetingtheplanex+2y+2z=0inacircleofradius3.

Ans.:x2+y2+z2–6z–4=0

44. Findtheequationofthespherethroughthecirclex2+y2+z2=1,2x+3y+4z=5andwhich

intersectthespherex2+y2+z2+3(x–y+z)–56=0orthogonally,x2+y2+z2–12x–18y–24z+29=0.

45. TheplaneABCwhoseequationis meetstheaxesinA,BandC.Findtheequationto

determinethecircumcircleofthetriangleABCandobtainthecoordinatesofitscentre.

46. Findtheequationofthecirclecircumscribingthetriangleformedbythethreepoints(a,0,0),(0,b,0),(0,0,c).Obtainthecoordinatesofthecentreofthecircle.

Ans.:x2+y2+z2–x–2y–3z=0,6x–3y–2z–6=0

Centre=

47. Findtheequationofthespherethroughthecirclex2+y2+z2+2x+3y+6=0,x–2y+4z–9=

0andthecentreofthespherex2+y2+z2–2x+4y–6z+5=0.

Ans.:x2+y2+z2+7y–8z+24=0

48. Findtheequationofthespherehavingitscentreontheplane4x–5y–z–3=0andpassing

throughthecirclex2+y2+z2–2x–3y+4z+8=0,x2+y2+z2+4x+5y–6z+2=0.

Ans.:x2+y2+z2+7x+9y–11z–1=0

49. Acirclewithcentre(2,3,0)andradiusunityisdrawnontheplanez=0.Findtheequationofthespherewhichpassesthroughthecircleandthepoint(1,1,1).

Ans.:x2+y2+z2–4x–6y–6z+12=0

50. Findtheequationofthespherewhichpassesthroughthecirclex2+y2=4,z=0andiscutbytheplanex+2y+2z=0inacircleofradius3.

Ans.:x2+y2+z2+6z–4=0,

x2+y2+z2–6z–4=0

51. Provethattheplanex+2y–z–4=0cutsthespherex2+y2+z2–x+z–2=0inacircleofradiusunityandalsofindtheequationofthespherewhichhasthiscircleasgreatcircle.

Ans.:x2+y2+z2–2x–2y+2z+2=0

52. Findtheequationofthespherehavingthecirclex2+y2+z2+10y–4z–3=0,x+y+z–3=0asagreatcircle.

Ans.:x2+y2+z2–4x+6y–8z+4=0

53. PisavariablepointonagivenlineandA,BandCareprojectionsontheaxes.ShowthatthesphereOABCpassesthroughafixedcircle.

Exercises3

1. Findtheequationsofthesphereswhichpassthroughthecirclex2+y2+z2=5,x+2y+3z=5andtouchtheplanez=0.

Ans.:x2+y2+z2–2x+y+5z+5=0

5(x2+y2+z2)–2x–4y+5z+1=0

2. Findtheequationsofthesphereswhichpassthroughthecirclex2+y2+z2–4x–y+3z+12=0,2x+3y–8z=10andtouchtheplanex–2y–2z=1.

Ans.:x2+y2+z2–2x+2y–4z+2=0

x2+y2+z2–6x–4y+10z+22=0

3. Showthatthetangentplaneat(1,2,3)tothespherex2+y2+z2+x+y+z–20=0is3x+5y+7z–34=0.

4. Findtheequationofthespherewhichtouchesthespherex2+y2+z2–x+3y+2z–3=0at(1,1,−1)andpassesthroughtheorigin.

Ans.:2x2+2y2+2z2–3x+y+4z=0

5. Findtheequationofthespherewhichtouchesthespherex2+y2+z2+2x–6y+1=0at(1,2,–2)andpassesthroughtheorigin.

Ans.:4(x2+y2+z2)+10x–28y–2z=0

6. Showthattheplane2x–2y+z+16=0touchesthespherex2+y2+z2+2x+4y+2z–3=0andfindthepointofcontact.

Ans.:(–3,4,–2)

7. Findtheequationofthetangentplaneattheorigintothespherex2+y2+z2–8x–6y+4z=0.Ans.:4x–3y+2z=0

8. Findtheequationofthetangentplanestothespherex2+y2+z2+2x–4y+6z–7=0whichpassesthroughtheline6x–3y–23=0,3z+2=0.

Ans.:8x−4y+z–34=0,4x–2y–z–16=0

9. Showthattheplane2x–2y+z–12=0touchesthespherex2+y2+z2–2x–4y+2z–3=0.

10. ShowthatthepointP(1,–3,1)liesonthespherex2+y2+z2+2x+2y–7=0andobtaintheequationofthetangentplaneatP.

Ans.:2x–2y+z=9

11. Ifthepoint(5,1,4)isoneextremitiesofadiameterofthespherex2+y2+z2–2x–2y–2z–22=0andfindthecoordinatesoftheotherextremity.Findtheequationtothetangentplanesatthetwoextremitiesandshowthattheyareparallel.

Ans.:(–3,1,–2);4x+3y–22=0,4x+3y+8=0

12. Findthevalueofkforwhichtheplanex+y+z=ktouchesthespherex2+y2+z2–2x–4y–6z+11=0.Findthepointofcontactineachcase.

Ans.:k=3or9;(0,1,2),(2,3,4)

13. Findtheequationtothetangentplanestothespherex2+y2+z2–2x–4y–6z–2=0whichareparalleltotheplanex+2y+2z–20=0.

Ans.:x+2y+2z–23=0x+2y+2z–1=0

14. Aspheretouchestheplanex–2y–2z–7=0atthepoint(3,–1,–1)andpassesthroughthepoint(1,1,–3).Findtheequation.

Ans.:x2+y2+z2–10y–10z–31=0

15. Showthattheline touchesthespherex2+y2+z2–6x+2y–4z+5=0.Find

thepointofcontact.Ans.:(5,–3,3)

16. Tangentplanesatanypointofthespherex2+y2+z2=r2meetsthecoordinateaxesatA,BandC.Showthatthelocusofthepointofintersectionoftheplanesdrawnparalleltothecoordinate

planesthroughthepointsA,BandCisthesurfacex–2+y–2+z–2=r–2.

17. Findtheconditionthattheline wherel,mandnarethedirectioncosinesofa

line,shouldtouchthespherex2+y2+z2+2ux+2vy+2wz+d=0.Showthattherearetwospheresthroughthepoints(0,0,0),(2a,0,0),(0,2b,0)and(0,0,2c)whichtouchtheaboveline

andthatthedistancebetweentheircentresis

18. Findtheequationofthespherewhichtouchesthespherex2+y2+z2+3y–x+2z–3=0at(1,1,–1)andpassesthroughtheorigin.

Ans.:2x2+2y2+2z2–3x+y+4z=0

19. Findtheequationofthetangentlineinsymmetricalformtothecirclex2+y2+z2+5x–7y+2z–8=0,3x–2y+4z+3=0.

20. Showthattheplane2x–2y–z+12=0touchesthespherex2+y2+z2–2x–4y+2z–3=0andfindthepointofcontact.

Ans.:(–1,4,–2)

21. Findtheequationofthespherewhichtouchesthespherex2+y2+z2+2x–6y+1=0atthepoint(1,2,–2)andpassesthroughtheorigin.

Ans.:4(x2+y2+z2)+10x–25y–22=0

22. Findtheequationsofthesphereswhichpassthroughthecirclex2+y2+z2=1,2x+4y+5z–6=0andtouchtheplanez=0.

Ans.:5(x2+y2+z2)–2x–4y–5z+6=0

5(x2+y2+z2)–2x–4y–5z+1=0

23. Findtheequationsofthespherepassingthroughthecirclex2+y2+z2–5=0,2x+3y+z–3=0andtouchingtheplane3x+4z−15=0.

Ans.:x2+y2+z2+4x+6y–2z–11=0

5(x2+y2+z2)–8x–12y–4z–37=0

24. Findthepointofintersectionofthelineandthespherex2+y2+z2–4x+6y–2z+5=0.Ans.:(4,–1,2),(0,–2,3)

25. Provethatthesumofthesquaresoftheinterceptsmadebyagivenlineonanythreemutuallyperpendicularlinesthroughafixedpointisconstant.

Exercises4

1. Provethatthespheresx2+y2+z2+6y+2z+8=0andx2+y2+z2+6x+8y+4z+2=0intersectorthogonally.

2. Findtheequationofthespherewhichpassesthroughthecirclex2+y2+z2–2x+3y–4z+6=0,

3x–4y+5z–15=0andwhichcutsorthogonallythespherex2+y2+z2+2x+4y–6z+11=0.

Ans.:5(x2+y2+z2)–13x+19y–25z+45=0

3. Findtheequationofthespherethatpassesthroughthecirclex2+y2+z2–2x+3y–4z+6=0,

3x–4y+5z–15=0andwhichcutsthespherex2+y2+z2+2x+4y+6z+11=0orthogonally.

Ans.:x2+y2+z2+x–y+z–9=0

4. Provethateveryspherethroughthecirclex2+y2–2ax+r2=0,z=0cutsorthogonallyevery

spherethroughthecirclex2+z2=r2,y=0.5. Findtheequationofthespherewhichtouchestheplane3x+2y–z+7=0atthepoint(1,−2,1)

andcutsthespherex2+y2+z2–4x+6y+4=0orthogonally.

Ans.:3(x2+y2+z2)+6x+20y–10z+36=0

6. Findtheequationofthespherethatpassesthroughthepoints(a,b,c)and(–2,1,–4)andcuts

orthogonallythetwospheresx2+y2+z2+x–3y+2=0and(x2+y2+z2)+x+3y+4=0.

Ans.:x2+y2+z2+2x–2y+4z–3=0

7. Findtheequationofthespherewhichtouchestheplane3x+2y–z+2=0atthepointP(1,–2,1)andalsocutsorthogonally.

Ans.:x2+y2+z2+7x+10y–5z+12=0

8. Ifdisthedistancebetweenthecentresofthetwospheresofradiir1andr2thenprovethatthe

anglebetweenthemis

9. Findtheconditionthatthespherea(x2+y2+z2)+2lx+z–y+2nz+p=0andb(x2+y2+z2)k2

maycutorthogonally.

Ans.:ak2=bp2

10. Findtheequationoftheradicalplanesofthespheresx2+y2+z2+2x+2y+2z–2=0,x2+y2+

z2+4y=0,x2+y2+z2+3x–2y+8z–6=0.Ans.:x–y+z–1=0,3x–6y+8z–6=0

x–4y+6z+4=0

11. Findtheequationoftheradicallineofthespheres(x–2)2+y2+z2=1,x2+(y–3)2+z2=6

and(x+2)2+(y+1)2+(z–2)2=6.

12. Findtheequationoftheradicallineofthespheresx2+y2+z2+2x+2y+2z+2=0,x2+y2+

z2+4y=0,x2+y2+z2+3x–2y+8z+6=0.Ans.:x–y+z+1=0,3x–6y+8z+6=0

13. Findtheradicalplaneofthespheresx2+y2+z2–8x+4y+4z+12=0,x2+y2+z2–6x+3y+3z+9=0.

Ans.:2x–y–z–3=0

14. Findthespherescoaxalwiththespheresx2+y2+z2+2x+y+3z–8=0andx2+y2+z2–5=0andtouchingtheplane3x+4y=15.

Ans.:5(x2+y2+z2)–8x–4y–12z–13=0

15. Findthelimitingpointsofthecoaxalsystemdefinedbythespheresx2+y2+z2+3x–3y+6=0,

x2+y2+z2–6y–6z+6=0.Ans.:(–1,2,1),(–2,1,–1)

16. Findthelimitingpointsofthecoaxalsystemdeterminedbythetwosphereswhoseequationsare

x2+y2+z2–8x+2y–2z+32=0,x2+y2+z2–7x+z+23=0.Ans.:(3,1,–2),(5,–3,4)

17. Findtheequationsofthesphereswhoselimitingpointsare(–1,2,1)and(–2,1,–1)andwhichtouchestheplane2x+3y+6z+7=0.

18. Findtheequationofthespherewhichtouchestheplane3x+2y–z+2=0atthepoint(1,–2,1)

andalsocutorthogonallythespherex2+y2+z2–4x+6y+4=0.

Ans.:x2+y2+z2+7x+10y–5z+12=0

19. Findthelimitingpointsofthecoaxalsystemtwoofwhosemembersarex2+y2+z2–3x–3y+6

=0,x2+y2+z2–4y–6z+6=0.Ans.:(2,–3,4)and(–2,3,–4)

20. Thepoint(–1,2,1)isalimitingpointofacoaxalsystemofspheresofwhichthespherex2+y2+

z2+3x–2y+6=0isamember.Findthecoordinatesoftheotherlimitingpoint.Ans.:(–2,1,–1)

Chapter15

Cone

15.1DEFINITIONOFCONE

Aconeisasurfacegeneratedbyastraightlinewhichpassesthroughafixedpointandintersectsafixedcurveortouchesagivencurve.Thefixedpointiscalledthevertexoftheconeandthefixedcurveiscalleda

guidingcurveofthecone.Thestraightlineiscalledagenerator.

15.2EQUATIONOFACONEWITHAGIVENVERTEXANDAGIVENGUIDINGCURVE

Let(α,β,γ)bethegivenvertexandax2+2hxy+by2+2gx+2fy+c=0,z=0betheguidingcurve.

Theequationsofanylinepassingthroughthepoint(α,β,γ)are

Whenthislinemeetstheplaneatz=0weget,

Thispointliesonthegivencurveax2+2hxy+by2+2gx+2fy+c=0,z=0.

Eliminatingl,m,nfrom(15.1)and(15.2)wegettheequationofthecone.

From(15.1)

Multiplyingthroughoutby(z−γ)2,weget

a(αz−γx)2+2h(αz−γx)(βz−γy)+b(βz−γy)2+2g(αz−γx)(z−γ)+zf(βz−γy)(z−γ)+c(z−γ)2=0.

Thisistherequiredequationofthecone.

Example15.2.1

Findtheequationoftheconewithitsvertexat(1,1,1)andwhichpassesthroughthecurvex2+y2=4,z=2.

Solution

LetVbethevertexoftheconeandPbeanypointonthesurfaceofthecone.LettheequationsofthegeneratorVPbe

Thislineintersectstheplanez=2.

Thispointliesonthecurvex2+y2=4.

Eliminatingl,m,nfrom(15.3)and(15.4)weget

or

Thisisrequiredequationofthecone.

Example15.2.2

Findtheequationoftheconewhosevertexis(a,b,c)andwhosebaseisthecurveαx2+βy2=1,z=0.

Solution

ThevertexoftheconeisV(a,b,c).

Theguidingcurveis

Letl,m,nbethedirectioncosinesofthegeneratorVP.

ThentheequationsofVPare

Whenthislinemeetsz=0wehave

Thepointliesonthecurveαx2+βy2=1.

Wehavetoeliminatel,m,nfrom(15.6)and(15.7)

Thisistheequationoftherequiredcone.

15.3EQUATIONOFACONEWITHITSVERTEXATTHEORIGIN

Toshowthattheequationofaconewithitsvertexattheoriginishomogeneous,let

betheequationofaconewithitsvertexattheorigin.LetP(α,β,γ)beanypointonthesurface.Then,OPisageneratorofthecone.

Since(α,β,γ)liesonthecone

TheequationsofOPare

Anypointonthislineis(tα,tβ,tγ).Thepointliesontheconef(x,y,z)=0.

Fromequations(15.8)and(15.11),weobservethattheequationf(x,y,z)=0ishomogeneous.

Conversely,everyhomogeneousequationin(x,y,z)representsaconewithitsvertexattheorigin.

Letf(x,y,z)=0beahomogeneousequationinx,y,z.Sincef(x,y,z)=0isahomogeneousequation,f(x,y,z)=0foranyreal

number.Inparticularf(0,0,0)=0.

Therefore,theoriginliesonthelocusoftheequation(15.8).Asf(tx,ty,tz)=0,anypointonthelinethroughtheoriginliesontheequation(15.8).Inotherwords,thelocusof(15.8)isasurfacegeneratedbythelinethroughtheorigin.Henceequationrepresentsaconewithitsvertexattheorigin.

Note15.3.1:Iff(x,y,z)canbeexpressedastheproductofnlinearfactorsthenf(x,y,z)=0representsnplanesthroughtheorigin.

Note15.3.2:Iff(x,y,z)=0isahomogeneousequationofseconddegreeinx,yandzthenf(x,y,z)=0isaquadriccone.Ifitcanbefactoredintotwolinearfactorsthenitrepresentsapairofplanesthroughtheorigin;thenweregardequationf(x,y,z)=0asdegeneratecone,thevertexbeinganypointonthelineofintersectionofthetwoplanes.

Generators:Theline isageneratoroftheconef(x,y,z)=0withits

vertexattheoriginifandonlyiff(l,m,n)=0.

Let beageneratoroftheconef(x,y,z)=0thenthepoint(lr,

mr,nr)liesonthecone.Takingr=1,thepoint(l,m,n)liesontheconef(x,y,z)=0.

∴f(l,m,n)=0

Converse:Letf(x,y,z)=0betheequationoftheconewithitsvertexatthe

originand bealinethroughtheoriginsuchthatf(l,m,n)=0.

Sincethevertexisattheorigin,f(x,y,z)=0isahomogeneousequationinx,yandz.

Nowwewillprovethat isageneratorofthecone.

Anypointonthegeneratoris(lr,mr,nr).Sincef(x,y,z)=0andf(l,m,n)=0,itfollowsthatf(lr,mr,nr)=0.

∴ isageneratoroftheconef(x,y,z)=0.

Example15.3.1

Findtheequationoftheconewithitsvertexattheoriginandwhichpassesthroughthecurveax2+by2+cz2−1=0=αx2+βy2−2z.

Solution

Lettheequationofthegeneratorbe

Anypointonthislineis(lr,mr,nr).Thispointliesonthecurve

From(15.14),

Substitutingthisin(15.13)weget

Asl,m,nareproportionaltox,y,ztheequationoftheconeis4z2(ax2+by2+cz2)=(αx2+βy2)2.

Example15.3.2

Findtheequationoftheconewhosevertexisattheoriginandtheguidingcurve

is

Solution

Sincethevertexoftheconeistheoriginitsequationmustbeahomogeneousequationofseconddegree.Theequationsoftheguidingcurveare

Homogenizingtheequation(15.15)withthehelpof(15.16)wegettheequation

oftherequiredcone.Hencetheequationoftheconeis

(i.e.)27x2+32y2+7z2(xy+yz+zx)=0

Example15.3.3

Theplane meetsthecoordinateaxesatA,BandC.Provethatthe

equationoftheconegeneratedbylinesdrawnfromOtomeetthecircleABCis

Solution

ThepointsA,B,Care(a,0,0),(0,b,0)and(0,0,c),respectively.TheequationofthesphereOABCisx2+y2+z2−ax−by−cz=0.TheequationsofthecircleABCare

Homogenizingequation(15.17)withthehelpof(15.18)wegettheequationoftherequiredcone.

15.4CONDITIONFORTHEGENERALEQUATIONOFTHESECONDDEGREETOREPRESENTACONE

Letthegeneralequationoftheseconddegreebe

Let(x1,y1,z1)bethevertexofthecone.Shifttheorigintothepoint(x1,y1,z1).Then

Thentheequation(15.19)becomes

SincethisequationhastobeahomogeneousequationinX,YandZ.CoefficientofX=0CoefficientofY=0CoefficientofZ=0andconstantterm=0.

Eliminatingx,y,zfrom(15.21),(15.22),(15.23)and(15.24),weget,


Note15.4.1:Ifthegivenequationoftheseconddegreeisf(x,y,z)=0thenmakeithomogeneousbyintroducingthevariabletwheret=1.Then

Solvinganythreeofthesefourequations,wegetthevertexofthecone.Testwhetherthesevaluesofx,y,zsatisfythefourthequation.

Example15.4.1

Findtheequationoftheconeoftheseconddegreewhichpassesthroughtheaxes.

Solution

Theconepassesthroughtheaxes.Therefore,thevertexoftheconeistheorigin.Theequationsoftheconeisahomogeneousequationofseconddegreeinx,y

andz.

Giventhatx-axisisagenerator.Theny=0,z=0mustsatisfytheequation(15.25)

∴a=0Sincey-axisisageneratorb=0.Sincez-axisisageneratorc=0.

Hencetheequationoftheconeisfyz+gzx+hxy=0.

Example15.4.2

Showthatthelinesthroughthepoint(α,β,γ)whosedirectioncosinessatisfytherelational2+bm2+cn2=0,generatestheconea(x−α)2+b(y−β)2+c(z−γ)2

=0.

Solution

Theequationsofanylinethrough(α,β,γ)are

where

Eliminatingl,m,nfrom(15.26)and(15.27)

weget,

a(x−α)2+b(y−β)2+c(z−γ)2=0Example15.4.3

Findtheequationtothequadricconewhichpassesthroughthethreecoordinate

axesandthethreemutuallyperpendicularlines ,

Solution

Wehaveseenthattheequationoftheconepassingthroughtheaxesis

Thisconepassesthroughline

or

Sincetheconealsopassesthroughtheline wehave

From(15.29)and(15.30)weget

From(15.28)and(15.31)weget5yz+8zx−3xy=0.

Sincetheconepassesthroughtheline

weget4f+5g+20h=0and4(5)+5(8)+20(−3)=0isalsotrue.Therefore,theequationoftherequiredconeis5yz+8zx−3xy=0.

Example15.4.4

Provethattheequation2x2+2y2+7z2−10yz−10zx+2x+2y+26z−17=0representsaconewhosevertexis(2,2,1).

Solution

LetF(x,y,z,t)=2x2+2y2+7z2−10yz−10zx−2xt+2yt+26zt−17t2=0

givetheequations

Solvingthefirstthreeequationswegetx=2,y=2,z=1.Thesevaluesalsosatisfythefourthequation.Therefore,thegivenequationrepresentsaconewithitsvertexatthepoint(2,2,1).

Exercises1

1. Findtheequationoftheconewhosevertexisattheoriginandwhichpassesthroughthecurveof

intersectionoftheplanelx+my+nz=pandthesurface,ax2+by2+cz2=1.

Ans.:p2(ax2+by2+cz2)=(lx+my+nz)2

2. Findtheequationoftheconewhosevertexis(α,β,γ)andwhoseguidingcurveistheparabolay2

=4ax,z=0.

Ans.:(ry−βz)2=4a(z−γ)(αz−rx)

3. Provethattheline where2l2+3m2−5n2=0isageneratorofthecone2x2+3y2−5z2

=0.

4. Findtheequationoftheconewhosevertexisatthepoint(1,1,0)andwhoseguidingcurveisx2+

z2=4,y=0.

Ans.:x2−3y2+z2−2xy+8y−4=0

5. Findtheequationoftheconewhosevertexisthepoint(0,0,1)andwhoseguidingcurveisthe

ellipse ,z=3.Alsoobtainsectionoftheconebytheplaney=0andidentifyitstype.

Ans.:36x2+100y2−225z2+450z−225=0

pairofstraightlines

6. Findtheequationsoftheconeswithvertexattheoriginandpassingthroughthecurvesof

intersectiongivenbytheequations:

i.

ii.

iii.

iv.

v.

7. Theplanex+y+z=1meetsthecoordinateaxesinA,BandC.Provethattheequationtothe

conegeneratedbythelinesthroughO,tomeetthecircleABCisyz+zx+xy=0.

8. Avariableplaneisparalleltotheplane andmeetstheaxesinA,BandC.Provethat,

thecircleABCliesonthecone

9.i. Findtheequationofthequadricconewhichpassesthroughthethreecoordinateaxesand

threemutuallyperpendicularlines

Ans.:16yz−33zx−25xy=0

ii. Provethattheequationoftheconewhosevertexis(0,0,0)andthebasecurvez=k,f(x,

y)=0is wheref(x,y)=ax2+2hxy+by2+2gx+2fy+c=0.

10. Findtheequationtotheconewhosevertexistheoriginandthebasecirclex=a,y2+z2=b2andshowthatthesectionoftheconebyaplaneparalleltothexy-planeishyperbola.

Ans.:a2(y2+z2)=b2x2

11. PlanesthroughOXandOYincludeanangleα.Showthatthelineofintersectionliesonthecone

z2(x2+y2+z2)=x2y2tan2α.12. Provethataconeofseconddegreecanbefoundtopassthroughtwosetsofrectangularaxes

throughthesameorigin.

13. Provethattheequationx2−2y2+3z2+5yz−6zx−4xy+8x−19y−2z−20=0representsaconewithitsvertexat(1,−2,3).

14. Provethattheequation2y2−8yz−4zx−8xy+6x−4y−2z+5=0representsaconewhose

vertexis .

15. Provethattheequationax2+by2+cz2+2ux+2vy+2wz+d=0representsaconeif

15.5RIGHTCIRCULARCONE

Arightcircularconeisasurfacegeneratedbyastraightlinewhichpassesthroughafixedpoint,andmakesaconstantanglewithafixedstraightline

throughthefixedpoint.Thefixedpointiscalledthevertexoftheconeandtheconstantangleiscalledthesemiverticalangleandfixedstraightlineiscalledtheaxisofthecone.Thesectionofrightcircularconebyanyplaneperpendiculartoitsaxisisa

circle.

15.5.1EquationofaRightCircularConewithVertexV(α,β,γ),AxisVLwithDirectionRatiosl,m,nandSemiverticalAngleθ

LetP(x,y,z)beanypointonthesurfaceofthecone.ThendirectionratiosofVParex−α,y−β,z−γ.

ThedirectionratiosoftheperpendicularVLarel,m,n.


Note15.5.1.1:

i. Ifthevertexisattheoriginthentheequationoftheconebecomes(lx+my+nz)2=[(x2+y2+

z2)(l2+m2+n2)]cos2θ.ii. Ifl,m,narethedirectioncosinesofthelinethen

iii. Ifaxisofconeisthez-axisthentheequation(15.33)becomes

or

15.5.2EnvelopingCone

Ithasbeenseeninthetwo-dimensionalanalyticalgeometrythattwotangentscanbedrawnfromagivenpointtoaconic.Inanalogywiththataninfinitenumberoftangentlinescanbedrawnfromagivenpointtoaconicoid,inparticulartoasphere.Allsuchtangentlinesgenerateaconewiththegivenpointasvertex.Suchaconeiscalledanenvelopingcone.

Definition15.5.2.1:Thelocusoftangentlinesdrawnfromagivenpointtoagivensurfaceiscalledanenvelopingconeofthesurface.Thegivenpointiscalledthevertexofthecone.

Equationofenvelopingcone:Letusfindtheequationoftheenvelopingconeofthespherex2+y2+z2=a2withthevertexat(x,y,z).

LetP(x,y,z)beanypointonthetangentdrawnfromV(x1,y1,z1)tothegivensphere.LetQbethepointthatdividesPQintheratio1:λ.Thenthecoordinates

ofQare

Ifthispointliesonthespherethen,

Thisisaquadraticequationinλ.TherearetwovaluesofλindicatingthattherearetwopointsonVPwhichdividesPQintheratio1:λandlieonthesphere.IfPQisatangenttothespherethenthesetwopointscoincideandthepointisthepointofcontact.Therefore,thetwovaluesofλofequation(15.34)mustbeequalandhencethe

discriminantmustbezero.

Thisistheequationofthe

envelopingcone.

Note15.5.2.2:Theequationoftheenvelopingconecanbeexpressedintheform

Example15.5.1

Findtheequationoftherightcircularconewhosevertexisattheorigin,whose

axisistheline andwhichhasaverticalangleof60°.

Solution

Theaxisoftheconeis .

Therefore,thedirectionratiosoftheaxisoftheconeare1,2,3.

Thedirectioncosinesare .

LetP(x,y,z)beanypointonthesurfaceofthecone.LetPLbeperpendiculartoOA.

Example15.5.2

Findtheequationoftherightcircularconewithitsvertexattheorigin,axisalongthez-axisandsemiverticalangleα.

Solution

Thedirectioncosinesoftheaxisoftheconeare0,0,1.

LetP(x,y,z)beanypointonthecone.

Then,


Example15.5.3

Findthesemiverticalangleandtheequationoftherightcircularconehavingitsvertexatoriginandpassingthroughthecircley2+z2=b2,x=a.

Solution

Theguidingcircleoftherightcircularconeisy2+z2=b2,x=a.Therefore,theaxisoftheconeisalongx-axis.

Ifθisthesemiverticalangle,then .

LetP(x,y,z)beanypointonthesurfaceofthecone.ThedirectionratiosofOParex,y,z.

Thedirectioncosinesofthex-axisare1,0,0.

whichistherequiredequationofthecone.

Example15.5.4

Arightcircularconehasitsvertexat(2,−3,5).ItsaxispassesthroughA(3,−2,6)anditssemiverticalangleis30°.Finditsequation.

Solution

Theaxisisthelinejoiningthepoints(2,−3,5)and(3,−2,6).

Therefore,itsequationsare .

Thedirectionratiosoftheaxesare1,1,1.

Thedirectionratiosoftheaxesare1,1,1.

Thedirectioncosinesare

LetP(x,y,z)beanypointonthecone.

Squaring,crossmultiplyingandsimplifyingweget,5x2+5y2+5z2−8xy−8yz−8zx−4x+86y−58z+278=0.

Example15.5.5

Arightcircularconehasthreemutuallyperpendiculargenerators.Provethatthe

semiverticalangleoftheconeis

Solution

Theequationoftherightcircularconewithvertexattheorigin,semiverticalangleαandaxisalongz-axisisgivenbyx2+y2=z2tan2α.Thisconewillhavethreemutuallyperpendiculargeneratorsifcoefficientof

x2+coefficientofy2+coefficientofz2=0.

Example15.5.6

TheaxisofarightconevertexO,makesequalangleswiththecoordinateaxesandtheconepassesthroughthelinedrawnfromOwithdirectioncosinesproportionalto(1,−2,2).Findtheequationtothecone.

Solution

Lettheaxisoftheconemakeanangleβwiththeaxes.Thenthedirectioncosinesoftheaxesarecosβ,cosβ,cosβ.(i.e.)1,1,1.Letαbethesemiverticalangleoftheaxisofthecone.Thedirectionratiosofoneofthegeneratorsare1,−2,2.

LetP(x,y,z)beanypointonthecone.ThenthedirectionratiosofOParex,y,z.Thedirectionratiosoftheaxisare1,1,1.

Squaringandcrossmultiplyingweget,

9(x+y+z)2=x2+y2+z2or4x2+4y2+4z2+9xy+9yz+9zx=0

Example15.5.7

Provethatthelinedrawnfromtheoriginsoastotouchthespherex2+y2+z2+2ux+2vy+2wz+d=0lieontheconed(x2+y2+z2)=(ux+vy+wz)2.

Solution

Thelinesdrawnfromtheorigintotouchthespheregeneratestheenvelopingcone.

TheequationoftheenvelopingconeofthegivensphereisT2=SS1.

Example15.5.8

Showthattheplanez=0cutstheenvelopingconeofthespherex2+y2+z2=11whichhasitsvertexat(2,4,1)inarectangularhyperbola.

Solution

Theequationoftheenvelopingconewithitsvertexat(2,4,1)isT2=SS1.

Thesectionofthisconebytheplanez=0is(2x+4y−11)2=10(x2+y2−11).Coefficientofx2+coefficientofy2=6−6=0Hence,theplanez=0cutstheenvelopingconeinarectangularhyperbola.

Exercises2

1. Findtheequationoftherightcircularconewhosevertexistheline andwhichhasa

verticalangleof60°.

Ans.:19x2+13y2+3z2−8xy−24yz−12zx=0

2. If(x,y,z)isanypointontheconewhosevertexis(1,0,2)andsemiverticalangleis30°andthe

equationtotheaxisis ,showthattheequationoftheconeis27[(x−1)2+y2+(z−

2)2]=4(x+2y−2z+3)2.3. Findtheequationtotherightcircularconeofsemiverticalangle30°,whosevertexis(1,2,3)and

whoseaxisisparalleltothelinex=y=z.

Ans.:5(x2+y2+z2)−8(yz+zx+xy)+30x+12y−6z−18=0

4. Findtheequationtotherightcircularconewhosevertexis(3,2,1),semiverticalangleis30°and

axisistheline

Ans.:7x2+37y2+21z2−16xy−12yz−48zx+38x−88y+126z−32=0

5. Findtheequationoftherightcircularconewithvertexat(1,−2,−1),semiverticalangle60°and

axis

Ans.:5[(5x+4y+14)2+(3z−5x+8)2+(4x+3y+2)2]

=75[(x−1)2+(y+2)2+(z+1)2]

6. Findtheequationoftherightcircularconewhichpassesthroughthethreelinesdrawnfromthe

originwithdirectionratios(1,2,2)(2,1,−2)(2,−2,1).

Ans.:8x2−5y2−4z2+yz+5zx+5xy=0

7. Linesaredrawnthroughtheoriginwithdirectioncosinesproportionalto(1,2,2),(2,3,6),(3,4,12).Findtheequationoftherightcircularconethroughthem.Alsofindthesemiverticalangleofthecone.

Ans:

8. Findtheequationoftheconegeneratedwhenthestraightline2y+3z=6,x=0revolvesaboutthez-axis.

Ans.:4x2+4y2−9z2+36z−36=0

9. Findtheequationtotherightcircularconewhichhasthethreecoordinateaxesasgenerators.Ans.:xy+yz+zx=0

10. Findtheequationoftherightcircularconewithitsvertexatthepoint(0,0,0),itsaxisalongthey-axisandsemiverticalangleθ.

Ans.:x2+z2=y2tan2θ

11. Ifαisthesemiverticalangleoftherightcircularconewhichpassesthroughthelinesox,oy,x=y

=z,showthat

12. Provethatx2+y2+z2−2x+4y+6z+6=0representsarightcircularconewhosevertexisthepoint(1,2,−3),whoseaxisisparalleltooyandwhosesemiverticalangleis45°.

13. Provethatthesemiverticalangleofarightcircularconewhichhasthreemutuallyperpendicular

tangentplanesis

14. Findtheenvelopingconeofthespherex2+y2+z2−2x−2y=2withitsvertexat(1,1,1).

Ans.:3x2−y2+4zx−10x+2y−4z+6=0

15. Findtheenvelopingconeofthespherex2+y2+z2=11whichhasitsvertexat(2,4,1)andshowthattheplanez=0cutstheenvelopingconeinarectangularhyperbola.

15.6TANGENTPLANE

Tangentplanefromthepoint(x1,y1,z1)totheconeax2+by2+cz2+2fyz+2gzx+2hxy=0.Theequationsofanylinethroughthepoint(x1,y1,z1)are

Anypointonthelineis(x1+lr,y1+mr,z1+nr).

Ifthispointliesonthegivenconethen

Thisequationisaquadraticinr.Since(x1,y1,z1)isapointonthecone

Therefore,onerootoftheequation(15.36)iszero.Ifline(15.35)isatangenttothecurve,thenboththerootsareequalandhencetheotherrootmustbezero.Theconditionforthisisthecoefficientofr=0.

Hencethisistheconditionfortheline(15.35)tobeatangenttothecurveatthepoint(x1,y1,z1).Sinceequation(15.38)canbesatisfiedforinfinitelymanyvaluesofl,m,n

thereareinfinitelymanytangentlinesatanypointofthecone.Thelocusofallsuchtangentlinesisobtainedbyeliminatingl,m,nfrom

(15.35)and(15.38).

Therefore,theequationofthetangentplaneat(x1,y1,z1)is

Note15.6.1:(0,0,0)satisfiestheaboveequationandhencethetangentplaneatanypointofaconepassesthroughthevertex.

ThetangentplaneatanypointofaconetouchestheconealongthegeneratorthroughP.

Proof:Lettheequationoftheconebeax2+by2+cz2+2fyz+2gzx+2hxy=0.LetP(x1,y1,z1)beanypointonthecone.TheequationofthetangentplaneatPis(ax1+hy1+gz1)x+(hx1+by1+

fz1)y+(gx1+fy1+cz1)z=0.

TheequationsofthegeneratorthroughPare

Anypointonthislineis(rx1,ry1,rz1).Theequationofthetangentplaneat(rx1,ry1,rz1)is(arx1+hry1+grz1)x+(hrx1+bry1+frz1)y+(grx1+fry1+crz1)z=0.

Dividingbyr,(ax1+hy1+gz1)x+(hx1+by1+fz1)y+(gx1+fy1+cz1)z=0whichisalsotheequationofthetangentplaneat(x1,y1,z1).Therefore,thetangentplaneatPtouchestheconealongthegeneratorthroughP.

15.6.1ConditionfortheTangencyofaPlaneandaCone

Lettheequationoftheconebe


Lettheplane(15.40)touchtheconeat(x1,y1,z1).Theequationofthetangentplaneat(x1,y1,z1)is

Iftheplane(15.40)touchesthecone(15.39)thenequations(15.40)and(15.41)areidentical.Thereforeidentifying(15.40)and(15.41)weget,

Alsoas(x1,y1,z1)liesontheplane

Eliminating(x1,y1,z1),r1from(15.43),(15.44),(15.45)and(15.46)weget,

Simplifyingthisweget,

whereA,b,C,F,G,Harethecofactorsofa,b,c,f,g,hinthedeterminant

Hence(15.47)istherequiredconditionfortheplane(15.40)totouchthecone.

15.7RECIPROCALCONE

15.7.1EquationoftheReciprocalCone

Letusnowfindtheequationoftheconereciprocaltothecone

Letatangentplanetotheconebe

Thenwehavethecondition

whereA,b,C,F,G,Harethecofactorsofa,b,c,f,g,hinthedeterminant

Theequationofthelinethroughthevertex(0,0,0)ofthecone(15.48)andnormaltothetangentplane(15.49)are

Thelocusof(4)whichisgotbyeliminatingl,m,nfrom(15.50)and(15.51)is

Thisistheequationofthereciprocalcone.

Note15.7.1.1:Ifwefindthereciprocalconeof(15.52)wegettheequationofconeasax2+by2+cz2+2fyz+2gzx+2hxy=0.

Definition15.7.1.2:Twoconesaresaidtobereciprocalconesofeachotherifeachoneisthelocusofthenormalthroughthevertextothetangentplanesoftheother.

15.7.2AnglebetweenTwoGeneratingLinesinWhichaPlaneCutsaCone

Lettheequationoftheconeandtheplanebe

Letoneofthelinesofthesectionbe

Sincethislineliesontheplanewehave,

from(15.56)

Substitutingthisin(15.55)weget,

Thisisaquadraticequationin

Therearetwovaluesfor

Hencethegivenplaneintersectstheconeintwolinesnamely,

Alsosince aretherootsoftheequation(15.57)

whereP2=−(Au2+Bv2+Cw2+2Fvw+2Gwu+2Huv)andA,B,C,F,G,H

arethecofactorsofa,b,c,fg,hinthedeterminant

Itfollowsbysymmetry

Ifθistheacuteanglebetweenthelinesthen

Note15.7.2.1:Ifthetwolinesareperpendicularthen(a+b+c)(u2+v2+w2)−f(u,v,w)=0.

(i.e.)f(u,v,w)=(a+b+c)(u2+v2+w2)Note15.7.2.2:Ifthelinesofintersectionarecoincidentthenθ=0.

Thisistheconditionfortheplaneux+vy+wz=0tobeatangentplanetothecone.

15.7.3ConditionforMutuallyPerpendicularGeneratorsoftheCone

Wehaveseenthattheconditionfortheplaneux+vy+wz=0cuttheconeintwoperpendiculargeneratorsisthat

Ifthereisathirdgeneratorwhichisperpendiculartotheabovetwolinesofintersectionthenitmustbeanormaltotheplaneux+vy+wz=0.Therefore,itsequationsare

Since(15.59)isageneratoroftheconef(x,y,z)=0,weget

(15.58)and(15.60)holdsifandonlyifa+b+c=0.Therefore,theconditionfortheconetohavethreemutuallyperpendicular

generatorsisa+b+c=0.

Example15.7.1

Findtheanglebetweenthelinesofsectionoftheplane3x+y+5z=0andthecone6yz−2zx+5xy=0.

Solution

Lettheequationsofthelineofsectionbe

Asthislineliesontheplaneandalsoontheconeweget

From(15.62)m=−(3l+5n)Substitutingthisin(15.63)weget,

Solvingl+n=0and3l+m+5n=0weget,

Solvingl+2n=0and3l+m+5n=0weget,

Therefore,thedirectionratiosofthetwolinesare1,2,−1and2,−1,−1.

Ifθistheanglebetweenthelines

Therefore,theacuteanglebetweenthelinesis

Example15.7.2

Provethattheanglebetweenthelinesgivenbyx+y+z=0,ayz+bzx+cxy=

0is ifa+b+c=0.

Solution

Theplane

willcutthecone

intwolinesthroughthevertex(0,0,0).Theequationsofthelinesofthesectionare

wherel,m,narethedirectionratiosofthelines.Sincethelinegivenby(15.66)liesontheplaneandthecone

Substitutingn=−(l+m)in(15.67)weget

Ifl1,m1,n1andl2,m2,n2arethedirectionratiosofthetwolinesweget,

Similarlywecanshowthat

Ifθistheanglebetweenthelines

Example15.7.3

Provethattheconesax2+by2+cz2=0and arereciprocal.

Solution

Theequationofthereciprocalconeax2+by2+cz2=0is

whereA,B,Carethecofactorsofa,b,cin

Theequationofthereciprocalconeisbcx2+cay2+abz2=0or .

Similarly,wecanshowthatthereciprocalconeof isax2+by2+

cz2=0.Therefore,thetwogivenconesarereciprocaltoeachother.

Example15.7.4

Showthattheequation representsaconewhichtouchesthe

coordinateplanes.

Solution

Squaring(fx+gy−hz)2=4fgxy

Thisbeingahomogeneousequationofseconddegreeinx,y,z,itrepresentsacone.Whenthisconemeetstheplanex=0weget,(gy−hz)2=0.Hencetheaboveconeiscutbytheplanex=0incoincidentlinesandhencex

=0touchesthecone.Similarly,y=0,z=0alsotouchthecone.

Exercises3

1. Findtheanglebetweenthelinesofthesectionoftheplanesandcones:

i. x+3y−2z=0,x2+9y2−4z2=0

ii. 6x−10y−7z=0,108x2−20y2−7z2=0

Ans:

2. Showthattheanglebetweenthelinesinwhichtheplanex+y+z=0cutstheconeayz+bzx+

cxy=0is

3. Provethattheequationa2x2+b2y2+c2z2−2bcyz−2cazx−2abxy=0representsaconewhichtouchesthecoordinateplane.

4. If representsoneofthegeneratorsofthethreemutuallyperpendiculargeneratorsofthe

cone5yz−8zx−3xy=0thenfindtheothertwo.

Ans:

5. If representsoneofthethreemutuallyperpendiculargeneratorsofthecone11yz+6zx−

14xy=0thenfindtheothertwo.

Ans:

Chapter16

Cylinder

16.1DEFINITION

Thesurfacegeneratedbyavariablelinewhichremainsparalleltoafixedlineandintersectsagivencurve(ortouchesagivensurface)iscalledacylinder.Thevariablelineiscalledthegenerator,thefixedstraightlineiscalledthe

axisofthecylinderandthegivencurveiscalledtheguidingcurveofthecylinder.

16.2EQUATIONOFACYLINDERWITHAGIVENGENERATORANDAGIVENGUIDINGCURVE

Letusfindtheequationofthecylinderwhosegeneratorsareparalleltotheline

andwhoseguidingcurveistheconic

Let(α,β,γ)beanypointonthecylinder.Thentheequationsofageneratorare

Letusfindthepointwherethislinemeetstheplanez=0.Whenz=0,

Thispointis

Whenthegeneratormeetstheconic,thispointliesontheconic.

Thelocusofthepoint(α,β,γ)is

Thisistherequiredequationofthecylinder.

Note1:Ifthegeneratorsareparalleltoz-axisl=0,m=0,n=lthentheequationofthecylinderbecomes,

Note2:Theequationf(x,y)=0inspacerepresentsacylinderwhosegeneratorsareparalleltoz-axis.

16.3ENVELOPINGCLINDER

Thelocusofthetangentlinesdrawntoasphereandparalleltoagivenline

Letthegivenspherebe

Letthegivenlinebe

Let(α,β,γ)beanypointonthelocus.Thenanylinethrough(α,β,γ)paralleltoline(16.5)is

Anypointonthislineis(α+lr,β+mr,γ+nr)Ifthepointliesonthesphere(16.4),then

Thisisaquadraticequationinrgivingthetwovaluesforrcorrespondingtotwopointscommontothesphereandtheline.Ifthelineisatangentthenthetwovaluesofrmustbeequalandhencethediscriminantmustbezero.

Thelocus(α,β,γ)is

whichisacylinder.Thiscylinderiscalledtheenvelopingcylinderofthesphere.

Envelopingcylinderasalimitingformofanenvelopingcone

Let betheaxisoftheenvelopingcylinder.Anypointonthislineis

(lr,mr,nr).Letthispointbethevertexoftheenvelopingcone.Thentheequationofthe

envelopingconeisT2=SS1.

whichistheequationtotheenvelopingcylinder.

16.4RIGHTCIRCULARCYLINDER

Arightcircularcylinderisasurfacegeneratedbyastraightlinewhichremainsparalleltoafixedstraightlineataconstantdistancefromit.Thefixedstraightlineiscalledtheaxisofthecylinderandtheconstantdistanceiscalledtheradiusofthecylinder.Theequationofarightcircularcylinderwhoseaxisisthestraightline

andwhoseradiusisa.

LetP(x,y,z)beanypointonthecylinder.LetAA′betheaxisofthecylinder.DrawPLperpendiculartotheaxisandPL=a.LetQ(α,β,γ)beapointontheaxisofthecylinder.

Thisistherequiredequationoftherightcylinder.


Example16.1

Findtheequationofthecylinderwhosegeneratorsareparalleltotheline

andwhoseguidingcurveisx2+y2=9,z=1.

Solution

LetP(x,y,z)beapointonthecylinder.

TheequationsofthegeneratorthroughPandparalleltotheline are

Theguidingcurveis

WhenthegeneratorthroughPmeetstheguidingcurve,

Sincethispointliesonthecurve(16.9),

Thisistheequationoftherequiredcylinder.

Example16.2

Findtheequationofthecylinderwhichintersectsthecurveax2+by2+cz2=1,lx+my+nz=pandwhosegeneratorsareparalleltoz-axis.

Solution

Theequationoftheguidingcurveis

Sincethegeneratorsareparalleltoz-axistheequationofthecylinderisoftheformf(x,y)=0.Theequationofthecylinderisobtainedbyeliminatingzinequation(16.10)

Substitutingthisinax2+by2+cz2=1,weget,


Example16.3

Findtheequationofthecylinderwhosegeneratorsareparalleltotheline

andwhoseguidingcurveistheellipsex2+2y2=1,z=3.

Solution

Theequationtotheguidingcurveis

Let(x1,y1,z1)beapointonthecylinder.Thentheequationsofthegenerator

throughP(x1,y1,z1)are

Whenthislinemeetstheplanez=3,wehave,

Thispointliesonthecurvex2+2y2=1.


Example16.4

Findtheequationofthesurfacegeneratedbythestraightliney=mx,z=nxand

intersectingtheellipse

Solution

Thegivenliney=mx,z=nxcanbeexpressedinsymmetricalformas

LetP(x1,y1,z1)beanypointonthecylinder.ThentheequationsofthegeneratorthroughPare

Thismeetsthecurve

Whenz=0,

Thispointliesonthecurve


whichistheequationoftherequiredcylinder.

Example16.5

Findtheequationofthecircularcylinderwhosegeneratinglineshavethedirectioncosinesl,m,nandwhichpassesthroughthecircumferenceofthefixedcirclex2+y2=a2onthexozplane.

Solution

LetP(x1,y1,z1)beanypointonthecylinder.Thentheequationsofthegenerator

throughPare

Thismeetstheplaney=0.

Thispointliesonthecurve


whichistherequiredequationofthecylinder.

Example16.6

Findtheequationsoftherightcircularcylinderofradius3withequationsofaxis

as

Solution

Theequationsoftheaxisare

(1,3,5)isapointontheaxis.2,2,–1arethedirectionratiosoftheaxis.

∴directioncosinesare

LetP(x1,y1,z1)beanypointonthecylinder.

Also,PQ2=QL2+LP2



Example16.7

Findtheequationoftherightcircularcylinderwhoseaxisisx=2y=–zandradius4.

Solution

Theequationsoftheaxisofthecylinderare

LetP(x1,y1,z1)beanypointonthecylinder.TheequationsofthegeneratorthroughPare

Thedirectioncosinesoftheaxisare .

Also,PQ2=QL2+LP2


Example16.8

Findtheequationofthecylinderwhosegeneratorshavedirectioncosinesl,m,nandwhichpassesthroughthecirclex2+z2=a2,y=0.

Solution

LetP(x1,y1,z1)beanypointonthecylinder.TheequationofthegeneratorsthroughPare

Thislinemeetsthecurvey=0,x2+z2=a2

Thispointlieson



Example16.9

Findtheequationoftherightcircularcylinderwhoseaxisis and

passesthroughthepoint(0,0,3).

Solution

Theequationsoftheaxisofthecylinderare

LetP(x1,y1,z1)beanypointonthecylinder,then



Example16.10

Findtheequationtotherightcircularcylinderwhichpassesthroughcirclex2+y2+z2=9,x–y+z=3.

Solution

Fortherightcircularcylinder,theguidingcurveisthecirclex2+y2+z2=9,x–y+z=3.Therefore,thedirectionratiosoftheaxisofthecylinderare1,–1,1.LetP(x1,y1,z1)beanypointonthecylinder.ThentheequationsofthegeneratorthroughPare

Anypointonthislineis(r–x1,–r+y1,r+z1).Ifthispointliesonthecircle,then

Eliminatingrfrom(16.16)and(16.17)weget

Simplifying,thelocusof(x1,y1,z1)is

whichistheequationoftherequiredcylinder.

Example16.11

Findtheequationtotherightcircularcylinderofradiusawhoseaxispassesthroughtheoriginandmakesequalangleswiththecoordinateaxes.

Solution

Letl,m,nbethedirectioncosinesoftheaxisofthecylinder.

Theaxispassesthroughtheorigin.LetP(x1,y1,z1)beanypointonthecylinder.



Example16.12

FindtheequationtotherightcircularcylinderdescribedonthecirclethroughthepointsA(1,0,0),B(0,1,0)andC(0,0,1)astheguidingcurvex2+y2+z2–yz–zx–xy=1.

Solution

TheequationofthesphereOABCis

TheequationoftheplaneABCis

Therefore,theequationofthecircleABCis

Thecentreofthesphereis .

TheequationsofthelineCNare

whichistheaxisofthecylinder.Thedirectionratiosoftheaxisare1,1,1.

Thedirectioncosinesoftheaxisare


whichistherequiredequationofthecylinder.

Example16.13

Findtheequationoftheenvelopingcylinderofthespherex2+y2+z2–2x+4y=1havingitsgeneratorsparalleltothelinex=y=z.

Solution

LetP(x1,y1,z1)beanypointonatangentwhichisparalleltotheline

Therefore,theequationofthetangentlinesare

Anypointonthislineis(x1+r,y1+r,z1+r).Thispointliesinthissphere

Ifequation(16.22)touchesthesphereofequation(16.23),thenthetwovaluesofrofthisequationareequal.

Onsimplifyingwegetthelocusof(x1,y1,z1)as

whichistherequiredequation.

Exercises

1. Findtheequationofthecylinder,whoseguidingcurveisx2+z2–4x–2z+4=0,y=0andwhoseaxiscontainsthepoint(0,3,0).Findalsotheareaofthesectionofthecylinderbyaplaneparalleltothexzplane.

2. Findtheequationofthecylinder,whosegeneratorsareparalleltotheline andpassing

throughthecurvex2+2y2=1,z=0.3. Provethattheequationofthecylinderwithgeneratorsparalleltoz-axisandpassingthroughthe

curveax2+by2=2cz,lx+my+nz=pisn(ax2+by2)+2c(lx+my)–2pc=0.

4. Findtheequationofthecylinder,whosegeneratorsareparalleltotheline andpasses

throughthecurvex2+y2=16,z=0.5. Findtheequationtothecylinderwithgeneratorsparalleltoz-axiswhichpassesthroughthecurve

ofintersectionofthesurfacerepresentedbyx2+y2+2z2=12andlx+y+z=1.

6. Findtheequationofthecylinder,whosegeneratorsintersecttheconicax2+2hxy+by2=1,z=0andareparalleltothelinewithdirectioncosinesl,m,n.

7. Acylindercutstheplanez=0withcurve andhasitsaxisparallelto3x=–6y=2z.

Finditsequation.

8. Astraightlineisalwaysparalleltotheyzplaneandintersectsthecurvesx2+y2=a2,z=0andx2

=az,y=0.Provethatitgeneratesthesurfacex4y2=(x2–az)2(a2–x2).9. Findtheequationofarightcircularcylinderofradius2whoseaxispassesthrough(1,2,3)and

hasdirectioncosinesproportionalto2,1,2.10. Findtheequationoftherightcircularcylinderofradius2whoseaxispassesthrough(1,2,3)and

hasdirectioncosinesproportional2,–3,6.

11. Findtheequationoftherightcircularcylinderofradius1withaxisas

12. Findtheequationoftherightcircularcylinderwhosegeneratorsareparallelto and

whichpassesthroughthecurve3x2+4y2=12,z=0.13. Findtheequationoftherightcircularcylinderofradius4whoseaxisisx=2y=–z.14. Findtheequationoftherightcircularcylinderwhoseguidingcurveisthecirclethroughthepoint

(a,0,0),(0,b,0),(0,0,c).

15. Findtheequationoftheenvelopingcylinderofthespherex2+y2+z2–2x+4y=1havingitsgeneratorsparalleltothelinex=y=z.

16. Findtheenvelopingcylinderofthespherex2+y2+z2–2y–4z=11havingitsgeneratorsparalleltothelinex=–2y=2z.

17. Findtheequationoftherightcylinderwhichenvelopesasphereofcentre(a,b,c)andradiusranditsgeneratorsparalleltothedirectionl,m,n.

Answers

1.10x2+5y2+13z2+12xy+4xz+6yz–36x–30y–18z+36=0

2.3x2+6y2+3z2–2xz+8yz−3=0

4.9x2+9y2+5z2–6xz–12yz–144=0

5.3x2+3y2+4xy–4x–4y–10=0

7.36x2+9y2+17z2+6yz–48xz–9=0

9.5x2+8y2+5z2–4yz–8zx–4xy+22x–16y–14z–10=0

10.45x2+40y2+13z2+36yz–24zx+12xy–42x–280y–126z+294=0

11.10x2+5y2+13z2–12xy–6yz–4zx–8x+30y–74z+59=0

12.27x2+36y2+112z2–36xz–120yz–180=0

13.5x2+8y2+5z2–4xy+4yz+8zx–144=0

14.

15.x2+y2+z2–xy–yz–zx–4x+5y–z–2=0

16.5x2+8y2+8z2+4xy+2yz–4xz+4x–18y–36z=99

17.(l2+m2+n2)[(x–a)2+(y–b)2+(z–c)2–r2]

=[l(x–a)+m(y–b)+n(z–c)]2

Acknowledgements

IexpressmysincerethankstoPearsonEducation,India,especiallytoK.Srinivas,Sojan,Charles,andRameshfortheirconstantencouragementandforsuccessfullybringingoutthisbook.

P.R.Vittal

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analytical geometry: 2d and 3d

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