co-ordinate geometry- questions set

Co-ordinate Geometry: Questions Set 1

Co-ordinate Geometry [Book: A Text Book on Co-ordinate Geometry with Vector Analysis by Rahman and Bhattacharjee]

Co-ordinate Geometry: Transformation of co-ordinates axes and its uses; Equation of conics and its reduction to

standard forms; Pair of straight lines; Homogeneous equations of second degree; Angle between a pair of straight

lines; Pair of lines joining the origin to the point of intersection of two given curves, circles; System of circles;

Orthogonal circles; Radical axis, radical center, properties of radical axes; Coaxial circles and limiting points;

Equations of parabola, ellipse and hyperbola in Cartesian and polar co-ordinates; Tangents and normals, pair of

tangents; Chord of contact; Chord in terms of its middle points; Pole and polar parametric co-ordinates; Diameters;

Conjugate diameters and their properties; Director circles and asymptotes.

Rh20/Chap4, KhoshMd(KM)14/Chap2, Isa6/Chap2

2 Change of Axes

1. Transformation of Co-ordinates 6, 2. Translation of axes 6, 3. Rotation of axes 7,

4. Simplification of the equation of a curve by transformation of coordinates 8, 5. Invariants 10,

6. Illustrative examples 12, Exercise-II 19.

Transformation of coordinates

The process of changing the coordinate of a point or equation of a curve by changing the origin or

the direction of axes is called transformation of coordinates. We proceed to establish the

fundamental formulae for such transformation of coordinates.

2.2. Translation of axes (change of origin).

To find the change in the coordinates of a point when the origin is shifted to another point

but the direction of axes remain unaltered.

Fig.2.1

(x - 1)2 + (y - 2)

2 = 1 becomes 2 2 1x y when 1 and 1x x y y

Thus, to transfer the origin to the point (,), the formulae of transformation are x x and y y . Which implies x x and y y .

N

y y

O N X

X L

O

P

y y

x

x O

O(,)

Coordinate Geometry: Lecs

2

2.3. Rotation of axes: (Origin fixed).

To rotate the axes through an angle , the formulae of transformation are

sincos yxx and cossin yxy

To find the change in the coordinates of a point when the directions of axes is turned through an

angle but the origin of coordinates remains the same.

Fig.2.2

It is to be noted that when the direction of axes are turned through an angle , the transformed

equation of a curve is obtained by substituting )sincos( yx and )cossin( yx for x and

y respectively in the equation of the curve.

Using polar coordinates to show the rotation formula

We know cos , sinx r y r . If rotation angle is , then

cos( ) cos sin

sin( ) cos sin

x r x y

y r y x

** If the angle of rotation is then will be replaced by .

cos sin

cos sin

x x y

y y x

; In matrix form it can be written as X = AX where The vectors

, x x

X Xy y

and the -rotation matrix is cos sin

cos sinA

Solving for x and y we get,

cos sin

sin cos

x x y

y x y

The transforming equations for x, y and x , y may be conveniently remembered from the following scheme:

x y

x cos sin

y sin cos

y y

P

L

x

x M N

M


3

2.5. Invariants.

If by the rotation of the rectangular coordinate axes about the origin through an angle the

expression 22 2 byhxyax changes to 22 2 ybxyhxa then

a b a b and 2 2ab h a b h

Proof: Replacing x and y in the expression 22 2 byhxyax we have

2 2

2 2

( cos sin ) 2 ( cos sin )( sin cos ) ( sin cos )

2

a x y h x y x y b x y

a x h xy b y

or, cossin2sincos 22 hbaa

2 2sin cos 2 sin cosb a b h

and 2 22 2 (cos sin ) 2( )sin cosh h a b

Now, (i)

2 2 2

2

cos sin 2 sin cos sin

cos 2 sin cos

a b a b h a

b h

)cos(sin)sin(cos 2222 ba

a b (proved)

(ii)

2 22 2 cos 4 sin cos 2 sin

(1 cos 2 ) 2 sin 2 (1 cos 2 )

( ) 2 sin 2 ( )cos 2

a a h b

a h b

a b h a b

similarly, 2 ( ) 2 sin 2 ( )cos2b a b h a b

Now, 2 24( ) (2 )(2 ) (2 )a b h a b h

= 2 2 2( ) {2 sin 2 ( )cos2 } {2 cos2 ( )sin 2 }a b h a b h a b

[used 2 2( )( )A B A B A B ]

= 2 2 2 2( ) 4 ( ) 4 4a b h a b ab h

2 2 ab h a b h (proved)

Hence, the quantities ba and 2hab of a second degree expression are invariants due to rotation of axes.

Removal of xy-term ( product term)

Now suppose we want to remove the product term ( x y term) then the coefficients of yx should

vanish and hence we get 0cossin)(2)sin(cos2 22 bah ,

i.e. 02sin)(2cos2 bah


4

i.e., ba

h

22tan i.e.,

ba

h2tan

2

1 1 .

Hence if the axes are be rotated through an angle

ba

h2tan

2

1 1 then xy term in the

expression 22 2 byhxyax vanishes .

So if it is necessary to remove the product term of an equation, the axes should be rotated through

an angle such that

ba

h2tan

2

1 1 .

[ So if it is necessary to remove the product term of an equation, the axes should be rotated

through an angle such that

ba

h2tan

2

1 1 .]

1. Rh23/Ex1

Determine the equation of the curve 2 22 3 8 6 7 0x y x y when the origin is transferred to

the point (2, 1). The equation will be reduced to 2 22 3 18x y .

Removing the dashes the equation is 2 22 3 18x y (Ans).

2. Rh23/Ex2

Determine the equation parabola 2 22 2 4 3 0x xy y x y after rotating of axes through

45o .

Hints. For 45o rotation ,

1 1( ) , ( )

2 2x x y y x y , put them in the given equation

and after simplification the given equation reduces to 22 2 3 2 3 0]y x y

Now dropping the dashes, the equation is 22 2 3 2 3 0y x y (Ans).

3. Rh24/Ex4, Isa20/Exr5

By transforming to parallel axes through a properly chosen point (h, k), prove that the equation

0251121012 22 yxyxyx

can be reduced to one containing only terms of the second degree.

Ans. 021012,2

5,

2

3 22

yxyxkh

4. KhM22/Ex


5

Transform the equation 2 29 24 2 6 20 41 0x xy y x y in rectangular coordinates so as to

remove the terms in x, y and xy. [Ans. 2 218 7 54 0x y ]

5. Rh25/Exr11, Isa19/Exr3

The equation 0502218323 22 yxyxyx is transformed to 124 22 yx when

referred to rectangular axes through the point (2, 3). Find the inclination of the latter axes to the

former. [Ans. 45].

6. Isa21/Exr14(iii)

Simply the following equations by suitable translation and rotation of axes:

0502218323 22 yxyxyx [Ans. 124 22 yx ]

Transfer the origin to (1, 1) then equation becomes 2 29 24 2 54 0x xy y

To remove xy term let axes be rotated through the angle then

12 24 3 3tan 2 or , tan or , tan7 4 4

h

a b

so, 3 4

sin , cos5 5

substituting these and simplifying the transformed equation is 2 218 7 54 0x y (Ans)

Rh51/Art48

3.1. To find the condition that the general equation of second degree in x and y may represent a

pair of straight lines.

1. The general equation of second degree 022222 cfygxbyhxyax ... ... .. (1)

will represent a pair of straight lines if = 0 where the symbol represents the determinant

a h g

h b f

g f c

(i) Lines are parallel if ab = h2 and (ii) Lines are perpendicular if a + b = 0.

0:

2. a circle if a = b , h = 0

3. a parabola if ab = h2

4. an ellipse if ab > h2

5. a hyperbola if ab < h2 , (i) if a + b = 0 then rectangular hyperbola.


6

2 2 22

a h g

h b f abc fgh af bg ch

g f c

Rh51/Art49

Centre of a Conic

Let 2 22 2 2F ax hxy by gx fy c

Then 2( ) 0F

ax hy gx

and 2( ) 0

Fhx by f

y

.

Solving these two equations 0ax hy g and 0hx by f , we get

2

1x y

hf bg gh af h ab

or

2 2( , ) ,

hf bg gh afx y

h ab h ab

.

7. Rh25/Exr11

Transform the equation 0502218323 22 yxyxyx to its standard form and then

identify the conic.

Hints. Centre is (2, 3) , Find angle then use invariants

a b a b and 2ab h a b (as 0h ), calculate 4 and 2a b . Finally, get the

transformed equation in standard form as 2 24 2 1x y .

Now dropping the primes, the equation is 12422 yx , which is an ellipse.

Rh53/

Working rule for the reduction of Conic to its standard form

Equation of conic is 022222 cfygxbyhxyax

Let 2 22 2 2F ax hxy by gx fy c

Then 2( )F

ax hy gx

and 2( )

Fhx by f

y

.

If (x1, y1) be the centre of the conic then 1 1 0ax hy g and 1 1 0hx by f (3)

Solving, we get

1 1

2

1x y

hf bg gh af h ab

or 1 1 2 2( , ) ,

hf bg gh afx y

h ab h ab

. (4)

If we transfer the origin to (x1, y1) the equation becomes

2 2

12 0ax hxy by c

where,


7

2 2

1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1

2 2 2

( ) ( )

(0) (0) [by (3)]

c ax hx y by gx fy c

x ax hy g y hx by f gx fy c

x y gx fy c

So, New constant is 1 1 1c gx fy c

Put the values of x1 and y1 from (4) then 2 2 2

1 2 2

2abc fgh af bg chc

ab h ab h

Calculate 1 1 1 2 from g or from c x fy c ab h

.

Then new equation 2 2 12 0ax hxy by c (5)

Which can be written as, 2 22 1Ax Hxy By [when c1 0 that is 0]

** If = 0, then c1 = 0 then equation (5) will represent two straight lines.

Find angle from

ba

h2tan

2

1 1

So, if the axes are be rotated through an angle then xy term will disappear. Then equation (5)

will be transferred to 2 21 1 1 0a x b y c

We can calculate 1 1, a b from the invariants 1 1 a b a b and 2

1 1a b ab h [as h1 = 0]

8. Rh54/Ex

Reduce the equation 2 232 52 7 64 52 148 0x xy y x y to the standard form and then

identify the conic.

Let 2 2( , ) 32 52 7 64 52 148 0f x y x xy y x y

Then 0f

x

or, 64 52 64 0x y

and 0f

y

or, 52 14 52 0x y

Solving, x = 1, y = 0 that is centre is at (x1 , y1) = (1, 0).

New constant, 1 1 1c gx fy c = 32(1) 26(0) 148 = 180

The equation of conic referred as origin is 2 232 52 7 180 0x xy y

When the xy-term is removed by the rotation of axes, let the reduced equation be 2 2

1 1 180a x b y ... ... ... (3)

Then 1 1 a b = 327 = 25 ; [Here , a = 32 , b = 7 , h = 26]

and 21 1a b ab h = 32(7) (26)2 = 900


8

2 2

1 1 1 1 1 1( ) ( ) 4a b a b a b = (25)2 4( 900) = 4225

or , 1 1 a b = 65

So, we have 1 145, 20a b

The equation (3) is 2 245 20 180x y or, 2 2

14 9

x y , which is a hyperbola

Therefore the given equation represents a hyperbola.

PSL1 (Hand note)

Rh26/Chap5, KhoshMd(KM)49/Chap4, Isa23/Chap3

3 Pair of Straight Lines

y = 2x , y = 3x or, y 2x = 0 , y 3x = 0

(y 2x)( y 3x) = 0

2 25 6 0y xy x which represents a pair of straight lines.

Homogeneous equation of the second degree

02 22 byhxyax ... ... ... (1)

always represents a pair of straight lines through the origin.

If (1) represents the straight lines y m1 x = 0 , y m2 x = 0 then

(y m1x)( y m2 x) = 2 22h ay xy x

b b

or, 1 2 1 22

, h a

m m m mb b

Angle between two lines Represented by

3.7. To find the angle between the lines represented by the equation

02 22 byhxyax ... ... ... ... (1)

Let the separate equations of the lines given by (1) be 01 xmy and 02 xmy so that

b

hmm

221 and .21

b

amm If be the angle between the lines, then

21

212

21

21

21

1

4)(

1

~tan

mm

mmmm

mm

mm


9

b

a

b

a

b

h

1

44

2

2

,

b

hmm

221 and

b

amm 21

ba

abh

bab

babh

2

2

2

)(1

12

.2

tan2

1

ba

abh

The general equation of second degree in x and y

0222 22 cfygxbyhxyax ... ... ... (1)

represents a pair of straight lines if

0

a h g

h b f

g f c

Expanding the determinant, the condition can be written as

2 2 22 0abc fgh af bg ch

If equation (1) represents a pair of straight lines, then the equation

02 22 byhxyax ... ... ... (2)

represents a pair of straight lines through the origin, parallel to those represented by equation (1),

then the equation

22tan

h ab

a b

;

Parallel if: = 0 or, tan = 0 or, 2 0h ab or, 2h ab

Perpendicular if: a + b = 0 [2 or, cot = 0 that is a + b = 0 ]

PSL2 (Hand note)

Point of intersection

The coordinates of the point of intersection of the two lines given by (1) are

22

,hab

afgh

hab

bghf i.e.

C

F

C

G, where C, G, F are the cofactors of the elements c, g, f of the

determinant

a h g

a b f

g f c


10

Also, the point of intersection is obtained by solving 0S

x

and 0

S

y

where

2 2( , ) 2 2 2S x y ax hxy by gx fy c

[See Rh51/Art49 / Centre of a Conic]

Bisectors of the angles between the lines

KhM 56/Art. 33, Isa28/Art.3.9

To find the equations of the bisectors of the angles between the lines given by

02 22 byhxyax ... ... ... ... (1)

Let the separate equations of the lines given by (1) be

01 xmy and 02 xmy

where

b

hmm

221 and

b

amm 21

the equations of the bisectors are

22

2

21

1

11 m

xmy

m

xmy

(Squaring or)

Their joint equation is

01111 22

2

21

1

22

2

21

1

m

xmy

m

xmy

m

xmy

m

xmy

or, 01

)(

1

)(22

22

21

21

m

xmy

m

xmy

or, )2)(1(22

1122

2 xmxymym

0)2)(1( 222222 xmxymym

or, 0))(1(2))(( 2121222

221 xymmxmyxmm

or, 0)1(2))(( 2122

21 xymmyxmm

or, 012)(2 22

xy

b

ayx

b

h

or, 0)()(22 xybayxh

or, h

xy

ba

yx

22 ... ... ... ... ... ... ... (2)


11

Observation:

The bisectors given by (2) are perpendicular to each other.

KhM 58/Cor. 33, Isa30/Art.3.11

To find the equations of the bisectors of the angles between the lines given by

0222 22 cfygxbyhxyax (1)

Let (, ) be the point of intersection of the lines given by (1). Referred to parallel axes though

(, ), as origin,

so that x X and y Y . Which implies X x and Y y .

the equation of the given lines reduces to the homogeneous form 2 22 0aX hXY bY ,

Whose bisectors angles are given by 2 2X Y XY

a b h

Reverting now to the old axes, this equation becomes h

yx

ba

yx ))(()()( 22

which is the required equations of bisectors of the angles between the lines given by (1).

Pair of lines joining the origin to the points of intersection of the curve

KhM 59/Art. 34, Isa32/Art.3.14

A special pair of lines

To find the equation of the pair of lines joining the origin to the points of intersection of the curve

0222 22 cfygxbyhxyax ... ... ... ... (1)

with the line, 0 nmylx ... ... ... ... ... .... (2)

Making equation (1) homogeneous with the help of the equation (2) which is written in the form

1

n

mylxwe get,

0)(22

2

22

n

mylxc

n

mylxfygxbyhxyax ... ... (3)

On simplification, this equation assumes the form 2 22 0Ax Hxy By which is a second

degree homogenous equation. The equation (3) being a homogeneous second degree equation,

represents a pair of straight lines through the origin. Moreover, it is satisfied by the coordinates of

points which satisfy (1) and (2).

Alternate

For circle


12

2 2 1x y in (1)

Making equation (1) homogeneous with the help of the equation (2)

2

2 2 lx myx yn

which contains terms with 2 2, and x y xy and is of the form

2 22 0Ax Hxy By ; where 2

21

lA

n ,

lmH

n ,

2

21

mB

n

Which represents a pair of straight lines through the origin.

PSL3 (Hand note)

Rh39/Exr5Q8-15, KhM67/Exr4,Q1b Isa37/Ex2

9. KhM60/Ex1

Find for what values of the equation 2 212 36 6 6 3 0x xy y x y

represents a pair of straight lines.

Solution: we have 12, , 3, 18, 3, 3a b c h g f .

2 2 22 0abc fgh af bg ch

gives 12(3) + 2(3)(3)(18) 12(3)2 (3)2 3(18)2 = 0

4 + 36 12 108 = 0 [dividing by 9]

3 = 84 or, = 28

The required values of = 28 (Ans).

10. Isa37/Ex2

Find the values of k so that the equation

021268103 22 ykxyxyx

represents a pair of straight lines and then find the angle between the lines.

Solution: The given equation is

021268103 22 ykxyxyx ... ... ... ... (1)

Comparing the given equation (1) with the equation

0222 22 cfygxbyhxyax

we have 13,2

,5,21,8,3 fk

ghcba .

The given equation, represents a pair of straight lines, if

02 222 chbgaffghabc ... ... ... ... ... ... (2)


13

Putting the values of cba ,, etc. in (2), we get

0)5(212

8)13(352

)13(22183 22

2

kk

or, 0525250765504 2 kk

or, 0528652 2 kk

or, 052832332 2 kkk

or, 0)332(16)332( kkk

or, 0)16)(332( kk

0332 k and 016 k i.e. 2

33k and k = 16 which are the required values of k.

11. KhM60/Ex2

Show that the equation 2 22 7 10 0x xy y x y

represents a pair of straight lines and then find the angle between the lines. Find also their point

of intersection and the equation of the bisectors of the angles between the two lines given by the

equation.

Solution: We have 1 1 7

2, 1, 10, , ,2 2 2

a b c h g f .

2 2 2 7 49 1 102 20 04 2 4 4

abc fgh af bg ch

The given equation represents a pair of straight lines.

Let 2 22 7 10F x xy y x y

Then 4 1F

x yx

and 2 7

Fx y

y

If (x1, y1 ) is the point of intersection then 1 14 1 0x y and 1 12 7 0x y

Solving these two equations, we get

1 1 1

7 2 1 28 8 1

x y

or 1 11, 3x y

The point of intersection is 1 1( , ) (1, 3)x y .

The equation of the bisectors of the angles between two lines is

h

yx

ba

yx ))(()()( 22

or, 2 2( 1) ( 3) ( 1)( 3)

2 1 1/ 2

x y x y

or, 2 26 20 10 0x xy y x (Ans)


14

(Contd.)

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co-ordinate geometry- questions set

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