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MAT1035Analytic Geometry

Lecture Notes

R.A. Sabri Kaan Gurbuzer

Dokuz Eylul University

2016

Contents

1 Review of Trigonometry 5

2 Polar Coordinates 7

3 Vectors in Rn 93.1 Located Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.2 The Algebra of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.3 The Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.4 The Norm of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.5 Projection of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.6 The Direction Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.7 Vector Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.8 Mixed Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

4 Straight Line 174.1 Direction of a Straight Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.2 Equations of a Straight Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.3 The Normal Form of a Straight Line in Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.4 Intersection of Two Straight Lines in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

5 The Plane 255.1 The Plane Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255.2 The Angle between two Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275.3 Other Forms of the Equation of a Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285.4 Distance from a Point to a Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.5 Intersection of two Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.6 Projecting Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325.7 Intersection of three Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325.8 Specialized Distance Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6 Transformation of Axes 376.1 Translation and Rotation of Axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

7 The Circle 397.1 The Equation of a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397.2 Intersections Involving Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407.3 Systems of Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427.4 The Angle between two Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

8 Ellipse, Hyperbola and Parabola 438.1 Ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438.2 Hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448.3 Parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3

4 CONTENTS

9 The Equations of Second Degree 49

10 Conic through Five Points and Family of Conics 51

Chapter 1

Review of Trigonometry

P (cos θ, sin θ)

P (cos−θ, sin−θ)

x2 + y2 = 1

θ

−θ adj

opp

hyp

Figure 1.1: The unit circle and trigonometric functions.

cos θ =adj

hyp, sin θ =

opp

hyp, tan θ =

opp

adj, cot θ =

adj

opp, sec θ =

hyp

adj, csc θ =

hyp

opp

Let P (x, y) be a point at which the terminal side of θ intersects the unit circle x2 + y2 = 1. Then we havecos2 θ + sin2 θ = 1. Dividing each term of this fundamental identity by cos2 θ gives the identity 1 + tan2 θ = sec2 θ.Dividing each term of the same identity by sin2 θ gives 1 + cot2 θ = csc2 θ. The followings are the addition formulas

5

6 CHAPTER 1. REVIEW OF TRIGONOMETRY

of the trigonometric functions.

sin(α± β) = sinα cosβ ± cosα sinβ

cos(α± β) = cosα cosβ ∓ sinα sinβ

sin 2θ = 2 sin θ cos θ

cos 2θ = cos2 θ − sin2 θ = 1− 2 sin2 θ = 2 cos2 θ − 1

cos2 θ =1

2(1 + cos 2θ)

sin2 θ =1

2(1− cos 2θ)

Angles frequently are measured in degrees with 360◦ in one complete revolution. In calsulus it is more convenientto measure angles in radians. The radian measure of an angle is the length of the arc it subtends in the unit circlewhen the vertex of the angle is at the center of the circle. The area of the unit disk with radius r is A = πr2 and thecircumference of the circle with radius r is C = 2πr. The circumference of the unit circle is 2π and its central angle

is 360◦. It follows that 2πrad = 360◦ ⇒ 180◦ = πrad and 1 =180◦

π. Consider an angle of θ radians at the center

of a circle of radius r. Then the proportionsS

2πr=

A

πr2=

θ

2πgive S = rθ and A = 1

2r2θ. The relationship between

the functions sin t and sin t◦ is sin t◦ = sinπt

180.

Remember the properties of the trigonometric functions sin t+2nπ = sin t and cos t+2nπ = cos t for all n ∈ Z.To find the values of trigonometric functions of angles larger than

π

2, we can use their periodicity and the following

identities.

sin(π ± θ) = ∓ sin θ

cos(π ± θ) = − cos θ

tan(π ± θ) = ± tan θ

Chapter 2

Polar Coordinates

Let O be a fixed reference point called the pole and let L be a given ray called the polar axis beginning at 0. Nowlet us begin with a given xy-coordinate system and take the origin as the pole and the nonnegative x-axis as the polaraxis.

Given the pole O and the polar axis, the point P with polar coordinates r and θ, written as the orderedpair (r, θ) is loocated as follows: First find the terminal side of θ given in radians, where θ is measuredcounterclockwise (ifθ > 0) from the x-axis (the polar axis) as its initial side.

Definition 1.

If r ≥ 0, then P is on the terminal side of this angle at the distance r from the origin. If r < 0, then P lies on theray opposite the terminal side at the distance |r| = −r > 0 from the pole.

θ θ

P

P

r > 0 r < 0

|r|

r

Figure 2.1: Polar coordinates of a point with r > 0 and r < 0.

The radial coordinate r can be described as the directed distance of P from the pole along the terminal side of θ.If r = 0, the polar coordinates (0, θ) represent the origin whatever the angular coordinate θ might be.

7

8 CHAPTER 2. POLAR COORDINATES

Polar coordinates differ from rectangular coordinates in that any point has more than one representationin polar coordinates.

Remark 1.

The polar coordinates (r, θ+ 2nπ) and (−r, θ+ (2n+ 1)π)), n ∈ Z, are the polar coordinates of the samepoint.

Remark 2.

The relation between polar coordinates and rectangular coordinates is given as follows:

[xy

]= r

[cos θsin θ

]r =

√x2 + y2 tan θ =

y

x, x 6= 0.

The correct choice of θ : If x > 0, then (x, y) lies in either the first or fourth quadrant, so −π2< θ <

π

2which is the range or the inverse tangent function. If x < 0 then (x, y) lies in the second or the third quadrant. Soθ = arctan

y

x, x > 0 and θ = π + arctan

y

x, x < 0.

Some curve have simpler equations in polar coordinates than in rectangular coordinates. The graph of an equationin the polar coordinate variables r and θ is the set of all those points P such that P has same pair of polar coordinates(r, θ) that satisfy the given equation {(r, θ)|F (r, θ) = 0}.

The polar coordinate equation of the circle with center (0, 0) and radius a > 0 is r = a. If we start withthe rectangular coordinate equation x2 + y2 = a2 of this circle and transform it using x2 + y2 = r2, weget r2 = a2.

Example 1.

Let us transform the equation r = 2 sin θ into rectangular coordinates. Now r2 = 2r sin θ and so x2+y2 =2y ⇒ x2 + (y − 1)2 = 1 is a circle whose center is (0, 1) with radius is 1. More generally, the graphsof the equations r = 2a sin θ and r = 2a cos θ are circles of radius a centered, respectively, at (0, a) and(a, 0).

Example 2.

We can transform the rectangular equation ax+ by = c of a straight line into ar cos θ + br sin θ = c. Letus take a = 1, b = 0. The polar equation of the vertical line x = c is r = c sec θ where −π

2 < θ < π2 .

Example 3.

Sketch the graph of the polar equation r = 2 + 2 sin θ.

Example 4.

Chapter 3

Vectors in Rn

The quantities, such as distance, area, volume, temprature, time densty, etc., posses only ”magnitude”. These canbe represented by real numbers and are called scalars. On the other hand, the quantities, such as force, velocity,acceleration, etc., posses both ”magnitude” and ”direction”. These quantities can be represented by arrows and arecalled vectors. Moreover, the elements of a vector space are called the vectors.

3.1 Located Vectors

Let A and B be the point in the space R3. We can denote a vector in R3 by an ordered triple. Hence, A =(a1, a2, a3), B = (b1, b2, b3). We define a located vector to be an ordered pair of points, for example, (A,B) anddenote it by

−−→AB. We visualize this an arrow from A to B. We call the point A the beginning point and the point B the

end point of the located vector−−→AB.

z

x

y

A

B−−→AB

Figure 3.1: Located vector from the point A to the point B in the three dimensional space.

Now, let us denote the located vector−−→AB by the order triple

(b1 − a1, b2 − a2, b3 − a3).

But this correspondence is not one to one. An other located vector, such as−−→CD, can also be identified by the same

ordered triple.It is clear that, the set of located vectors forms an equivalence class with the equivalence relation having the

following properties:

i. reflexive, that is,−−→AB ∼

−−→AB

ii. symmetric, that is,−−→AB ∼

−−→CD ⇒

−−→CD ∼

−−→AB.

iii. transitive, that is,−−→AB ∼

−−→CD ∧

−−→CD ∼

−−→EF ⇒

−−→AB ∼

−−→EF.

Each equivalence class called a vector. Therefore−−→AB = B − A =

−−−−−−−→O(B −A) and bi − ai are called the

coordinates of−−→AB.

9

10 CHAPTER 3. VECTORS IN RN

Determine the beginning point of the vector (2, 6, 0) whose end point is (1,−1, 3).

Question 1.

For what values of a, b and c are the vectors (2a− b, a− 2b, 6) and (−2, 2, a+ b− 2c) equal ?

Question 2.

3.2 The Algebra of Vectors

Two vectors −→a = (a1, a2, a3) and−→b = (b1, b2, b3) are equivalent if and only if a1 = b1, a2 = b2, a3 = b3.

Geometrically, −→a and−→b have the same direction and their magnitudes (or lengths) are equal.

The addition of vectors is defined by coordinate-wise, that is,−→a +

−→b = (a1 + b1, a2 + b2, a3 + b3)

and addition can geometrically be considered by the following triangle or parallelogram laws.

−→a−→a −→a

−→b

−→b

−→b

−→a +−→b

−→a +−→b

Triangle Law Prallelogram Law

Figure 3.2: Addition of vectors

Since the vectors are the elements of a vector space, we can also define−−→b . Geometrically, −

−→b has the opposite

direction of−→b but equal size.

Two vectors −→a and−→b in Rn are parallel if and only if −→a = k ·

−→b , where k ∈ R.

Compute ~u+ ~v, ~u− ~v, 2~u− 3~v, 3~u+ 12~v and sketch them if ~u = (2,−1, 4) and ~v = (1, 2,−3).

Question 3.

Show that−−→AB +

−−→BC +

−→CA = ~0 for a triangle

4ABC.

Question 4.

Show that−−→AD = 1

2(−→AC +

−−→AB) for the median vector

−−→AD of a triangle

4ABC.

Question 5.

3.3. THE SCALAR PRODUCT 11

Let the points A,B and C be collinear and let the point D do not lie on the line containing the points A,Band C. If

−→AC = 3

−−→AB and

−−→DC = k

−−→AD + p

−−→DB then k · p =?

Question 6.

Let G be a barycenter of a triangle4

ABC. Show that

a)−→GA+

−−→GB +

−−→GC = ~0

b) for any point K,−−→KA+

−−→KB +

−−→KC = 3

−−→KG.

Question 7.

Let4

ABC be a triangle, and D and E be the midpoints of AB and AC. Then show that|−−→DE||−−→BC|

=1

2.

Question 8.

If the vectors ~u = (1, 2−m,n−m) and ~v = (3, 3m, 8m) are parallel, then find the values of m and n.

Question 9.

3.3 The Scalar Product

Let −→a ,−→b ∈ Rn. We define their scalar (or dot or inner) product −→a ·

−→b to be a bilinear map Rn × Rn → R such that

(−→a ,−→b )→ −→a ·

−→b := a1b1 + a2b2 + · · ·+ anbn =

n∑i=1

aibi (3.1)

We define two vectors−→a and−→b to be orthogonal (or perpendicular) denoted by−→a ⊥

−→b if and only if−→a ·

−→b =

0.

Let V be a vector space and <,> an inner product on V .

a) Show that < ~0, ~u >= 0 for all ~u in V .

b) Show that < ~u,~v >= 0 for all ~v in V , then ~u = ~0.

Question 10.


Let <,> be standard inner product on R2.

a) Let ~α = (1, 2) and ~β = (−1, 1). Find a vector ~γ such that < ~α,~γ >= −1 and < ~β,~γ >= 3.

b) Show that ~α =< ~α,~e1 > ~e1+ < ~α,~e2 > ~e2 for any ~α in R2. (Here, the vectors ~e1 and ~e2 arestandard basis vectors of R2.)

Question 11.

3.4 The Norm of a Vector

The norm of length of a vector −→a is denoted by ‖−→a ‖ and is defined by the number√−→a · −→a , that is,

‖−→a ‖ :=√−→a · −→a =

√a1a1 + · · ·+ anan =

√a21 + · · ·+ a2n =

√√√√ n∑i=1

a2i (3.2)

Let the vectors −→u = (3,−1),−→v = (2, 3),−→w = (4,−5) be given.

(‖−→w ‖−→u ) · −→v =√

42 + (−5)2(3 · 2 + (−1) · 3) = 3√

41

‖−→v −−→w ‖ = ‖(2− 4, 3− (−5))‖ =√

(−2)2 + 82 =√

68

Example 5.

We shall say that a vector −→e is a unit vector if ‖−→e ‖ = 1. Also we can normalize a vector −→a dividing it by its

norm, that is, −→e =−→a‖−→a ‖

.

Find the norm of ~α = (3, 4) with respect to

a) the usual inner product.

b) the inner product given as < ~x, ~y >= x1y1 − x1y2 − x2y1 + 3x2y2.

Question 12.

Find the vector−−→AB and

−−→CD, and their lengths, and also sketch them in R3 if A(−2, 1, 0) and B(−1, 1, 7)

and C(5,−2, 3) and D(2,−1, 0).

Question 13.

For −→u ,−→v ∈ Rn, |−→u ‖ · −→v | ≤ ‖−→u ‖‖−→v ‖.

Theorem 1.

3.5. PROJECTION OF A VECTOR 13

Proof. Let f : R → R be a function such that f(t) = ‖−→u + t−→v |2 = −→u · −→u + 2−→u · t−→v + t2(−→v · −→v ). Clearly, thefunction f is a second degree polynomial of t and since we consider a square of a reel number, f has at most one root.So

4(−→u · −→v )2 − 4(−→u · −→u )(−→v · −→v ) ≤ 0⇔ (−→u · −→v )2 ≤ (−→u · −→u )(−→v · −→v )⇔ |−→u · −→v | ≤ ‖−→u ‖‖−→v ‖

For −→u ,−→v ∈ Rn, ‖−→u +−→v ‖ ≤ ‖−→u ‖+ ‖−→v ‖.

Theorem 2.

Proof. By using Cauchy-Schwartz inequality, we get

‖−→u +−→v ‖2 = −→u · −→u + 2−→u · −→v +−→v · −→v ≤ −→u · −→u + 2‖−→u ‖‖−→v ‖+−→v · −→v = (‖−→u ‖+ ‖−→v ‖)2

Since both sides of inequality are square of positive numbers, we have

‖−→u +−→v ‖ ≤ ‖−→u ‖+ ‖−→v ‖.

3.5 Projection of a Vector

Let −→a and−→b be two vectors in the same space and

−→b 6= 0. We wish to find the projection of −→a along

−→b , that is, the

vector −→p in the following figure.

−→a −−→p−→a

−→p −→b

Figure 3.3: Projection of a vector

It is clear that −→p = c−→b , for some c ∈ R. On the other hand, −→a − −→p ⊥

−→b . Hence we have the following

computations.

(−→a − c−→b ) ·−→b = 0⇒ −→a ·

−→b − c

−→b−→b = 0⇒ −→a

−→b = c

−→b−→b ⇒ c =

−→a ·−→b

‖−→b ‖2

(3.3)

The vector −→p := c−→b =

(−→a ·−→b )

‖−→b ‖2

−→b is called the projection of −→a along

−→b and the scalar c is called the

component of −→a along−→b .

Our construction gives us a geometric interpretation for the scalar product. Let θ be the acute angle between −→aand−→b . The cosine of θ is computed as follows:

cos θ =‖c−→b ‖

‖−→a ‖=c‖−→b ‖

‖−→a ‖=

(−→a ·−→b )

‖−→b ‖2

‖−→b ‖‖−→a ‖

=−→a ·−→b

‖−→a ‖‖−→b ‖

(3.4)


−→a − c−→b

−→a

c−→b −→

b

θ

Figure 3.4: Angle between two vectors

Let −→u = (3, 1) and −→v = (−1,−2). Then

cosα =3 · (−1) + 1 · (−2)√

32 + 12√

(−1)2 + (−2)2=−5√50

=−1√

2

Example 6.

Find the projection of the vector ~u = (1, 2, 2) onto ~v = (−3, 4, 5)).

Question 14.

Find the angle between the vectors ~u and ~v if ~u = (1, 1, 2) and ~v = (−1, 1,−1).

Question 15.

Let4

ABC be a triangle. Prove the cosine theorem.

Question 16.

3.6 The Direction Angles

The unit vectors ei = (0, 0, . . . , 0, 1, 0, . . . , 0, 0), where 1 sits in the i-th coordinate, are called the natural basevectors. The natural base vectors are mutually perpendicular.

Consider the vector−→a making the angles α1, . . . , αn with the natural base vectors−→e1 , . . . ,−→en, respectively. Theseangles are called the direction angles of−→a and their cosines are called the direction cosines of−→a . Since each vectoris a linear combination of the natural base vectors, −→a · −→ei = ai. So the cosine of αi is computed by cosαi =

ai‖−→a ‖

.

The direction cosines are related by the equation∑n

i=1 cos2 αi = 1.

For 1 ≤ i ≤ n, the coordinate −→a · −→ei = ai is called the direction number of −→a . Any set proportional to thecoordinates are also called the direction numbers of −→a , such as ka1, . . . , kan, where k ∈ R.

3.7. VECTOR PRODUCT 15

3.7 Vector Product

Let −→a and−→b be two vectors in R3. The vector product or cross product of −→a and

−→b is a antisymmetric bilinear

map R3 × R3 → R3 defined by

−→a ×−→b = (a2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1) (3.5)

The cross product of the natural base vectors in R3 is given as follows:

−→e1 ×−→e2 = −→e3 −→e2 ×−→e1 = −−→e3−→e2 ×−→e3 = −→e1 −→e3 ×−→e2 = −−→e1−→e3 ×−→e1 = −→e2 −→e1 ×−→e3 = −−→e2

We can simply get the same result by the Sarrus Rule using the following determinant.

−→a ×−→b =

∣∣∣∣∣∣−→e1 −→e2 −→e3a1 a2 a3b1 b2 b3

∣∣∣∣∣∣ (3.6)

Compute the followings:

• ~x× ~y if ~x = (1, 2, 1) and ~y = (1,−1, 1).

• ~u× ~v if ~u = (1,−5, 8) and ~v = (−2, 4, 1).

Question 17.

If ~u× ~v = ~u× ~ω then can we necessarily say that ~v = ~ω?

Question 18.

Let −→a ,−→b ,−→c ,

−→d ∈ R3 and k ∈ R. The vector product satisfies the following properties.

1. −→a ×−→b = −

−→b ×−→a .

2. −→a × (−→b +−→c ) = (−→a ×

−→b ) + (−→a ×−→c ) and (−→a +

−→b )×−→c = (−→a ×−→c ) + (

−→b ×−→c ).

3. (k−→a )×−→b = k(−→a ×

−→b ) = −→a × (k

−→b ).

4. (−→a ×−→b )×−→c = (−→a · −→c )

−→b − (

−→b · −→c )−→a .

5. −→a ×−→b is perpendicular to both −→a and

−→b .

6. (−→a ×−→b ) · (−→c ×

−→d ) = (−→a · −→c )(

−→b ·−→d )− (−→a ·

−→d )(−→b · −→c ).

7. −→a //−→b in R3 if and only if −→a ×

−→b = 0.

8. ‖−→a ×−→b ‖ = ‖−→a ‖‖

−→b ‖ sin θ, where θ is the acute angle between the vectors −→a and

−→b .

Let ~u,~v and ~ω be arbitrary vectors in 3-space. Is there any difference between (~u×~v)×~ω and ~u×(~v×~ω)?

Question 19.

The geometric interpretation of the last property may be given as the area of the parallelogram formed with thevectors −→a and

−→b as two adjacent sides.


θ

−→a

−→b

h

Figure 3.5: The parallelogram formed by two vectors

3.8 Mixed Product

The combination of a vector product and a scalar product is called mixed product (or triple product). The mixedproduct of the vectors −→a ,

−→b ,−→c ∈ R3 is defined by

(−→a ,−→b ,−→c ) := −→a · (

−→b ×−→c ) =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ (3.7)

Since interchanging the adjacent rows of a determinant changes the sign of the determinant, we have the followingequalities.

(−→a ,−→b ,−→c ) = −(

−→b ,−→a ,−→c ) = (

−→b ,−→c ,−→a ) = −(−→c ,

−→b ,−→a ) = (−→c ,−→a ,

−→b ) = −(−→a ,−→c ,

−→b )

The geometric interpretation of triple product may be given as the volume of the parallelepiped formed with thevectors −→a ,

−→b and −→c as adjacent edges.

−→a

−→b

−→c

−→b ×−→c

hθ

Figure 3.6: The Parallelepiped formed with the vectors in three dimensional space.

Chapter 4

Straight Line

4.1 Direction of a Straight Line

The direction angles of a straight line are the direction angles of a vector lying on that line. If α1, . . . , αn are directionangles of a line then for some λ ∈ R different form zero, we have a1 = λ cosα1, . . . , an = λ cosαn. The numbersa1, . . . , an are called a set of direction numbers of the line and the vector −→a = (a1, . . . , an) is called the directionvector of the line. It is clear that the component of any vector are a set of direction numbers of any line on which thevector lies. Conversely, if a1, . . . , an are direction numbers of a line then the vectors (a1, . . . , an) and (−a1, . . . ,−an)are oppositely directed vectors on that line.

Let a1, . . . , an and b1, . . . , bn be the the direction numbers of two lines l1 and l2, respectively.

i. The lines l1 and l2 are perpendicular if and only ifn∑i=1

aibi = 0.

ii. The lines l1 and l2 are parallel if and only ifa1b1

=a2b2

= · · · = anbn.

• The angle θ between two lines l1 and l2 is given by cos θ = ∓∑n

i=1 aibi√∑ni=1 a

2i

√∑ni=1 b

2i

.

Theorem 3.

4.2 Equations of a Straight Line

Let a1, . . . , an be a set of direction numbers of a line containing the point P0(p1, . . . , pn), then −→a = (a1, . . . , an)is a vector parallel to the line. Our discussion is to find the locus of the points lying on the straight line. We take anarbitrary point to determine all the points lying on the straight line. Let the arbitrary point be P (x1, . . . , xn).

According to the Figure 4.1, we have the equation−−→OP =

−−→OP0 + t−→a , where the parameter t ∈ R. Hence we have

the vector equation (vector form) of the straight line.

(x1, . . . , xn) = ((p1, . . . , pn) + t(a1, . . . , an) (4.1)

which is also written as follows.

x1 = p1 + ta1... (4.2)

xn = pn + tan

17

18 CHAPTER 4. STRAIGHT LINE

x

l

x2

xn

−→a

O

P0(p1, . . . , pn)

P (x1, . . . , xn)

Figure 4.1: Straight Line in n-Space

The equations 4.2 are called the parametric equations of the line in terms of a point P0(p1, . . . , pn) and a set ofdirection numbers a1, . . . , an.

If none of the quantities a1, . . . , an vanishes we may solve each of the equations in parametric equation for t andobtain

x1 − p1a1

= · · · = xn − pnan

(= t). (4.3)

The equation 4.3 is called the symmetric equation of the line passing through the point P0(p1, . . . , pn) with direction−→a = (a1, . . . , an).

Under which condition the points A = (x1, y1, z1), B = (x2, y2, z2) and C = (x3, y3, z3) are collinear?

Question 20.

Let P1 and P2 be two distinct points in 3-space.

a) Show that ~P = t ~P1 + (1− t)~P2, t ∈ R is a vector equation of the line passing through P1 and P2.

b) Show that the line segment joining P1 to P2 is given by [P1, P2] = t ~P1 + (1− t)~P2, t ∈ [0, 1] .

Question 21.

Write the equations of the line whose parametric equations are x = 5 − 2t, y = 3 + t, z = −2 − 3t insymmetric form.

Question 22.

Write the equations of the line through the points (7,−1, 2) and (3, 2,−4).

Question 23.

Write the equations of the line parallel to x−12 = y + 4 = − z

3 and through (4, 2,−3).

Question 24.

4.2. EQUATIONS OF A STRAIGHT LINE 19

In 2-space we can express the equations of a line passing through the point P0(p1, p2) with direction numbers a, bas

x− p1a

=y − p2b

. (4.4)

We can express this equation as

y =b

a(x− p1) + p2 (4.5)

letting m = ba , y − p2 = m(x1 + p1) and letting n = p2 −mp1,

y = mx+ n. (4.6)

The ratio m = tanα = ba is called the slope of the straight line. The equation 4.5 is called the point-slope form

of the straight line and the equation 4.6 is called the slope-intercept form of the straight line. In equation 4.6, they-intercept is (0, n).

On the other hand, given two points define a direction, hence a straight line passing through these points. Forinstance in R2 let the points P0(p1, p2) and Q0(q1, q2) be given. Then we have a straight line passing through P0 andQ0 with direction

−→d =

−−−→P0Q0 and the equation

−−→OP =

−−→OP0 + t

−−−→P0Q0. One can also use the equality of the slope

tanα =q2 − p2q1 − p1

=y − p2x− p1

. (4.7)

The equation 4.7 is called the two point form of a line.The equation of the straight line whose x and y-intercepts are respectively (a, 0) and (0, b) is

x

a+y

b= 1. (4.8)

The equation 4.8 is called the intercept form of the line.Every equation of the first degree in x and y, that is, a straight line in R2, may be reduced to the form

ax+ by + c = 0, (4.9)

where a, b, c are arbitrary constants. The equation 4.9 is called general form of a line in R2.

Two nonvertical lines l1 and l2 in R2 are parallel if and only if their slopes are equal. Two nonvertical linesare perpendicular if and only if the slpoe of one is the negative of the reciprocal of the slope of the other,

that is, two lines with slopes m1 and m2 are perpendicular to each other if and only if m2 =−1

m1.

Theorem 4.

Find the coordinates of midpointM(x1, x2) of the line segment joining the points P1(3, 7) and P2(−2, 3).

Question 25.

Find the coordinates of the point P (x, y) which divides the line segment joining the points P1(−2, 5) andP2(4,−1), in the ratio at 6

5 and −2, respectively.

Question 26.


Find the slope and direction cosines of the line which is perpendicular to the line joining the points P1(2, 4)and P2(−2, 1).

Question 27.

Find the angle between the directed lines joining P1(1, 3), P2(−4,−3), and P3(2, 0) and P4(−5, 6) byslopes and direction cosines.

Question 28.

Given the triangle whose vertices are A(−5, 6), B(−1,−4) and C(3, 2) derive the equations of the threemedians and solve algebraically for their point of intersection.

Question 29.

Find the intercepts of the line perpendicular to 2x+ 3y − 7 = 0 and passing through the point (1, 6).

Question 30.

Find the equations of the lines through (1,−6) if the product of the intercepts for each line is 1.

Question 31.

What are the direction cosines of a line perpendicular to x− 5y + 3 = 0.

Question 32.

Find the equations of all the lines with slope m = −34 such that each line will occur an area 24 unit2 with

the coordinate axes.

Question 33.

Determine the parameter k such that

a) the line 3kx+ 5y + k − 2 = 0 passes through A(−1, 4).

b) the line 4x− ky − 7 has the slope 3.

c) the line whose equation 2x+ ky + 3 = 0 shall make an angle 45o with the line 2x+ 5y − 17 = 0.

Question 34.

4.3. THE NORMAL FORM OF A STRAIGHT LINE IN PLANE 21

4.3 The Normal Form of a Straight Line in Plane

Let l be a straight line which does not pass through the origin. If the length of the perpendicular drawn from the originto the line l is p and the angle that the perpendicular makes with the x-axis is ω then we find the equation of the line l.

x

y

O

ω

p

x1

y1 A(x1, y1)

Q(x, y)

Figure 4.2: Normal form of a straight line

LetA(x1, y1) be the intersection of the perpendicular line segment and the line l. Our aim is to define the arbitrary

point Q(x, y) on the line. Form the Figure, x1 = p cosω, y1 = p sinω and the slope of l is−1

tanω. Then by the

equation 4.5, y − y1 =−1

tanω(x − x1) or y − p sinω =

−1

tanω(x − p cosω). This equation reduces to the equation

4.10.

y sinω + x cosω = p (4.10)

The equation 4.10 with p > 0 is called the normal form of a line not passing through the origin.If ax + by + c = and x cosω + y sinω − p = 0 are the general and normal form of the same straight line l then

the coefficients of the two equations are proportional. Hence

cosω

a=

sinω

b=−pc

= k,

where k is the constant ratio. Then we have cosω = ka, sinω = kb and k = ± 1√a2 + b2

. Hence

cosω =±a√a2 + b2

∧ sinω =±b√a2 + b2

∧ p =∓c√a2 + b2

.

Then the normal form of ax+ by + c = 0 is

a

±√a2 + b2

x+b

±√a2 + b2

+c

±√a2 + b2

= 0. (4.11)

Since p > 0, the sign before the square root is chosen the opposite sign of c.To find the perpendicular distance d from the line l to the point x1, y1, draw the line L1 through the point (x1, y1)

which is parallel to l.If the normal equation of l is x cosω + y sinω = p then the normal equation of L1 is x cosω + y sinω = p+ d.

Since (x1, y1) ∈ L1, the coordinates (x1, y1) satisfy the equation of L1, x1 cosω + y1 sinω = p. Hence the distanced is obtained by

d = x1 cosω + y1 sinω − p. (4.12)

If (x1, y1) and the origin are on opposite sides of the line l then the distance d is positive and if they are on the sameside of the line l then d is negative.


xO

y

L1

l

p d(x1, y1)

Figure 4.3: Distance from a point to a line

Find the perpendicular distances from origin to the lines

a) x− 3y + 6 = 0

b) 15x− 8y − 25=0

Question 35.

Find the equation of the line through the point of intersection of the lines x−3y+1 = 0 and 2x+5y−9 = 0and whose distance from the origin is

√5.

Question 36.

Calculate the distance d from the line 5x − 12y − 3 = 0 to point (2,−3). Are the point (2,−3) and theorigin on the same side?

Question 37.

Determine the equations of the bisectors of the angles between the lines l1 : 3x + 4y − 2 = 0 andl2 : 12x− 5y + 5 = 0.

Question 38.

4.4 Intersection of Two Straight Lines in Space

Two straight line in space may be parallel or not, which is easily determined by their direction numbers. If they arenot parallel then they may still have no point of intersection, in which case they are called skew lines.

We shall illustrate the method of intersection of straight lines by using the parametric form. Let us discuss the

intersection of the two lines l1 :x− x1a1

=y − y1b1

=z − z1c1

(= t) and l2 :x− x2a2

=y − y2b2

=z − z2c2

(= u). The

4.4. INTERSECTION OF TWO STRAIGHT LINES IN SPACE 23

−→d1

−→d1

−→d1

l1 l1 l1l2 l2 l2

−→d2 =

−→d1

−→d2 =

−→d1

l1 = l2

−→d2

−→d2

Figure 4.4: Position of two lines in 3-space

parametric equations are

x = x1 + ta1 x = x2 + ua2

y = y1 + tb1 y = y2 + ub2

z = z1 + tc1 z = z2 + uc2

We have to discuss whether they have common point(s) or not. Hence

x = x1 + ta1 = x2 + ua2 a1t+ a2u = x1 − x2y = y1 + tb1 = y2 + ub2 ⇒ b1t+ b2u = y1 − y2z = z1 + tc1 = z2 + uc2 c1t+ c2u = z1 − z2

There are 3 equations and 2 unknowns (u, t). Let A =

a1 a2b1 b2c1 c2

and b =

x1 − x2y1 − y2z1 − z2

The system is consistent ifRank(A) = Rank(A : b), that is, we have solution forRank(A) = Rank(A :b).

Remark 3.

Here we have two cases If Rank(A) = Rank(A : b) = 2 then the system has a unique solution and hence thelines intersect at a unique point. If Rank(A) = Rank(A : b) = 1 we have infinite solutions, hence the lines intersectat infinitely many points, that is, the lines are coincide. If Rank(A) < Rank(A : b) then the system has no solutionand hence the straight lines have no point in common, which means that they are parallel or skew.

Discuss the intersection of the lines l1 and l2 if

a) l1 : x+ 1 = y − 1 = z−3−2 and l2 : −2x+ 2 = 6− 2y = z + 1.

b) l1 : x+ 1 = y − 1 = z−3−2 and l2 : −2x− 4 = −2y − 2 = z − 5.

c) l1 : x+ 1 = y − 1 = z−3−2 and l2 : x−3

2 = y − 4 = z−12 .

Question 39.

Given l = {(x, y, z) : x−12 = y+2

−1 = z3}, find a line l′ through (1, 2, 3) such that l and l′ are skew.

Question 40.

Chapter 5

The Plane

5.1 The Plane Equation

Let P0(x0, y0, z0) be a fixed point and let P (x, y, z) be an arbitrary point in the plane. Let a, b, c be a set of directionnumbers of any line perpendicular to the plane. Let −→n be the vector −→n = (a, b, c) which is direction vector of theline perpendicular to the plane. Since

−−→P0P lies in the plane, −→n is perpendicular to

−−→P0P . Hence

−→n ·−−→P0P = −→n · (

−→P −

−→P0) = 0. (5.1)

x

y

z

O

P0 P

−→n

−−→P0P

Figure 5.1: Vector form of a plane

The equation 5.1 is called the vector form of the equation of the plane which contains the point P0 and is perpen-dicular to the direction −→n . The vector −→n is called the normal to the plane. If the vectors −→n ,

−→P and

−→P0 are expressed

in terms of their coordinates then we obtain the nonvector forms of the plane given in the following equations

−→n ·−−→P0P = 0⇒ −→n · (

−→P −

−→P0) = 0

(a, b, c) · (x− x0, y − y0, z − z0) = 0

a(x− x0) + b(y − y0) + c(z − z0) = 0 (5.2)

Let d = −(ax0 + by0 + cz0)⇒ ax+ by + cz + d = 0 (5.3)

25

26 CHAPTER 5. THE PLANE

In each of the equations 5.2 and 5.3, the coefficients a, b and c are a set of direction numbers of a normal to theplane. The equation 5.2 is called the standard form and the equation 5.3 is called the general form of the equation ofthe plane.

Any plane may be represented by a linear equation and conversely any linear equation in R3 represents aplane.

Theorem 5.

Proof. The direct statement has been proved by driving the equation 5.3. To prove the converse, suppose that we have

a linear equation ax + by + cz + d = 0. We may assume that c 6= 0 and then write ax + by + c(z − d

c) = 0 or

a(x− 0) + b(y − 0) + c(z − d

c) = 0. It follows that the vector through P (x, y, z) and P0(0, 0,

d

c) is perpendicular to

the fixed direction −→n = (a, b, c). This is just a plane equation passing through P0 with normal −→n .

Under which condition, four points A = (x1, y1, z1), B = (x2, y2, z2), C = (x3, y3, z3) and D =(x4, y4, z4) are coplanar, i.e. lie on a same plane?

Question 41.

For which value of λ, the points A(2,−3, λ), B(1, 0, 2), C(2,−1, 2) and D(1,−1, 2) are coplanar ?

Question 42.

Find the equation of the plane perpendicular to the line joining (1, 3, 5) and (4, 3, 2) at the midpoint ofthese two points.

Question 43.

Write equation of the plane passing through (1, 1, 2) and (3, 5, 4) which is perpendicular to the xy-plane.

Question 44.

Find the equation of the plane perpendicular to the plane x − y + 2z − 5 = 0, parallel to the line whosedirection cosines are 1

5 ,−25 ,

2√5

5 and passing through (1, 4, 1).

Question 45.

Find the equation of the plane parallel to the 3x− 2y + 6z + 5 = 0 and passing through (1, 4, 1).

Question 46.

5.2. THE ANGLE BETWEEN TWO PLANES 27

Find the equation of the plane parallel to the xz-plane and through (1, 4, 6).

Question 47.

Find an equation of the plane through (1, 2, 3) and perpendicular to the line l = {(x, y, z) : x−12 = y+2

−1 =z3}.

Question 48.

Find an equation of the plane through (1, 2, 3) and l = {(x, y, z) : x−12 = y+2

−1 = z3}.

Question 49.

Find an equation of the plane determined by the lines l1 = {(x, y, z) : x + 1 = −y + 1 = z} andl2 = {(t− 2, 2t− 1,−3t+ 3) : t ∈ R}.

Question 50.

5.2 The Angle between two Planes

By definition, the angle between two planes is the angle between their normals. For the given planes

P1 : a1x+ b1y + c1z + d1 = 0

P2 : a2x+ b2y + c2z + d2 = 0

the direction numbers to the normals are −→n1 = (a1, b1, c1) and −→n2 = (a2, b2, c2), respectively. Then we have

cosα =−→n1 · −→n2‖−→n1‖‖−→n2‖

=a1a2 + b1b2 + c1c2√

a21 + b21 + c21√a22 + b22 + c22

. (5.4)

P1 P2

−→n1−→n2α

Figure 5.2: Angle between two planes

If α =π

2in 5.4 then the planes are perpendicular to each other. If α = 0 or α = π then the normals to the plane

are parallel to each other and so the planes are parallel.


Determine the angle between the planes x+ y − z + 7 = 0 and 3x− y − 6 = 0.

Question 51.

Find the cosine of the angle between the two planes with equations

a) P1 : x+ y + z − 1 = 0 and P2 : 2x− y + 2z + 4 = 0,

b) P3 : x+ 2y + z − 1 = 0 and P4 : 4x− 2y − z − 3 = 0,

c) P5 : x− 6y + 2z − 1 = 0 and P6 : x− 2y + 2z − 3 = 0.

Question 52.

5.3 Other Forms of the Equation of a Plane

If a plane intersects a coordinate plane in a straight line, that line is called a trace of that plane. If a plane intersects acoordinate axis in a point, that point is called an intercept of the plane. The intercepts may be found from the equationsof the traces and the general form of the equation of the plane. Suppose a, b, c and d are all different from zero in the

equation ax+by+cz+d = 0. Calling x, y and z-interceptsA,B and C, respectively, we find thatA = −da,B = −d

b

and C = −dc. Dividing each coefficient in the general form ax + by + cz + d = 0 by −d and taking the reciprocals

we getx

(−da)

+y

(−db )

+z

(−dc )

+ 1 = 0. Hence the general form may be written in the form

x

A+y

B+z

C= 1. (5.5)

The equation 5.5 is called the intercept form of the equation of the plane. Let us write the vector form of theequation of a plane as

−→n ·−→P = t, (5.6)

where −→n is the normal vector of the plane and t = −→n ·−→P0. Dividing both sides by ±‖−→n ‖ and using the sign of t we

have−→n±‖−→n ‖

·−→P =

t

±‖−→n ‖. If we set

−→N =

−→n±‖−→n ‖

and p =t

±‖−→n ‖, it follows that p > 0 and

−→N is a unit vector.

The equation 5.6 may be written as

−→N ·−→P = p, p > 0. (5.7)

The equation 5.7 is the vector normal form of the equation of a plane. Since−→N is a unit vector, it may be written

in terms of its direction angles as−→N = (cosα, cosβ, cos γ). It is easily seen that the vector

−→R0 = p

−→N satisfies the

equation 5.7. The corresponding point p−→N = (p cosα, p cosβ, p cos γ) lies in the plane, p

−→N is perpendicular to the

plane and ‖p−→N ‖ = p. In nonvector form, the equation 5.7 becomes

(cosα, cosβ, cos γ) · (x, y, z) = p, p > 0

x cosα+ y cosβ + z cos γ = p, p > 0. (5.8)

The equation 5.8 is called the normal form of the equation of a plane, where α, β and γ are the direction anglesof the normal vector

−→N and p is the directed distance from origin to the plane. One can convert the general form

ax+ by+ cz+ d = 0 of the equation of a plane to the normal form by dividing the equation by ∓√a2 + b2 + c2 and

using the opposite sign of d.

5.4. DISTANCE FROM A POINT TO A PLANE 29

x

y

z

O

α

γ

β

−→R0 = p

−→N

Figure 5.3: Vector normal form of a plane

Reduce 3x+ 2y − z + 5 = 0 to normal form.

Question 53.

Find the equation of the plane parallel to the xy-plane and 2 units from it.

Question 54.

Reduce x− 5y + 2z − 3 = 0 to intercept form and write the coordinates of the intercepts.

Question 55.

Find the equation of the plane with intercepts 1,−2,−4.

Question 56.

5.4 Distance from a Point to a Plane

Similar to the normal form of the straight line in R2 one can simply obtain the distance from a point to a plane bynormal form of its equation. We want to find the distance l of the point P0 to the plane ax+ by + cz + d = 0. Let P1

be the foot of the perpendicular drawn from P0 to the plane then−−→OP0 =

−−→OP1 +

−−−→P1P0. Since −→n //

−−−→P1P0, taking the

inner product of both sides by −→n , we get


x

y

z

O

−→n = (a, b, c)

P0(x0, y0, z0)

P1(x1, y1, z1)

−−→OP0

−−→OP1

−−−→P1P0

ax+ by + cz + d = 0

Figure 5.4: Distance from a point to a plane.

−→n ·−−→OP0 = −→n ·

−−→OP1 +−→n ·

−−−→P1P0

(a, b, c) · (x0, y0, z0) = (a, b, c) · (x1, y1, z1) + ‖−→n ‖‖−−−→P1P0‖ cosα

ax0 + by0 + cz0 = ax1 + by1 + cz1 ± ‖−→n ‖‖−−−→P1P0‖

ax0 + by0 + cz0 = −d± ‖−→n ‖.lax0 + by0 + cz0 + d = ±‖−→n ‖.l

l = ±ax0 + by0 + cz0 + d

‖−→n ‖= ±ax0 + by0 + cz0 + d√

a2 + b2 + c2.

Two points lie on the same side of the plane ax + by + cz + d = 0 if the left member of ax + by + cz + d = 0has the same sign for both points and they are opposite sides if the left member has opposite signs.

Find the distance from the plane x− y + z + 2 = 0 to the point (0,-2,3).

Question 57.

Find the distance from P0 = (3,−3,−2) to the plane P = {(x, y, z) : x+ 2y − 2z + 8 = 0}.

Question 58.

Determine the locations (i.e. whether they are on the same side with the origin or not) of the pointsA(4, 5, 4), B(−5, 8,−6) and C(3, 25, 4) with respect to the plane P : 2x− 3y + z + 7 = 0.

Question 59.

5.5. INTERSECTION OF TWO PLANES 31

5.5 Intersection of two Planes

Let two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 be given. These two planes are parallel

if and only ifa1a2

=b1b2

=c1c2

and they are coinciding if and only ifa1a2

=b1b2

=c1c2

=d1d2. If these two planes are

not parallel, they intersect in a straight line which consists of those points satisfying both equations of the planes. Wehave three methods for finding the intersection line of these planes.

1. First Method: Find any two points from the intersection of the planes and then find the equation of the inter-section line.

2. Second Method: Use the normals to the planes to find the direction of the straight line. Intersection line isperpendicular to the normals of both planes (−→n1,−→n2). Hence the direction of the intersection line is −→n1 × −→n2.After finding a common point satisfying both of the equations of planes, one can easily find the equation of theintersection line.

3. Third Method: We eliminate one variable from the equations and then eliminate an other one. Hence we maywrite the symmetric equations of the straight line.

Determine l ∩ P if l and P are given by l = {(t + 1, t − 2, 3t + 1) : t ∈ R} and P = {(x, y, z) :x− y + 2z + 1 = 0}.

Question 60.

Determine l ∩ P if l and P have equations:

a) l = {(−t− 1, t+ 2, 2t− 1) : t ∈ R} and P = {(x, y, z) : x− 2y + 3z + 4 = 0}.

b) l = {(x, y, z) : −x = y+22 = z

3} and P = {(x, y, z) : −x+ 2y + 3z − 4 = 0}.

c) l = {(x, y, z) : x+63 = y − 2 = −z − 1} and P = {(x, y, z) : x− 2y + z − 4 = 0}.

Question 61.

Find the vector formulation for the line of intersection of the planes P1 : x + 2y + 3z + 4 = 0 andP2 : x− 2y + 2z − 1 = 0.

Question 62.

Find a vector equation of the line of intersection of the following planes:

a) P1 : x+ y + z − 1 = 0 and P2 : 2x− y + 2z + 4 = 0,

b) P3 : x+ 2y + z − 1 = 0 and P4 : 4x− 2y − z − 3 = 0,

c) P5 : x− 6y + 2z − 1 = 0 and P6 : x− 2y + 2z − 3 = 0.

Question 63.


Discuss the intersection of given planes with respect to the parameter λ.P1 : (2λ+ 5)x+ (λ+ 10)y + 3z − 7 = 0, P2 : 2x+ (λ+ 7)y + 2z + 15 = 0.

Question 64.

5.6 Projecting Planes

A straight line lies in infinitely many planes. Any two of them determine the line. By eliminating one of the variablesfrom two such equations we obtain an equation of a plane which contains the line and perpendicular to a coordinateplane. Thus if a line is given by the equations a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 then theelimination of y (for example) gives the plane Ax + Cz + D = 0 which is clearly perpendicular to xz-plane. Theplane Ax + Cz + D = 0 is called the projecting plane for the given line. The equations Ax + Cz + D = 0, y = 0determines a straight line in the projecting plane an xz-plane. This straight line is the projection of the given straightline on the xz-plane.

x

z

yO

a1x+ b1y + c1z + d1 = 0, a2x+ b2y + c2z + d2 = 0

intersection line

projection of the

intersection line

xz-plane

Figure 5.5: Projection of a line onto projecting plane

5.7 Intersection of three Planes

We have the following possibilities:

1. All three planes are parallel. This includes the cases all planes coincide or two of them coincide. Generallyparallelism of three planes means they have no point in common.

2. Two planes are parallel but the third one intersects these planes. If two planes coincide, there is a line ofintersection of three planes.

3. No two planes are parallel but the pairwise intersection lines are parallel. This includes the case where theplanes intersect in a single point.

4. The planes intersect in a single point.

5.7. INTERSECTION OF THREE PLANES 33

Three planes coincide. Two planes coincideand the third on isparallel to others.

Three planes are parallel.

−→n1 −→n2 −→n3−→n1 −→n2

−→n3 −→n1−→n2

−→n3

Figure 5.6: First case

−→n1 −→n2

−→n3

−→n1

−→n2

−→n3

Figure 5.7: Second case

Discuss the intersection of given planes

a) P1 : 7x+ 4y + 7z + 1 = 0, P2 : 2x− y − z + 2 = 0, P3 : x+ 2y + 3z − 1 = 0.

b) P1 : 2x− y + 3z − 5 = 0, P2 : 3x+ y + 2z − 1 = 0, P3 : 4x+ 3y + z + 2 = 0.

Question 65.

Discuss the intersection of given planes with respect to the parameters a and b.P1 : x+ 2y − z + b = 0, P2 : 2x− y + 3z − 1 = 0, P3 : x+ ay − 6z + 10 = 0.

Question 66.


−→n3

−→n2

−→n1

−→n3

−→n2 −→n1

Figure 5.8: Third case

−→n1−→n2

−→n3

Figure 5.9: Fourth case

Determine α such that the following four planes P1 : x + 2y − 3 = 0, P2 : x + y − 2z − 9 = 0,P3 : 3y + 5z + 15 = 0 and P4 : 3x+ αz − 15 = 0 pass through the same point. Find the coordinates ofthis point.

Question 67.

5.8 Specialized Distance Formula

In this section, we obtain rather specialized formulas for the distance from a point to a line and for distance betweentwo skew lines. In both cases the distances are undirected.

Let P1 be a point on the line l and let −→m be the direction of l. Let P0 be a point in R3 not containing by the line l.Let α be the acute angle between −→m and

−−−→P1P0. It is clear that the distance d from P0 to the line l is computed by the

equation d = ‖−−−→P1P0‖ sinα. Using the properties of vector product we have ‖

−−−→P1P0 × −→m‖ = ‖

−−−→P1P0‖‖−→m‖ sinα and

hence

d =‖−−−→P1P0 ×−→m‖‖−→m‖

. (5.9)

The equation 5.9 is called the specialized distance formula from a point to a line.

5.8. SPECIALIZED DISTANCE FORMULA 35

P0

P1

−−−→P1P0

−→m

d

lα

Figure 5.10: Specialized distance formula from a point to a line.

The distance between parallel lines is similar to the above case. Now we want to obtain the smallest distancebetween two skew lines. Let P1 and P2 be any two points on the skew lines l1 and l2, respectively. Let −→m1 and −→m2 bedirections of l1 and l2, respectively. It is clear that the projection of

−−−→P1P2 on −→m1 ×−→m2 is the line segment which has

the length equal to the smallest distance between l1 and l2. The projection vector−→k is computed as

−→k =

−−−→P1P2 · (−→m1 ×−→m2)

‖−→m1 ×−→m2‖2(−→m1 ×−→m2).

The norm of the projection vector−→k is the perpendicular distance between l1 and l2. Hence the distance d is obtained

by

d =

−−−→P1P2 · (−→m1 ×−→m2)

‖−→m1 ×−→m2‖(5.10)

P1 P2

l1 l2

−→m1

−→m2

d

θ−→P

−→m1 ×−→m2

Figure 5.11: Specialized distance between two skew lines.

Find the distance from (1, 0, 1) to the line l = {(x, y, z) : (x, y, z) = (2t− 1,−2t+ 2, t+ 1), t ∈ R}.

Question 68.

Find the distance between the lines l1 = {(x, y, z) : x + y − z + 3 = 0, 2x − y − 4z + 2 = 0} andl2 = {(x, y, z) : 3x+ y + 5z − 1 = 0, y + z + 2 = 0}.

Question 69.

Chapter 6

Transformation of Axes

6.1 Translation and Rotation of Axes

If two sets of parallel coordinate axes are drawn, the origin O′ of the x′y′-coordinate plane has the coordinates α, βin the xy-coordinate system. Then a point P has the coordinates (x, y) in the xy-coordinate plane and (x′, y′) in thex′y′-coordinate plane. It may be seen from the figure that two point of coordinates, that is, (x, y) and (x′, y′) arerelated by the equations

x = x′ + α, y = y′ + β. (6.1)

x

y

O

O′

y′

x′

β

α

P (x, y)(x′, y′)

y

x

Figure 6.1: Translation of axes.

The equations 6.1 are called the equations of translation. One can obtain the coordinates of a point P in xy-coordinate system using the coordinates of P in x′y′-coordinate system and the position of the origin O′. Reverse isalso true.

Let us rotate the xy-coordinate axes about the origin through an angle θ to obtain the x′′y′′- coordinate axes. Thistime let us take a point P in the x′′y′′-coordinate plane and define its coordinates in the xy-plane. It is clear that thedistance taken on the x-axis is equal to the projection on the x-axis of the distance taken on the x′′-axis minus theprojection on the x-axis of the distance taken on the y′′-axis. Similarly, y-coordinate can be obtained and they can becombined in the equations

x = x′′ cos θ − y′′ sin θ, y = x′′ sin θ + y′′ cos θ. (6.2)

37

38 CHAPTER 6. TRANSFORMATION OF AXES

x

x′′

y

x

y

y′′

x′′ cos θy′′ sin θ

θθ

P (x, y)

Figure 6.2: Rotation of axes.

The equations 6.2 are the equations of the rotation that gives the coordinates of a point in the xy-system afterrotation about the origin through an angle θ.

a) Translate axes to the new origin (2, 3) and reduce the equation of the curve x2−4y2−4x+24y−36 =0.

b) For the equation x2 + 4xy + y2 = 1 rotate the axes through θ = 45◦.

Question 70.

The x, y− axes are translated to x′, y′− axes with origin O′(2,−3). Find x′, y′ coordinates of the point(4,−6) and transform the equation 4x2 − 12y2 − 16x− 72y − 128 = 0.

Question 71.

Find x′, y′ coordinates of the point (9, 5) under the rotation of axes through the obtuse angle θ for whichtanθ = −3

4 and transform the equation 3x− 5y = 2.

Question 72.

Chapter 7

The Circle

7.1 The Equation of a Circle

A circle is defined as a set of points in a plane which are at the same distance, the radius, from a fixed point, thecenter. Let C(α, β) be the center and r > 0 be the radius. Now let P (x, y) be an arbitrary (representative) point onthe circle. Then we have ‖

−−→CP‖ = r. Hence using the distance formula

√(x− α)2 + (y − β)2 = r or

(x− α)2 + (y − β)2 = r2. (7.1)

C(α, β)

P (x, y)

α

β

Figure 7.1: A circle with radius r and center C(α, β).

The equation 7.1 is called the standard form of the equation of a circle and exhibits directly the radius and thecenter (α, β). Expending the equation 7.1 we get

x2 + y2 − 2αx− 2yβ + α2 + β2 − r2 = 0

x2 + y2 +Dx+ Ey + F = 0. (7.2)

The equation 7.2 is called the general form of the equation of a circle. It is second degree equation in x and y with noxy term and with equal coefficients of x2 and y2 terms. We can solve α, β and r in terms of E,D and F and get

α =−D

2, β =

−E2, r =

1

2

√D2 + E2 − 4F . (7.3)

Therefore from a given general form we obtain the standard form as (x +D

2)2 + (y +

E

2)2 =

D2 + E2 − 4F

4.

The equation 7.2 represents a circle if and only if D2 + E2 − 4F > 0 and it represents a single point if and only ifD2 +E2 − 4F = 0 and it represents no real graph if and only if D2 +E2 − 4F < 0. If the general form represents asingle point, the graph is sometimes called a point circle.

In either equation (x − α)2 + (y − β)2 = r2 or x2 + y2 + Dx + Ey + F = 0 three essential constants appear,α, β, r or E,D,F, respectively. This means that in general a circle may be required to satisfy three conditions whichmay be of various types.

39

40 CHAPTER 7. THE CIRCLE

7.2 Intersections Involving Circles

We have three possible cases for the intersection of a straight line and a circle;

1. they have no point in common,

2. they have distinct two point in common,

3. they have exactly one point in common.

In the third case, the line is tangent to the circle. We illustrate these cases in Figure 7.2.

Figure 7.2: Possible intersections of a circle and a straight line.

These cases may be characterized algebraically. If the equations of the line and the circle have the general forms

ax+by+c = 0 and x2 +y2 +Dx+Ey+F = 0, respectively, then for the case b 6= 0 we can find y as y = −ax+ c

band substitute this value into the general form of circle gives that

x2 + (ax+ c

b)2 +Dx+ E(−ax+ c

b) + F = 0.

The result is a quadratic equation in x. If the roots of this quadratic equation in x are not real (imaginary) then the lineand the circle do not meet. If the roots are real and distinct then there are two distinct points of intersection. Finally,if the roots are real and equal then there is one point of intersection, where the tangency appears.

Let us now consider the intersection of two different circles C1 and C2 with general forms

C1 : x2 + y2 +D1x+ E1y + F1 = 0 and

C2 : x2 + y2 +D2x+ E2y + F2 = 0.

If one equation is subtracted from the other, we get a linear equation in x and y. This is the line equation passingthrough the intersection of the circles C1 and C2. The straight line (D1 − D2)x + (E1 − E2)y + (F1 − F2) = 0,which is obtained by subtracting the two circle equations of C1 and C2, is called the radical axis of the given circlesC1 and C2. In general we have three cases, they have no common points or they have two distinct common points orthey have exactly one point in common which is the point of tangency.

Let P (a, b) be a point on the radical axis of the circles C1 and C2. Then the lengths of the tangents fromP to the points of tangency of the circles are equal.

Theorem 6.

7.2. INTERSECTIONS INVOLVING CIRCLES 41

Figure 7.3: Possible intersections of two circles

x2 + y2 + E1x+D1y + F1 = 0,

x2 + y2 + E2x+D2y + F2 = 0

(D1 −D2)x+ (E1 − E2)y + (F1 − F2) = 0

l1 l2

r1

r2

C(α1, β1) C(α2, β2)

P (a, b)

Figure 7.4: Radical axis of two circles.

Proof. The figure 7.4 shows one possible case, the proof, however, applies to all cases.

l21 = (a− α1)2 + (b− β1)2 − r21 = a2 + b2 − 2aα1 − 2bβ1 + α2

1 + β21 − r21l22 = (a− α2)

2 + (b− β2)2 − r22 = a2 + b2 − 2aα2 − 2bβ2 + α22 + β22 − r22

l21 − l22 = 2(α1 − α2)a+ 2(β1 − β2)b+ (α21 + β21 − r21)− (α2

2 + β22 − r22)

l21 − l22 = (D1 −D2)a+ (E1 − E2)b+ (F1 − F2) = 0

⇒ l21 = l22 ⇒ l1 = l2.

Find any points of intersection of the given curves and an equation of radical axis, and draw a picture.

a) x2 + y2 + 2x = 0, x2 + y2 − 2y = 0.

b) x2 + y2 + 8x− 6y − 11 = 0, x2 + y2 + 3x− y − 4.

c) 2x2 + 2y2 + 4x− 2y − 1 = 0, x2 + y2 + 3x− y − 4 = 0.

Question 73.

42 CHAPTER 7. THE CIRCLE

Show that each of circles x2 + y2 = 9, x2 + y2 − 12x + 27 = 0 and x2 + y2 − 6x − 8y + 21 = 0 istangent to other two. Do the common tangents meet in a point? If they do, find the point.

Question 74.

Discuss the intersection followings with respect to the parameter λ.

a) l : λx− y + 1 = 0, C : x2 + y2 − 10x+ 24 = 0.

b) C1 : x2 + y2 − 3x+ 8y − 5 = 0, C2 : x2 + y2 + 6x− 2y + λ = 0.

Question 75.

7.3 Systems of Circles

Let the equations C1 : x2 + y2 + D1x + E1y + F1 = 0 and C2 : x2 + y2 + D2x + E2y + F2 = 0 represent twocircles. The equation

x2 + y2 +D1x+ E1y + F1 + k(x2 + y2 +D2x+ E2y + F2) = 0 (7.4)

represents for each value of k, except −1, a circle through the points of intersection of the circles C1 and C2. Theequation 7.4 also represents the family of circles passing through the intersection of C1 and C2. When k = −1, theequation 7.4 reduces to the equation of the radical axis.

7.4 The Angle between two Circles

The angle between two curves is the angle between their tangents at the point of intersection. If the angle betweentwo circles is

π

2then the circles ar called orthogonal.

θ

Figure 7.5: Angle between two circles.

Show that the circles C1 : x2 + y2− 6x− 2y+ 2 = 0 and C2 : x2 + y2− 4x+ 4y+ 6 = 0 intersect eachother under right angle.

Question 76.

Chapter 8

Ellipse, Hyperbola and Parabola

8.1 Ellipse

An ellipse is the locus of a point in a plane the sum of whose distances from two fixed points is constant. The distancebetween the two fixed points (called foci) is taken as 2c, where c > 0. The midpoint of the line segment joining fociis the center and the constant sum is taken as 2a, where a > 0.

A1(a, 0)A2(−a, 0)

B1(0, b)

B2(0,−b)

F1(c, 0)F2(−c, 0)

P (x, y)

Figure 8.1: A horizontal ellipse.

Let us discuss the ellipse which has the foci on x-axis centered at the origin. The foci are F1(−c, 0) and F2(0, c).

Letting P (x, y) denote any point on the locus. We have by definition ‖−−→F1P‖+‖

−−→F2P‖ = 2a.According to the distance

formula, we get√

(x+ c)2 + y2 +√

(x− c)2 + y2 = 2a which is equivalent to a2(a2 − c2) = x2(a2 − c2) + y2a2.

From the triangle4

PF1F2, it is clear that a > c. Let b2 = a2 − c2. Substituting b into the above equation, we obtain

x2

a2+y2

b2= 1. (8.1)

The equation 8.1 is called the standard form of the equation of an horizontal ellipse with foci at the x-axis and center atthe origin. There are two horizontal and two vertical vertices on the ellipse which can be obtained from the equation8.1 by substituting y = 0 and x = 0, respectively. Let us denote the horizontal vertives by A1 and A2 and let usdenote the vertical vertices B1 and B2. The line segment A1A2 of length 2a is called the major axis and the linesegment B1B2 of length 2b is called the minor axis of the ellipse. If P is a point of an ellipse, the vectors

−−→F1P

and−−→F2P are called focal vectors of P. Their lengths are called focal radii of P and they are ‖

−−→F1P‖ = a − c

ax and

‖−−→F2P‖ = a+

c

ax.

43

44 CHAPTER 8. ELLIPSE, HYPERBOLA AND PARABOLA

Find an equation of the indicated ellipse.

a) Foci (∓3, 0), a = 5. Also find the focal radii of a point for which x = 2.

b) Foci (0,∓√

2), a = 4.

c) Vertices (∓4, 0) and passing through the point (3,−√

5).

Question 77.

Find the coordinates of the foci and the vertices, and draw the graph of the equation

a) 9x2 + 4y2 = 36.

b) x2 + 4y2 = 16.

c) 2x2 + 6y2 = 9.

Question 78.

Find an equation of the specified curves

a) Of the locus of points the sum of whose distances from the points (0,∓2) is 6.

b) Of the locus of points the sum of whose distances from the points (2, 3) and (4, 1) is 8.

Question 79.

8.2 Hyperbola

A hyperbola is the locus of a point in a plane the difference of whose distances from two fixed points is constant.The fixed points are again foci, the distance between them is 2c, the midpoint of the line segment joining the foci isthe center and the constant difference of the distances is 2a. In contrast with the ellipse, we have here a < c. Let usagain choose the x-axis through the foci and the y-axis through the center. If P (x, y) is a point on the locus, we have‖−−→F1P‖ − ‖F2P‖ = ∓2a. Again expending and simplifying this equation by setting b2 = c2 − a2, we get

x2

a2− y2

b2= 1. (8.2)

B1(0, b)

B2(0,−b)

A1(−a, 0) A2(a, 0)

F1(−c, 0) F2(c, 0)

Figure 8.2: A horizontal hyperpola

8.2. HYPERBOLA 45

The equation 8.2 is the standard form of the equation of a horizontal hyperbola with foci on the x-axis and centerat the origin. The graph of the equation 8.2 is symmetric in both coordinate axes and in the origin has x-intercepts ∓abut no y-intercepts. There are no points for |x| < a. At the points (∓a, 0) there are two vertical tangents. The pointsA1(−a, 0) and A2(a, 0) are vertices of the hyperbola, the line segment A1A2 of length 2a is the transverse axis. Thesegment B1B2 of the length 2b is the conjugate axis even though B1 and B2 are not on the hyperbola. The hyperbolawith the equation 8.2 has two asymptotes which pass through the origin and have the equations

y = ∓ bax. (8.3)

If a rectangle is drawn with sides parallel to the coordinate axes and through the ends of the transverse axis andconjugate axis, the diagonals of the rectangle, extended, are the asymptotes. The focal radii of the hyperbola withthe equation 8.2 may be found as ‖

−−→F1P‖ =

c

ax − a and

−−→F2P =

c

ax + a if x > 0 and ‖

−−→F1P‖ = a − c

ax and

‖−−→F2P‖ = −a− c

ax if x < 0.

Find an equation of the indicated hyperbolas.

a) Foci (∓5, 0), a = 3. Also find the focal radii of a point for which x = 6.

b) Foci (∓3, 0), vertices (∓2, 0). Also find the focal radii of a point for which x = −4.

c) Foci (0,∓4), b =√

2.

d) Foci (0,∓5), ends of conjugate axis (∓√

3, 0).

e) Vertices (0,∓√

3) and passing through the point (23 , 3).

f) Foci (∓5, 0)and passing through the point (203 , 4).

g) Vertices (∓2, 0) and asymptotes with slopes ∓32 .

h) Foci (0,−4) and (0, 0), and passing through the point (3,−4).

Question 80.

Find the foci, vertices, and asymptotes, and draw the graph of the equations:

a) 16x2 − y2 = 16.

b) 8x2 − y2 = 8.

c) 4x2 − 9y2 + 36 = 0.

Question 81.

Find an equation of the specified curves:

a) Of the locus of points the difference of whose distances from the points (0,∓2) is 2.

b) Of the locus of points the difference of whose distances from the points (∓4, 0) is 6.

c) Of the locus of points the difference of whose distances from the points (2, 3) and (4, 1) is 2.

Question 82.


Find equations of asymptotes of the given hyperbolas in the form of single second degree equations:

a) 2x2 − 3y2 + 4 = 0.

b) x2 − y2 − 1 = 0.

c) x2

4 −y2

9 − 1 = 0.

Question 83.

8.3 Parabola

A parabola is the locus of a point in a plane whose distances from a fixed point, called focus, and a fixed line, calledthe directrix, are equal. Let us examine the parabola passing through origin (vertex at the origin) focus on the x-axis,

directrix perpendicular to the x-axis. Let the focus be F (p

2, 0). Now the directrix is x =

−p2. The line of symmetry

of any parabola is called its axis and the point of the curve which lies on the axis is called the vertex of the parabola.Let P (x, y) be an arbitrary point on the parabola and let M the intersection point of the riectrix and the line whichis perpendicular to directrix and passing through P . We have the equality ‖

−−→PM‖ = ‖

−−→PF‖ and hence this equality

becomes simply

2px = y2. (8.4)

F (p2 , 0)

P (x, y)

O

M

x = −p2 x = −p

2

Figure 8.3: A horizontal parabola.

The equation 8.4 represents the standard form of the horizontal parabola vertex at the origin, focus at F (p

2, 0) and

directrix at−p2. The parabola opens to the right if p > 0 and opens to the left if p < 0. The lotus rectum of a parabola

is the chord through the focus and perpendicular the axis.

8.3. PARABOLA 47

Find an equation of the indicated parabolas.

a) Directrix x = −2, focus (2, 0).

b) Directrix y = 4, focus (0,−4).

c) Vertex (0, 0), directrix y = 43 .

d) Vertex (0, 0), directrix x = 34 .

e) Vertex (0, 0), focus on the x−axis, and passing through the point (8,−4).

Question 84.

Find the coordinates of the focus and equation of the directrix for the given parabolas and sketch thegraphs.

a) y2 = 6x.

b) 3x2 + 4x = 0.

c) x2 − 8y = 0.

d) 3x2 + 2y = 0.

Question 85.

Find the specified equations.

a) Of the locus of points equidistant from the line y = 43 and the point (0,−4

3).

b) Of the locus of points equidistant from the line x = 52 and the point (−5

2 , 0).

c) Of the parabola with y = 2 as directrix and (0, 4) as focus.

d) Of the parabola with y = 3 as directrix and (−2, 0) as focus.

e) Of the parabola with x+ y + 1 = 0 as directrix and (2, 3) as focus.

f) Of the parabola with 3x− 4y = 2 as directrix and (3, 0) as focus.

Question 86.

Chapter 9

The Equations of Second Degree

Every equation of second degree in rectangular coordinates represents a conic.

Theorem 7.

Proof. The general equation in x and y is

ax2 + 2hxy + by2 + 2gx+ 2fy + c = 0, (9.1)

where a, h, b, g, f and c are arbitrary constants, but a, h and b are not all three equal to 0. By rotating the axes througha properly choosen angle we can eliminate the xy term and the equation becomes a′x2 + b′y2 + 2g′x+ 2f ′y+ c′ = 0,which represents an ellipse, a hyperbola or a parabola.

The equation of any conic referred to rectangular coordinates is of the second degree.

Theorem 8.

Proof. The equation of any conic referred to axes taken through any origin O parallel to certain lines is Ax2 +By2 +Cx + Dy + E = 0. The equation of the same conic referred to axes through O making any angle θ with the otheraxes is found by substitutions which do not change the degree.

Our problem is to classify all the conics. Let us find the conditions under which the general equation ax2+2hxy+by2 + 2gx+ 2fy + c = 0 represents a parabola, an ellipse or a hyperbola.

Rotate the axes through an angle θ such that tan 2θ =2h

a− b, thus eliminating the xy term. Putting x cos θ−y sin θ

for x and x sin θ + y cos θ for y, the general equation takes the form

a′x2 + b′y2 + (terms of lower degree) = 0,

where a′ = a cos2 θ + b sin2 θ + 2h sin θ cos θ and b = a sin2 θ + b cos2 θ − 2h sin θ cos θ. It is clear from theseequations that a′ + b′ = a + b and a′ − b′ = (a − b) cos 2θ + 2h sin 2θ. By using the construction of tan 2θ andthe last equation, we have (a′ − b′)2 = (a − b)2 + 4h2. Subtracting this equation from the square of the equationa′ + b′ = a+ b we get a′b′ = ab− h2. We may now list the following properties.

1. If ab − h2 = 0, the given conic is a parabola. For then either a′ or b′ is equal to zero. Also note that ax2 +2hxy + by2 is a perfect square if h2 − ab = 0.

2. If ab− h2 > 0, the given conic is an ellipse. For then a′ and b′ must have like sign.

3. If ab− h2 < 0, the given conic is a hyperbola. For then a′ and b′ must have unlike sign.

49

50 CHAPTER 9. THE EQUATIONS OF SECOND DEGREE

The ellipse and hyperbola have centers and hence called central conics. If ab − h2 6= 0, and equation of seconddegree represents a central conic. For central conics, we move the origin in order to eliminate the x terms and y termsand we then rotate the axes in order to eliminate the xy term. There are always two values of 2θ less than 360◦ for

which tan 2θ =2h

a− b, and one of them is less than 180◦, in which case θ < 90◦. In the treatment of central conics

we shall always take θ < 90◦.To move the origin to any point O′(α, β), we must write x′ + α and y′ + β for x and y, respectively, in the given

equation. Then an easy computation gives that α =hf − bgab− h2

and β =hg − afab− h2

if and only if the coefficient of x′

and y′ are zero in the new equation. Then the transformed equation becomes ax′2 + 2hx′y′ + by′2 + c′ = 0 and the

center of this conic has the coordinates O′(α, β). To rotate the axes through an angle θ such that tan 2θ =2h

a− b,

we write X cos θ − Y sin θ and X sin θ + Y cos θ for x′ and y′, respectively. This gives a′X2 + b′Y 2 + c′ = 0 orX2

− c′

b′+

Y 2

− c′

b′= 1. This completes the reduction of any equation of a central conic to the standard form. The explicit

form of the coefficients a′, b′ and c′ can be determined in terms of the coefficients of the general equation such that

a′ =1

2(a+ b±

√(a− b)2 + 4h2

b′ =1

2(a+ b)∓

√(a− b2 + 4h2)

c′ =abc− af2 − bg2 − ch2 + 2fgh

ab− h2.

Note that if the conic represents a parabola, first we rotate the coordinates and after that we use translation. Thegeneral form of the conic can represents also two straight lines. This time the conic is called degenerate. A generalform of a conic represents a degenerate conic if and only if abc − af2 − bg2 − ch2 + 2fgh = 0. This expression iscalled the discriminant of the equation of the second degree and denoted by ∆.

Draw the graph of the following equations of second degree

a) 2x2 + 4xy + 4y2 + 2x− 4y + 1 = 0.

b) x2 + 4xy − 2y2 + 2x+ 4y − 2 = 0.

c) 16x2 − 24xy + 9y2 + 4y = 0.

d) x2 + 2xy + y2 − 2x− 2y − 2 = 0.

Question 87.

Chapter 10

Conic through Five Points and Family of Conics

The general equation 9.1 has six terms. If we divide by c, the equation becomes

px2 + qxy + ry2 + sx+ ty + 1 = 0. (10.1)

The equation therefore involves essential constants and hence a conic is determined by five conditions.Let C and C1 be two conics given by the equations f(x, y) = 0 and f1(x, y) = 0, respectively. These conics have

4 points in common. Since five points determine a single conic, then four points determine infinite conics and theirequation is given

f(x, y) + λf1(x, y) = 0. (10.2)

P1

P2

P3

P4

f(x, y) = 0

f1(x, y) = 0

Figure 10.1: Intersection of two conics.

If C1 consists of the lines l1 = 0 and l2 = 0 then we have the equation of the family as

f(x, y) + λl1l2 = 0.

In this case we have three other possibilities.

1. l1 = 0 and l2 = 0 intersect each other at a point A on the conic C and they intersect the conic C at the point P1

and P2, respectively. Then the equation of family of conics becomes f(x, y) + λl1l2 = 0 and it represents thefamily of conics passing through the points P1 and P2 and tangent to the conic C at the point A of order 1.

51

52 CHAPTER 10. CONIC THROUGH FIVE POINTS AND FAMILY OF CONICS

P1 P2

A

Figure 10.2: Case 1 of two lines.

P1

A

Figure 10.3: Case 2 of two lines.

2. The line l1 = 0 or l2 = 0 tangent to the conic C at the point A. In this case the equation f(x, y) + λl1l2 = 0represents the family of the conics passing through the points P1 and P2 and tangent to the conic C of order 2.

3. The lines l1 = 0 and l2 = 0 are tangent to the conic C at the point A, the equation f(x, y) +λl21 = 0 representsthe family of conics through P1 and is tangent to the conic C at A of order 2.

If the conic C also consists of the lines l3 = 0 and l4 = 0, the equation of family of conics passing through the pointsP1, P2, P3 and P4 is in the form l1l2 + λl3l4 = 0. In this case we have other two possibilities.

53

d

A

Figure 10.4: Case 3 for two lines.

P1 P2

P3P4

Figure 10.5: Family of Conics of four lines.

1. The point of intersection of the lines l3 = 0 and l4 = 0 is on the line l1 = 0 or l2 = 0. The equationl1l2 + λl3l4 = 0 represents the family of conics through P3 and P4 and is tangent to l1 = 0 or l2 = 0 at thepoint A of order 1.

2. The lines l3 = 0 and l4 = 0 are coincident. In this case P1 = P2 and P3 = P4 and the equation l1l2 + λl23 = 0represents the family of conics which is tangent to the lines l1 = 0 and l2 = 0 at the point P1 and P3,respectively.


P1

P2

A

Figure 10.6: Family of conics of four lines case 1.

P1 = P2

P3 = P4

Figure 10.7: Family of conics of four lines case 2.

Find the member of the family of lines through the intersection of x − y + 2 = 0 and 2x + 3y − 5 = 0which passes through (1, 5).

Question 88.

55

Find the equation of the line which passes through the point of intersection of 2x + y − 2 = 0 andx− y + 7 = 0, and which is perpendicular to x+ 6y − 3 = 0.

Question 89.

What is the slope of the line joining the origin with the point of intersection of x − 4y + 1 = 0 and3x+ y + 2 = 0?

Question 90.

Show that the three lines x − y + 6 = 0, 2x + y − 5 = 0 and x + 2y − 11 = 0 are concurrent (i.e. theymeet in a common point).

Question 91.

Find k so that x+ y + 1 = 0, kx− y + 3 = 0 and 4x− 5y + k = 0 will be concurrent.

Question 92.

Find that member of the family of circles through the intersections of x2 + y2 − 5x + y − 4 = 0 andx2 + y2 + 2x− 3y − 1 = 0 which passes through (1,−5).

Question 93.

Determine the type of the following second degree equations depending on a parameter λ:

a) x2 + 2y2 + 4x+ λ = 0.

b) 2x2 − 4y2 + 2y + λ = 0.

c) x2 + 6x+ 2(λ− 1)y − 4 = 0.

d) 3λx2 + 2√

3(λ− 1)xy + (2λ+ 1)y2 + 2x− 2y = 0.

e) (λ− 1)(x+ y)2 − 2x+ y + 1 = 0.

f) x2 + 2λxy + 2λ(λ+ 2)y2 + 4x = 0.

g) 2λx2 + 4(λ− 1)xy + (λ+ 14)y2 + 2x− 2y = 0.

Question 94.

Find the equation of the locus of the center of the conics 3x2 + (m+ 8)xy + 4y2 − 12x− 8y = 0.

Question 95.


Find the locus of focus of the family of parabolas y = (m− 1)x2 − 2mx+m+ 1.

Question 96.

Let C1 : x2 − 4y2 + 4x− 2y = 0 and C2 : 2x2 − xy − y2 + x+ y = 0 be two conics.

a) Find the equation of the family of conics passing through the intersection of the conics C1 and C2.

b) Discuss the family of conics for the parameter λ.

c) Find the locus of the center of the conics in the family.

d) Draw the graphs of the conic for λ = −1.

Question 97.

Let C1 : x2 + 16y2 − 4x− 1 = 0 and C2 : x2 − 4xy + 4y2 = 0 be two conics.

a) Find the equation of the family of conics passing through the intersection of the conics C1 and C2.

b) Discuss the family of conics for the parameter λ.

c) Find equation and draw the graph of locus of the center of the conics in the family.

Question 98.

Let l1 : x− 2y = 0, l2 : 2x− y = 0 and C : x2 + 2xy + y2 + y − 1.

a) Draw the graph of l1, l2 and C.

b) Find the equation of the family of conics passing through the intersection of the conic C1 and linesl1, and l2.

c) Discuss the family of conics for the parameter λ.

d) Find equation and draw the graph of locus of the center of the conics in the family.

Question 99.

Let C be a circle passing through the points (0, 1), (1, 0) and (2, 0). Find the equation of the family conicspassing through the points (1, 0) and (2, 0), and also tangent to the circle C at the point (0, 1).

Question 100.

Find the equation of family of conics passing through the points O(0, 0), A(4, 0),B(0, 4) and whosenormal lines at the point O(0, 0) bisect the line segment AB. Also, determine the types of members of thefamily depending on a parameter λ.

Question 101.

57

Find the equation of family of conics passing through the points O(0, 0), A(1, 0),B(0, 2) and whosetangent lines at the point O(0, 0) passing through C(43 ,−

23). Also, determine the types of members of the

family depending on a parameter λ.

Question 102.

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