MAT1035Analytic Geometry

Lecture Notes

R.A. Sabri Kaan Gurbuzer

Dokuz Eylul University

2016

2

Contents

1 Review of Trigonometry 51.1 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Polar Coordinates 9

3 Vectors in Rn 133.1 Located Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2 The Algebra of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3 The Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.4 The Norm of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.5 Projection of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.6 The Direction Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.7 Vector Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.8 Mixed Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 Straight Line 234.1 Direction of a Straight Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.2 Equations of a Straight Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.3 The Normal Form of a Straight Line in Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.4 Intersection of Two Straight Lines in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

5 The Plane 315.1 The Plane Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.2 The Angle between two Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335.3 Other Forms of the Equation of a Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345.4 Distance from a Point to a Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.5 Intersection of two Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.6 Projecting Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.7 Intersection of three Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.8 Specialized Distance Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

6 Transformation of Axes 436.1 Translation and Rotation of Axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

7 The Circle 457.1 The Equation of a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457.2 Intersections Involving Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467.3 Systems of Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487.4 The Angle between two Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3

4 CONTENTS

8 Ellipse, Hyperbola and Parabola 498.1 Ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498.2 Hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508.3 Parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

9 The Equations of Second Degree 55

10 Conic through Five Points and Family of Conics 57

Chapter 1

Review of Trigonometry

1.1 Cartesian Coordinates

A Cartesian coordinate system in two dimensions (also called a rectangular coordinate system or an orthogonal coordinatesystem) is defined by

• an ordered pair of perpendicular lines (axes),

• a single unit of length for both axes,

• an orientation for each axis.

The point where the axes meet is taken as the origin for both, thus turning each axis into a number line. For any point P, aline is drawn through P perpendicular to each axis, and the position where it meets the axis is interpreted as a number. Thetwo numbers, in that chosen order, are the Cartesian coordinates of P. The reverse construction allows one to determinethe point P given its coordinates.

The first and second coordinates are called the abscissa and the ordinate of P, respectively. The point where the axesmeet is called the origin of the system.

In mathematics, physics, and engineering, the first axis is usually defined or depicted as horizontal and oriented to theright, and the second axis is vertical and oriented upwards. The origin is often labeled O, and the two coordinates are oftendenoted by the letters X and Y, or x and y. The axes may then be referred to as the X-axis and Y -axis.

A Euclidean plane with a chosen Cartesian coordinate system is called a Cartesian plane. In a Cartesian plane one candefine canonical representatives of certain geometric figures, such as the unit circle, the unit square and so on.

The two axes divide the plane into four right angles, called quadrants. The quadrants may be named or numbered invarious ways, but the quadrant where all coordinates are positive is usually called the first quadrant.

If the coordinates of a point are (x, y), then its distances from theX-axis and from the Y -axis are |y| and |x|, respectively;where | . . . | denotes the absolute value of the number.

Plot the points whose coordinates are: (1, 1), (2, 3), (0,−4), (−3, 0), (−2, 3), (3,−1), (√

3, 2), (−√

2,√

2), (3.2, 7.1).

Question 1.

Draw the triangle whose vertices are (2 +√

3,−2), (√

2, 2), (−1 +√

8).

Question 2.

5

6 CHAPTER 1. REVIEW OF TRIGONOMETRY

Draw the polygon whose vertices are (−5,−2), (−2, 5), (2, 7), (5, 1), (2,−4).

Question 3.

Find the distance between the pairs of points whose coordinates are:

• (4,−1), (−2, 3)

• (−2, 6), (2,−2)

Question 4.

Find the perimeter of the triangle whose vertices are (−1,−2), (−3, 5) and (4, 2).

Question 5.

Show that the triangle with vertices (2, 4), (5, 1) and (6, 5) is isoscales.

Question 6.

Show that the triangle with vertices (10, 5), (3, 2) and (6,−5) is a right triangle and find its area.

Question 7.

Find the coordinates of the point which is equidistant from the points (4, 3), (2, 7) and (−3,−8).

Question 8.

Show that the points (1, 2), (−3, 10) and (4,−4) lie in a straight line.

Question 9.

1.2. TRIGONOMETRY 7

1.2 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 have cos2 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 thesame identity by sin2 θ gives 1 + cot2 θ = csc2 θ. The followings are the addition formulas 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 convenient tomeasure angles in radians. The radian measure of an angle is the length of the arc it subtends in the unit circle when thevertex of the angle is at the center of the circle. The area of the unit disk with radius r is A = πr2 and the circumference ofthe 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.

8 CHAPTER 1. REVIEW OF TRIGONOMETRY

Remember the properties of the trigonometric functions sin t + 2nπ = sin t and cos t + 2nπ = cos t for all n ∈ Z. Tofind 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

Use a sketch of the trigonometric circle to determine each of the following numbers: sin π6 , sin

π4 , sin

π3 , sin

25π4 ,

sin −27π4 , cos π6 , cos π4 , cos π3 , tan 26π6 .

Question 10.

By using the trigonometric circle prove that sin 2x = 2 sinx cosx and cos 2x = cos2 x− sin2 x.

Question 11.

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. Now let usbegin with a given xy-coordinate system and take the origin as the pole and the nonnegative x-axis as the polar axis.

Given the pole O and the polar axis, the point P with polar coordinates r and θ, written as the ordered pair (r, θ) is loocatedas follows: First find the terminal side of θ given in radians, where θ is measured counterclockwise (if θ > 0) from thex-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 the rayopposite 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. Ifr = 0, the polar coordinates (0, θ) represent the origin whatever the angular coordinate θ might be.

9

10 CHAPTER 2. POLAR COORDINATES

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

Remark 1.

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

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. So θ = arctany

x, x > 0

and θ = π + arctany

x, x < 0.

Some curve have simpler equations in polar coordinates than in rectangular coordinates. The graph of an equation in thepolar coordinate variables r and θ is the set of all those points P such that P has same pair of polar coordinates (r, θ) thatsatisfy 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 with the rectangularcoordinate equation x2 + y2 = a2 of this circle and transform it using x2 + y2 = r2, we get 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 graphs of the equations r = 2a sin θand 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. Let us 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.

Find the cartesian coordinates and plot the points given in polar coordinates: (3, π6 ), (−2, π3 ), (2, −π6 ), (−3,−π4 ).

Question 12.

11

Find |PQ| where P (1, π6 ) and Q(−2, π3 ).

Question 13.

Sketch the graph of the following polar equations:

• r = −2 cos θ

• r = 2− 2 sin θ

• r2 = 4 cos 2θ

• r = 2 + 2 sin θ.

Question 14.

12 CHAPTER 2. POLAR COORDINATES

Chapter 3

Vectors in Rn

The quantities, such as distance, area, volume, temprature, time densty, etc., posses only ”magnitude”. These can be repre-sented 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 are called 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) and denote 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.

13

14 CHAPTER 3. VECTORS IN RN

Determine the coordinates of the point D(x, y, z) if ~AB ∼= ~CD where A(3, 5, 2), B(4, 6,−1) and C(−2, 5, 7).

Question 15.

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

Question 16.

Find the coordinates and the length of the vector starting at the point A(−1, 5, 7) and ending at the point B(2,−3, 4).

Question 17.

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

Question 18.

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 19.

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.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 20.

3.2. THE ALGEBRA OF VECTORS 15

−→a−→a −→a

−→b

−→b

−→b

−→a +−→b

−→a +−→b

Triangle Law Prallelogram Law

Figure 3.2: Addition of vectors

Show that−−→AB +

−−→BC +

−→CA = ~0 for a triangle

4ABC.

Question 21.

Show that−−→AD = 1

2(−→AC +

−−→AB) for the median vector

−−→AD of a triangle

4ABC.

Question 22.

Let the points A,B and C be collinear and let the point D do not lie on the line containing the points A,B and C. If−→AC = 3

−−→AB and

−−→DC = k

−−→AD + p

−−→DB then k · p =?

Question 23.

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 24.

Let4

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

=1

2.

Question 25.

16 CHAPTER 3. VECTORS IN RN

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 26.

Check whether or not the vectors ~u1 = (1, 2, 1), ~u2 = (−1, 2, 1) and ~u3 = (2, 1,−2) are linearly independent.

Question 27.

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 28.

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 are standard basis vectorsof R2.)

Question 29.

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)

3.4. THE NORM OF A VECTOR 17

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 4.

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 30.

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 31.

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

Theorem 1.

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, the functionf 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 ‖.

18 CHAPTER 3. VECTORS IN RN

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 −→a and−→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 5.

3.6. THE DIRECTION ANGLES 19

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

Question 32.

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

Question 33.

Find the cosine of the angle between the vectors ~u = (2,−1, 2) and ~v = (1, 4, 1).

Question 34.

Find a unit vector which is perpendicular to both ~u = (1, 4, 1) and ~v = (0,−1, 1).

Question 35.

Prove that ‖~u+ ~v‖ = ‖~u− ~v‖ ⇔ ~u ⊥ ~v.

Question 36.

Let4

ABC be a triangle. Prove the cosine theorem.

Question 37.

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 base vectors.The natural base vectors are mutually perpendicular.

Consider the vector−→a making the angles α1, . . . , αn with the natural base vectors−→e1 , . . . ,−→en, respectively. These anglesare called the direction angles of −→a and their cosines are called the direction cosines of −→a . Since each vector is a linearcombination 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 the coordinatesare also called the direction numbers of −→a , such as ka1, . . . , kan, where k ∈ R.

20 CHAPTER 3. VECTORS IN RN

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 38.

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

Question 39.

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 40.

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

−→b as two adjacent sides.

3.8. MIXED PRODUCT 21

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 mixed productof 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 the vectors−→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.

22 CHAPTER 3. VECTORS IN RN

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 direction anglesof a line then for some λ ∈ R different form zero, we have a1 = λ cosα1, . . . , an = λ cosαn. The numbers a1, . . . , an arecalled a set of direction numbers of the line and the vector −→a = (a1, . . . , an) is called the direction vector of the line. Itis clear that the component of any vector are a set of direction numbers of any line on which the vector lies. Conversely, ifa1, . . . , an are direction numbers of a line then the vectors (a1, . . . , an) and (−a1, . . . ,−an) are oppositely directed vectorson 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 if (a1, a2, . . . , an) = λ(b1, b2, . . . , bn) for a non zero λ ∈ R.

• 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 vectorparallel to the line. Our discussion is to find the locus of the points lying on the straight line. We take an arbitrary point todetermine 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

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

23

24 CHAPTER 4. STRAIGHT LINE

x

l

x2

xn

−→a

O

P0(p1, . . . , pn)

P (x1, . . . , xn)

Figure 4.1: Straight Line in n-Space

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

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 41.

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 42.

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

Question 43.

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

Question 44.

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

3 and through (4, 2,−3).

Question 45.

4.2. EQUATIONS OF A STRAIGHT LINE 25

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

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, the y-intercept is(0, n).

On the other hand, given two points define a direction, hence a straight line passing through these points. For instance inR2 let the points P0(p1, p2) and Q0(q1, q2) be given. Then we have a straight line passing through P0 and Q0 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 lines are perpendicularif 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 midpoint M(x1, x2) of the line segment joining the points P1(3, 7) and P2(−2, 3).

Question 46.

Find the coordinates of the point P (x, y) which divides the line segment joining the points P1(−2, 5) and P2(4,−1), in theratio at 6

5 and −2, respectively.

Question 47.

26 CHAPTER 4. STRAIGHT LINE

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 48.

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

Question 49.

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

Question 50.

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

Question 51.

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

Question 52.

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

Question 53.

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 54.

4.3. THE NORMAL FORM OF A STRAIGHT LINE IN PLANE 27

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 55.

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 origin to theline 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

Let A(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.

28 CHAPTER 4. STRAIGHT LINE

xO

y

L1

l

p d(x1, y1)

Figure 4.3: Distance from a point to a line

To find the perpendicular distance d from the line l to the point x1, y1, draw the line L1 through the point (x1, y1) whichis 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 distance d is obtainedby

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 same side ofthe line l then d is negative.

Find the perpendicular distances from origin to the lines

a) x− 3y + 6 = 0

b) 15x− 8y − 25=0

Question 56.

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

√5.

Question 57.

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

Question 58.

4.4. INTERSECTION OF TWO STRAIGHT LINES IN SPACE 29

Determine the equations of the bisectors of the angles between the lines l1 : 3x+ 4y − 2 = 0 and l2 : 12x− 5y + 5 = 0.

Question 59.

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 are notparallel then they may still have no point of intersection, in which case they are called skew lines.

−→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

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 parametric equationsare

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 if Rank(A) = Rank(A : b), that is, we have solution for Rank(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 the linesintersect at a unique point. If Rank(A) = Rank(A : b) = 1 we have infinite solutions, hence the lines intersect at infinitelymany points, that is, the lines are coincide. If Rank(A) < Rank(A : b) then the system has no solution and hence thestraight lines have no point in common, which means that they are parallel or skew.

30 CHAPTER 4. STRAIGHT LINE

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 60.

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 61.

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 direction numbersof any line perpendicular to the plane. Let −→n be the vector −→n = (a, b, c) which is direction vector of the line perpendicularto 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 perpendicularto 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)

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 the plane.The equation 5.2 is called the standard form and the equation 5.3 is called the general form of the equation of the plane.

31

32 CHAPTER 5. THE PLANE

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

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 62.

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 63.

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

Question 64.

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

Question 65.

Find the equation of the plane perpendicular to the plane x− y+ 2z− 5 = 0, parallel to the line whose direction cosines are15 ,−

25 ,

2√5

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

Question 66.

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

Question 67.

5.2. THE ANGLE BETWEEN TWO PLANES 33

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

Question 68.

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 69.

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

−1 = z3}.

Question 70.

Find an equation of the plane determined by the lines l1 = {(x, y, z) : x+1 = −y+1 = z} and l2 = {(t−2, 2t−1,−3t+3) :t ∈ R}.

Question 71.

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.

34 CHAPTER 5. THE PLANE

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

Question 72.

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 73.

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-intercepts A,B and C, respectively, we find that A = −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 get

x

(−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 the equation ofa 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 angles of thenormal 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 35

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 74.

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

Question 75.

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

Question 76.

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

Question 77.

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 by normalform 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 ofthe 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

36 CHAPTER 5. THE PLANE

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 = 0 has thesame 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 78.

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

Question 79.

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

Question 80.

5.5. INTERSECTION OF TWO PLANES 37

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. We have three methods forfinding 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 intersectionline.

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

3. Third Method: We eliminate one variable from the equations and then eliminate an other one. Hence we may writethe 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 81.

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 82.

Find the vector formulation for the line of intersection of the planes P1 : x+ 2y+ 3z+ 4 = 0 and P2 : x−2y+ 2z−1 = 0.

Question 83.

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 84.

38 CHAPTER 5. THE PLANE

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 85.

5.6 Projecting Planes

A straight line lies in infinitely many planes. Any two of them determine the line. By eliminating one of the variables fromtwo such equations we obtain an equation of a plane which contains the line and perpendicular to a coordinate plane. Thusif a line is given by the equations a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 then the elimination of y (forexample) gives the plane Ax + Cz + D = 0 which is clearly perpendicular to xz-plane. The plane Ax + Cz + D = 0is called the projecting plane for the given line. The equations Ax + Cz + D = 0, y = 0 determines a straight line in theprojecting plane an xz-plane. This straight line is the projection of the given straight line 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. Generally parallelismof 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 of intersectionof three planes.

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

4. The planes intersect in a single point.

5.7. INTERSECTION OF THREE PLANES 39

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 86.

40 CHAPTER 5. THE PLANE

−→n3

−→n2

−→n1

−→n3

−→n2 −→n1

Figure 5.8: Third case

−→n1−→n2

−→n3

Figure 5.9: Fourth case

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 87.

Determine α such that the following four planes P1 : x+ 2y− 3 = 0, P2 : x+ y− 2z − 9 = 0, P3 : 3y+ 5z + 15 = 0 andP4 : 3x+ αz − 15 = 0 pass through the same point. Find the coordinates of this point.

Question 88.

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 between twoskew 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. Letα 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

5.8. SPECIALIZED DISTANCE FORMULA 41

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.

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 distance betweentwo skew lines. Let P1 and P2 be any two points on the skew lines l1 and l2, respectively. Let −→m1 and −→m2 be directions of l1and 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 89.

42 CHAPTER 5. THE PLANE

Find the distance between the lines l1 = {(x, y, z) : x + y − z + 3 = 0, 2x − y − 4z + 2 = 0} and l2 = {(x, y, z) :3x+ y + 5z − 1 = 0, y + z + 2 = 0}.

Question 90.

Chapter 6

Transformation of Axes

6.1 Translation and Rotation of Axes

If two sets of parallel coordinate axes are drawn, the originO′ of the x′y′-coordinate plane has the coordinates α, β in the xy-coordinate system. Then a point P has the coordinates (x, y) in the xy-coordinate plane and (x′, y′) in the x′y′-coordinateplane. It may be seen from the figure that two point of coordinates, that is, (x, y) and (x′, y′) are related 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-coordinatesystem using the coordinates of P in x′y′-coordinate system and the position of the origin O′. Reverse is also true.

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

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

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

43

44 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.

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 91.

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

Question 92.

Find x′, y′ coordinates of the point (9, 5) under the rotation of axes through the obtuse angle θ for which tanθ = −34 and

transform the equation 3x− 5y = 2.

Question 93.

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, the center. LetC(α, β) be the center and r > 0 be the radius. Now let P (x, y) be an arbitrary (representative) point on the circle. Then wehave ‖

−−→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 the center(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 no xyterm 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 if D2 + E2 − 4F = 0and it represents no real graph if and only if D2 + E2 − 4F < 0. If the general form represents a single point, the graph issometimes 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, ror E,D,F, respectively. This means that in general a circle may be required to satisfy three conditions which may be ofvarious types.

45

46 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 line andthe circle do not meet. If the roots are real and distinct then there are two distinct points of intersection. Finally, if the rootsare 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 passing throughthe intersection of the circles C1 and C2. The straight line (D1 −D2)x+ (E1 − E2)y + (F1 − F2) = 0, which is obtainedby subtracting the two circle equations of C1 and C2, is called the radical axis of the given circles C1 and C2. In generalwe have three cases, they have no common points or they have two distinct common points or they have exactly one point incommon 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 from P to the points oftangency of the circles are equal.

Theorem 6.

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.

7.2. INTERSECTIONS INVOLVING CIRCLES 47

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.

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 94.

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

Question 95.

48 CHAPTER 7. THE CIRCLE

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 96.

7.3 Systems of Circles

Let the equations C1 : x2 + y2 + D1x + E1y + F1 = 0 and C2 : x2 + y2 + D2x + E2y + F2 = 0 represent two circles.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. The equation7.4 also represents the family of circles passing through the intersection of C1 and C2. When k = −1, the equation 7.4reduces 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 between twocircles 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 each other under rightangle.

Question 97.

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 foci is thecenter 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). LettingP (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 equation 8.1 bysubstituting y = 0 and x = 0, respectively. Let us denote the horizontal vertives by A1 and A2 and let us denote the verticalvertices B1 and B2. The line segment A1A2 of length 2a is called the major axis and the line segment 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.

49

50 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 98.

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 99.

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 100.

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 fixedpoints are again foci, the distance between them is 2c, the midpoint of the line segment joining the foci is the center and theconstant difference of the distances is 2a. In contrast with the ellipse, we have here a < c. Let us again choose the x-axisthrough 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)

The equation 8.2 is the standard form of the equation of a horizontal hyperbola with foci on the x-axis and center atthe origin. The graph of the equation 8.2 is symmetric in both coordinate axes and in the origin has x-intercepts ∓a but noy-intercepts. There are no points for |x| < a. At the points (∓a, 0) there are two vertical tangents. The points A1(−a, 0)and A2(a, 0) are vertices of the hyperbola, the line segment A1A2 of length 2a is the transverse axis. The segment B1B2

of the length 2b is the conjugate axis even though B1 and B2 are not on the hyperbola. The hyperbola with the equation 8.2has 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 and conjugateaxis, the diagonals of the rectangle, extended, are the asymptotes. The focal radii of the hyperbola with the equation 8.2 maybe 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.

8.2. HYPERBOLA 51

B1(0, b)

B2(0,−b)

A1(−a, 0) A2(a, 0)

F1(−c, 0) F2(c, 0)

Figure 8.2: A horizontal hyperpola

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 101.

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 102.

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 103.

52 CHAPTER 8. ELLIPSE, HYPERBOLA AND PARABOLA

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 104.

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, called thedirectrix, 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 anarbitrary point on the parabola and let M the intersection point of the riectrix and the line which is perpendicular to directrixand 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 53

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 105.

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

a) y2 = 6x.

b) 3x2 + 4x = 0.

c) x2 − 8y = 0.

d) 3x2 + 2y = 0.

Question 106.

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 107.

54 CHAPTER 8. ELLIPSE, HYPERBOLA AND PARABOLA

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 through aproperly choosen angle we can eliminate the xy term and the equation becomes a′x2 + b′y2 + 2g′x+ 2f ′y + c′ = 0, whichrepresents 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 other axes is found bysubstitutions 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 these equationsthat a′ + b′ = a + b and a′ − b′ = (a − b) cos 2θ + 2h sin 2θ. By using the construction of tan 2θ and the last equation,we have (a′ − b′)2 = (a − b)2 + 4h2. Subtracting this equation from the square of the equation a′ + b′ = a + b we geta′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.

55

56 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 second degreerepresents a central conic. For central conics, we move the origin in order to eliminate the x terms and y terms and 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 pointO′(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 bedetermined 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− b)2 + 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. The generalform of the conic can represents also two straight lines. This time the conic is called degenerate. A general form of a conicrepresents a degenerate conic if and only if abc− af2 − bg2 − ch2 + 2fgh = 0. This expression is called the discriminantof 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 108.

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 their equationis 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 andP2, respectively. Then the equation of family of conics becomes f(x, y) + λl1l2 = 0 and it represents the family ofconics passing through the points P1 and P2 and tangent to the conic C at the point A of order 1.

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

57

58 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.

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 represents thefamily 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.

1. The point of intersection of the lines l3 = 0 and l4 = 0 is on the line l1 = 0 or l2 = 0. The equation l1l2 + λl3l4 = 0represents the family of conics through P3 and P4 and is tangent to l1 = 0 or l2 = 0 at the point 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 = 0

59

d

A

Figure 10.4: Case 3 for two lines.

P1 P2

P3P4

Figure 10.5: Family of Conics of four lines.

represents the family of conics which is tangent to the lines l1 = 0 and l2 = 0 at the point P1 and P3, respectively.

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

Question 109.

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

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

Question 110.

61

What is the slope of the line joining the origin with the point of intersection of x− 4y + 1 = 0 and 3x+ y + 2 = 0?

Question 111.

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

Question 112.

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

Question 113.

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

Question 114.

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 115.

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

Question 116.

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

Question 117.

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

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 118.

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 119.

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 lines l1, 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 120.

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

Question 121.

Find the equation of family of conics passing through the pointsO(0, 0),A(4, 0),B(0, 4) and whose normal lines at the pointO(0, 0) bisect the line segment AB. Also, determine the types of members of the family depending on a parameter λ.

Question 122.

63

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

23). Also, determine the types of members of the family depending on a parameter λ.

Question 123.