UNIT 3 GENERAL THEORY O F CONICS

Structure 3.1 Introduction

Objectives

3.2 General Second Degree Equation

3.3 Central and Non-central Conics

3.4 Tracing a Conic Central Conics Parabola

3.5 Tangents

3.6 Intersection of Conics

3.7 Summary

3.8 Solutions/Answers

3.1 INTRODUCTION

So far you have studied the standard equations of a parabola, an ellipse and a hyperbola. We defined these curves and other conics by the focus-directrix property of a conic. This defining property was discovered by Pappus (approx. 320 AD) long after the definition of conic sections by the ancient Greeks. In his book "Conics", the ancient Greek mathematician Apollonius defined these curves to be the intersection of a plane and a cone. You will study cones later, in Unit 6, but let us show you how conics are planar sections of a cone, with the help of diagrams (see Fig. 1).

Fig. I: A planar section of a cone can be (a) m elllpse. (b) n parabola, (c) a hrpcrbolr (d) r pair of line& (e) a point.

In this unit we will prove a result that may surprise you. According to this result, the general second degree equation ax2 + hxy + by2 + gx + fy + c = 0 always represents a conic section. You will see how to identify it with the various conics, depending on the conditions satisfied by the coefficients.

In Unit 2 you saw one way of classifying conics. There is another way of doing so, which you will study in Sec. 3.3. We shall discuss the geometric properties of the different types of conics, and see how to trace them. After that, we shall discuss the tangents of a conic. And finally, we shall see what curves can be obtained when two conics interesect.

With this unit we end our discussion on conics. But in the next two blocks you will be coming across them again. So, the rest of the course will be easier for ;,nu to grasp if you ensure that you have achieved the unit objectives given below.

1 After studying this unit. you shorild be able to

1 identify the conic represented by a qmadratic expression;

1 @ find the centre (if it exists) and axes of i conic:

I cp trace any given conic;

I \ 0 find the tangent and normal to a given conic at a given point;

0 obtain the equations of conics which pass through the points of intersection of two given conics.

1 3.2 GENERAL SECOND DEGREE EQUATION

In Unit 2 you must have noticed that the standard equation of each conic is a. second degree equation of the form I

1 ax2 + hxy + by2 + gx + fy + c = 0,

I for some a,b,c,f,g,h E R and where at least one of a,h,b is non-zero.

In this section we will show you that the converse is also true. That is, we will pr.ove that the general second degree equation

where at least one of a,h,b is non-zero, can be transformed into a standard equation of a conic. We achieve this by translating and rotating the coordinate axes. Let us. see how.

we first get rid of the term containing xy by rotating the XY-system through a "suitable" angle 8 about 0. You will see how we choose 9 a little further on. Now, by (16) and (17) of Unit 1, we see that (1) becomes .

a(xlcos e -i y' sin 8)2 + ~ ~ ( X ' F O S 0 - y'sin 9) (x'sin e + y 'cos 8) +b(x'siaB + y'cos 8)2 + ~ ~ ( X ' C O S 9 - y'sin 8) + 2f(x1sin 8 + y'cos 8) + c = 0 * (a cosZ 9 + 2h cos 8 sin 8 + b sin2 8) x'2 - 2 {(a - b) sin 9 cos 8 -

h(cos2 9 - sin2 8)) x'y' + (a sin2 8 - 2h sin 8 cos 8 + b cos2 8)yl2 +.(2gcos 8 + 2f s inO)x ' + ( 2 f c o s 9 - - 2 g s i n 8 ) y ' + c = 0.

/ The x'y' term will disappear if (a - b) sin 8 cos 8 = h(cos2 8 - sin2 8), that is,

- (a - b) sin 28 = h cos 28. 1 , : lr 1 So, t o get rid of the x'y' term, if a = b we can choose 8 = f; otherwise, we

I 1 can choose 0 = - tan-' -

2 ( a ' - b ) . 7r

can always choose such a 0 lying between - - and 4 4

of 8 the .x'yl term becomes zero1.

So, if we rotate the axcs through anangle 8 = - tan-' (a:b), - then (1) transforms into the second degree equation 2

Ax" + + 2Gx' + 2Fy' + C = 0; . -42) . where A = a cos2 8 + 2hcos 0 sin 8 + b sin2 8 and

B = a sin29 - 2h sin 9 cos8 + b cos28.

Thus, A + B = a + b.

General Theory of Conics I

We write the coefficienls of xy, x and y RS 211, 28, and 2F to have simpler expressions later on, as you wi!l see.

sln 28 = 2sin 8 case cos 2B = ros28 - sin29

Also, with a bit of computation, you can check bhat ab - h2 = AB. Now various situations can arise.

Case 1 (ab - h2 = 0): In this case we see that either A = 0 or B = 0. So, let us I

assume that A = 0. Then we claim that B must be non-zero. Do you agree? What would happen i f A = 0 and B = O? In this case we would get a = 0, b = 0 and h = 0, which contradicts our assumption that (1) is a quadratic equation. I

So, let A = 0 and B # 0: Then (2) can be written as

Now, if G = 0, then the above equation is

F2 - BC , that is.

y + - = * B c.

This represents a pair of parallel lines if F2 zBC, and the empty set if F2 < BC. 1

I

On the other hand, if G # 0, then we write (3) as 1

where X, Y are the current coordinates, From Sec. 2.3 you know that this i

represents a parabola with focus G (- $, 0) and directrix X = - 2B '

Now let us look at the other case.

Case 2 lab - h2,# 0): Now both A and B are non-zero. We can write (2) as

+ - - C, which is a constant K, say. B

G Let us shift the origin to (- A, - z). Then this equation becomes

where X and Y are the current coordinates.

Now, what happens if K = O? Well, if both A and B have the same sign, that is, if AB = ah - h2 > 0, then (4) represents the point (0, 0).

And, if ab - h2 < 0, then (4) represents the pair of lines, I X = t E Y .

And, what happens if K # O? Then we can write (4) as

x2 --- + ---- Y 2 r: 1 . K/A K/B

. * . (5)

1 Does this equation look familiar? From.Sec. 2.4 you'can see that this represents an

K K ellipse if both - and - are positive, that is, if K > 0 and AB = ab - h2 > 0.

A . B

But, what if K/A < 0 and K/B < O? In this case K < 0 and ab - h2 > 0. And then (5) represents the empty set.

K K And, if - and - are of opposite signs, that is, if AB = ab - hZ c 0, then

A B what will (5) represent? A hyperbola. '

So we have covered all the possibilities for ab - h2, and hence for (1). Thus, we haye proved the following result.

Theorem 1: The general second degree equation ax2 + 2hxy + by2 + 2gx + 2fy + c .= 0 represents a conic.

While proving this. theorem you must have noticed the importance we gave the expression a b - h2. Let Gs tabulate the various types of non-degenerate and degenerate conics that ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents, according to the way ab - h2 behaves. (Recall from Unit 2 that a degenerate conic is a conic whose focus lies o n the corresponding directix.)

Table 1: Classification of Conics.

I i b - 112 z o 1 ellipse I point, or empty set I

Condition

ab - h2 = 0

I ab - h2 c O ( hyperbola I pair of intersecting lines 1

Table 1 tells us about all the possible conics that exist. This is what the following exercise is about.

Types of Conics

El) a) write down all the possible types o'f conics'there are. Which of them are

Non-degenerate

parabola

d~generate?

b) If (1) represents a circle, will ab - h2 = O?

Degenerate .

pair of parallel lines, or empty set

Now let us use the procedure in the proof above in some examples.

Example.1: Find the conic reprevented by

9x2 - 24xy + 16y2 - 1 2 4 ~ + 132y + 324 = 0.

Solution: The given equation is of the form (I), where a = 9, b = 16, h = - .12., Now let us rotate the axes through an angle 0 , where

2h . 24 2tan 0 tan 20 = - = - - 24 , that is, - - , that is,

a,- b 7 I - tan2 0 7

12 tan20 + 7 tan0 - 12 = 0. 3 3 4

So we can take tan 0 = - and then sin 0 = - and cos 0 = -. 4 ' 5 5

Then, in the new coordinate system the given equation becomes

General Theory of Conies

124 132 2sYl2 - - (4x1 - 3y1) + - (3x ' + 4y1) + 324 = 0, that is,

5 5

4 5

Now let us .shift the origin. to . Then the equation becomes

4 y2 q - X,

5

where X an4 Y zire the current coordinates.

Can you recognise the conic represented by this equation? From Unit 2 YOU know that this is a'parabola, Since the transformations we have applied do no1 alter the curve, the original. equation also represents a parabola.

Example 2: Identify the conic x2 - 2xy + y2 = 2. 53

Conics Sokutian: Over here, since a = b = 1, we choose 0 = 45". S q , let US rotate the axes through 4 5 "7 The new coordinates x ' and y' are given by

Then, in the new coordinate system,

x2 - 2xy + y2 = 2 transforms to y '2 = 1,

which represents the pair of straight lines y' = 1 and y' = - 1.

You can d o the following exercise on the same lines.

E2) Identify the conic

a) x2 - 2xy + y2 + j Z x = 2,

b) 9x2 - 6xy + y2 - 40x - 20y + 75 = 0.

So far you have seen that any second degree equation represents one of the following conics:

a parabola, an ellipse, a hyperbola, a pair of straight lines, a point, the empty set.

But, from Table 1 you can see that even if we know the value of a b - h2, we can't immediately say what the conic is. So, each time we have to go through the whole procedure of Theorem 1 to identify :he conic represented by a given equation. Is there a short cut? Yes, there is. We have a simple condition for ( I ) t o represent a pair of lines. It can be obtained from the proof of Theorem 1 after some calculations, or independently. We shall only state it, and then see how to use it to cut short our method for identifying a given conic.

Theorem 2: The quadratic equation

ax2 + 2hxy + by2 i- 2gx + i f y + c = 0

Recall the definition of a represents a pair.of straight lines if and only if abc + 2fgh - af2 - bg2 - ch2 = 0, determinant from Unit 5 of MTE-04. that is, the determinant

Further, if the condition is satisfied, then the angle between the lines is

2&Tz tan-' ( ).

a + b

The 3 x 3 determinant given above is called the discriminant of the given conic. You can see that the discrinlinant looks neater if we take 2h, 2g and 2f as coefficients, instead of h, g and f.

Let us consider some examples of the use of Theorem 2.

Example 3: Show that x2 - 5xy + 6 9 = 0 represents a pair of straight lines. Find the angle between these lines.

5 Solution: With reference to Theorem 2, in this case a = 1 , h = - -, b = 6, g = 0 = f = c. Thus, the related discriminant is 3 -

- 56

5 1 - - 0

2

5 - - 6 2

0

0 0.w. 0

, which is 0, as

you know from our course 'Elementary Algebra'. Thus, the given equation represents a pair of lines.

The angle between them is tan-' 7 '

Example 4: Find the conic represented by 2x2 + 5xy + y2 = 1

Solution : ln this case ab - h2 = - 23 < 0. So, from Table 1 we know that the equation represents a hyperbola or a pair of liiics. Further, in this case the discriminant becomes.

So, by Theorem 2 we know that the given equation doesn't represent a pair of lines. Hence, it represents a hyperbola.

Why don't you do these exercises now'?

E3) Check whether 3x2 t 7xy + 2y2 + 5x + 5y + 2 = 0 represents a pair of lines.

E4) Show that the real quadratic equation ax2 + 2hxy -I- by2 = 0 represents a pair of lines.

E5) Under what cchditions 011 a, b and h, will the equation in Theorem 2 represent

a) a pair of parallel lines?

b) a pair of perpendicular lines?

So ,far we have studied all the conics in a unified manner. Now we will categorise. them according'to the property'of centrality.

3.3 CENTRAL AND NON-CENTRAL CONICS ----

In our discussion on the ellipse in Unit 2, we said that the midpoint of the major axis was the centre of the ellipse. The reason that this point is called the centre is because of a property that we ask you to prove in the following exercise. .

x2 y2 E6) Consider the equation -- + -- = I . Let P(x,, y,) be a point on this ellipse

a2 b2 and 0 be (0, 0). Show that the line PO also meets the ellipse in P' (Lx,, -yl).

What you have just proved is thatO(0, 0) bisects every chord of the ellipse x2 - y2 + 7 = 1 that passes through it. Siqilarly, any chord of the hyperbola a2 b x - - - - y2 - I through O(0, 0) is bisected by 0. Hence, according to the following a 2 b2 definition, 0 is the centre of the ellipse and hyperbola given above.

Definition: The centre of a conic C is a point which bisects any chord of C,that passes through it.

Not all conics have centres, as you will see. A conic that has a centre is called a

Conics . central conic. For example, an ellipse and a hyperbola are central conics.

Now, can a central conic have more than one centre? Suppose it has two centres C, and C2. Then the chord of the conic intercepted by the line CICz must be bisected by both C, and C2, which is not possible. Thus,

a central conic has a unique centre.

Let us see how we can locate this point.

Consider the conic (1). Suppose it is central with centre at the origin. .Then we have the following result, which we will give without proof.

Theoiem.3: A central conic with centre at (0, 0) is of the form ax2 + 2hxy + by2 = 1,

for some a, h, b in R.

This result is used to prove the following theorem about any central. conic. We shall not prove the theorem in chis course but we will apply it very often.

Theorem 4: Let ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 be a central conic. Then its centre is the intersection of the lines

What this theorem tells us is that if ax + hy + g = 0 and hx + by + f = 0 intersect, then the conic is central; .and the point of intersection of these straight linesiis the centre of the conic.

But what if the lines don't intersect? Then the conic under consideration can't be central; that is, it is non-central. Thus, the conic is non-central if the slopes of these lines are equal, that is, if ab = h2.

So, we have the following result:

i) . central if ab f h2, and

Does this result and Table 1 tell you which conics are non-central? You can immediately tell that a parabola doesn't have a centre. * .

Let us see how we can apply the above results on centres of conics.

Example 5: Is the conic 17x2 - 12xy .+ By2 + 46x - 28y + 17 = 0 central? If it is, find its centre.

Solution: In this case a = 17, b = 8, h = -6, g = 23, f = -14. So, ab f h2. Hence, the conic is central. Its centre is the intersection of the lines 17x - 6y + 23 = 0 and 3x - 4y + 7 = 0, which is (-1, 1).

Why don't you try some exercises now?

-

E7) Is the conic in E3 central? .If yes, find its centre.

E8) Identify the conic x2 - 3xy + y2 + lox - 10y + 21 = 0. If it is central, find its centre.

E9) ~ h i ' c h degenerate conics are central, and which are not?

One point that has been made in this sub-section is that a parabola is a non-central conic, while an ellipse and a hyperbola are central conics, NOW-let us see if this fact helps us to trace a conic corresponding to a given quadratic equation.

General Theory of Conics ) I

3.4. TRACING A CONIC I

Suppose you are given a quadratic equation. Can you get enough geometric information from it to be able to draw its geometric representation? You are now in a position to check whether it is a pair of lines or not, You can also tell yhether it is a central conic or not. But there is still one piece of information that you would need before you could draw th'e required conic. You need to know the equation of its axis, or axes, as the case may be. So'let us see how to find the awes. We shall consider the centraI and non-central cases separately.

3.4.1 Central Conics

Suppose we are given the equation of a central conic. By translating the axes, if necessary, we can assume that its centre lies at (0, 0). Then, by Theorem 3, its equation is

ax2 + 2hxy + by2 = 1 . . . (6)

where a, h, b E R.

In Theorem 1 you saw that if we rotate the coordinate axes through an angle 1 2h

6 = - tan-' - , then the axes of the conic lie along the coordinate axes. 2 a - b

Therefore, the axes of the conic are inclined at the angle 8 to the coordinate axes. (Here if a = b, we take 8 = 45".) Now,

2h tan 28 =

a - b

a - b r tan2 8 + (?) tan 0 - 1 = 0. r

This is a quadratic equation in tan 8, and hence is satisfied by two values of 8, say 8, and 02. Then the sIopes of the axes of the conic are tan 8, and tan 02. Note that the axes are mutually perpendicular, since (tan 8') (tan B2) = -1.

Now, t? find the lengths of the axes of the conic, we write (6 ) in polar form (see Sec. 1.5). For this we substitute x = r cos 8, j r = r sin 8 in (6). Then we get

r2 (a cos2 fl + 2h cos 8 sin 8 + b sin2 0) = 1

cos2 0 + sin2 8 3 r2 = - , writing 1 = cos2 8 + sin2 8

a cos2 8 + 2h cos 0 sin 8 + b sin2 0

- - 1 + tan2 0 -4

a + 2h tan 8 + b tan2 8 '

I, we substitute tan 8, and tan B2 in (7), we will get the corresponding vakes of 'r, which will give the lengths of the corresponding semi-axes.

* A semi-axis is half the axis.

Let us use what we have just done to trace the conic in Example 5. Sincc ab - h 2

> 0, from Theorem 1 we know that the conic is an ellipse. You have already seen that its centre lies at (-1, 1): Now, we need to shift the axes to the centre (-1, I ) , to get the equation in the form (6). The equation becomes

Now we can obtain the directions of the axes from

3 tan2 0 - - tan 0 - 1 = 0.

2 1

This gives us tan B = 2, - - 2 -

Conics Tnerefore, we can take 8 , = tan-' 2 = 63.43" (approximately), .and

n- Bz = - + tan-' 2.

2

The lengths of the semi-axes, r l and r2, are given by substituting these values in (7). So

1 + 4 r: = I = 4 =, r l = 2, and

17 6 8 + - 20 5 5

Thus, the length of the major axis is 4, and that of the minor axis is 2.

So now we can trace the conic. We first draw a line O ' X ' through 0'(-1, 1) a t an angle of tan-' 2 to the x-axis (see Fig. 2). Then we draw O ' Y ' perpendicular to O'X'. Now we mark of A' and A on O ' X ' such that A ' O ' = 2 and O'A = 2. Similarly, we mark off B and B' on O ' Y ' such that O'B = 1 and O 'B ' = 1.

'The required ellipse has AA' and BB' as its axes. For further help in tracing the curve, we can check where it cuts the x and y axes. It cuts the x-axis in (-..I, O), (-2.2, O), alid the y-axis in (0; 2.7) and (0, 3). So the curve is what we have, drawn in Fig. 2.

1 Now why don't you see if you've understood what has been done in this section? Fig. 2: The ellipse 17x2 - 12xy + 8y2 +. 4 6 x - 28y + 17 = 0.

E10) Trace the conic in E8.

E l l ) Under what conditions on the coefficients, will x 2 + 2hxy + y 2 + 2fy = 0 be central ? And then, find its centre and axes.

So far you have seen how to trace a central conic. But what about a non-central conic? ' l e t us look at this case now. - 3.4.2 Parabola

In this sub-section we shall look at a method for finding the axis of a parabola, and hence tracing it. We will use the fact that if (1) is a parabola it can be written in the form

where Ax +' By + C = 0 is the axis of the parabola and A'x + B ' y + C' G 0 is the tangent at the vertex, and hence they are perpendicular to each other.

The vertex (x,, yl) of this parabola is the intersection of Ax + By + C = 0 and A'x + B'y + C' = 0, k is the length of its latus rectum, and

I I

k is its focus, where tan 0 is the slope of the axis.

I 4

Let us see the method with the help of an illustration.

Example 6: Show that the conic x2 + 2xy + y2 - 2x - 1 = 0 is a parabola. Find its axis and trace it.

Solution: Here a 7 1, b = 1, h = 1. .'. ab - h2 = 0. 1 1 -1

Further, the discriminant of the conic is 1 1 0

-1 0 -1 = -1 # 0.

Hence, by Theorem 2, the equation does not represent a pair of straight lines. Thus, by Theorem 1, we know that the given conic is a parabola.

We can write the given equation as (x + y)2 = 2x + 1.

Now we will introduce a constant c so that we ran write the equation in the form (8). So, let us rewrite the equation as

(x + y + c ) ~ = 2x + 1 + 2cx + 2cy + c2, that is, ' -..

(x + y + c)2 = 2(1 + c)x + 2cy + c2 + !. . . . (9)

We will choose c in such a way that the lines x + y + c = 0 and 2(1 + c)x + 2cy + c2 + 1 = 0 are perpendicular. From Equation (13) of Unit 1, you know that the condition is

Then (9) becomes

This is in the form (8). 1 Thus, the axis of the parabola is x + y - - = 0, and the tangent at the vertex is

5 2 x - y + - = 0.

A

The vertex is the intersection of these two lines, that is, 1 .

The length of the latus rectum of the parabola is I a' Thlls, the focus is at

angle that the, axis makes with the x-axis, that is, 0 = tan-' (-1).

1 .'. sin 0 = - - 1 . 6, cOS O- = 43' - .

Therefore, the focus is F - - - ( :, :)* What are the points of intersection of the parabola and the coordinate axes? They are(1 + \IZ, O)+(1 -a, O), (0, I), (0,-1).

S o , we can trace the parabola as in ~ i g . ' 3 .

Fig.

A

X

3: The parabola x2 + 2xy f y2 - 2x - 1 = 0

Genernl Theory of Conics

Conics

L

Has the example helped you to understand the method for tracing a parabola? The following exercise will help you to find out.

E12) Trace the conic 4x2 - 4xy + y 2 - 8x - 6y + 5 = 0.

Let us n o w see how to obtain the tangents of a general conic.

3.5 TANGENTS

In Unit I you studied the equations of tangents to the conics in standard form. Now we will discuss the equation of a tangent to the general conic (1).

So, consider two distinct points P(xl, y,) and Q(x2, y2) on the conic ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. If x, = x2 = u, say, then the line PQ is x = a. Similarly, if yl = yz = a, say, then the line PQ is y = a. Otherwise, the line PQ is

Y - Yl X - X I - =-

Y2 - Y1 x2 - X I

Since P and Q lie on the conic,

ax: + 2hxlyl + by: + 2gxI + i f y l + c = o and ax; + 2hx2y2 + by: + 2gx2 + 2fy2 + c ' = 0.

Then (12) - (11)

a(x: - x?) + 2h(x2~2 - X I Y I ) + b(y$ - Y:) + 2g(x2

= $- - - [ ~ ( X I + x2) + 2hy2 + 2gl ( ) - [b (y, + y2) + 2hx1 + 2fl

Putting this in (lo), we get

I

As (x2, y2) tends to (x,, y,), (13) gives us the equation of the tangent to the given conic at (xl, y,). Thus, the equation of the tangent at P(x,, yl) is

( Y - Y I ) ( ~ Y I + hxl + f ) + (x-x l ) (ax , + ~ Y I + g ) = 0 +

e x(ax, + hy, + g) + y(by, + hxl + f ) + (gxl + fy, + c) = 0, using (11).

w axx, + h(xyl + xly) + byy, + g(x + x,) + f(y + y,) + c = 0. ...( 14)

Thus, (14) is the equation of the tangent to the conic ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 at the point (x,, y,) lying on the conic. From (14) you can see that we can use the following rule of thumb to obtain the equation of a conic.

In the equation of the conic, replace x2 by xxl, y2 by yyl, 2x 1 by (x + x l ) , 2y by (y + Y,) and 2xy by (XYI + YXI), to get the equation of the tangent at (xl , y,).

For instance, the tangent to the parabola y 2 - 4ax = 0, at a point.(xl, y,) is - n I - 2a (x + xl) = 0. We have already derived this in Sec. 2.3:2. In fact, the equations of thgents to the ellipse and hyperbola given in standard form are also special cases of (14), as you can verify from Unit 2.

Now youmay like to try your hand at finding tangents at some points.

E l j ) Obtain the equations of the tangent and the normal to the conic in E8 at the points where it cuts the y-axis.

In Unit 2 you have seen that not every line can be a tangent to a given standard conic. Let us now iee which lines qualify for being tangents to the general conic ax2 + 2hxy + by2 +'2gx + 2fy t c = 0. With your experience in Unit 2, can

. you tell the conditions under which tlie line px t qy + r = O will be a tangent to this conic ?

supposk it is a tangent a t a point (x,, y,) to the conic. Now, either p # 0 or

(" + r, in the q Z 0. Let us suppose p # 0. Then we can substitute x = - - equation of the conic, to get P

* (aq2 - 2hpq i- bp2) y 2 - 2y (prh + pqg - aqr - ~ ' f ) + (ar2 - 2gpr + cp 2)=0

The roots of this quadratic equation in y give us the y-coordinates of the points of intersection of the given line and conic. The line will be a tangent if these points coincide, that is, if the quadratic equation has coincident roots, that is, i f (prh + pqg - aqr - p2f)2 = (aq' - 2hpq + bp2) (ar2 - 2gpr + cp2). ...( 1 5 ) .

In terms of determinants (see MTE-04, Unit 5), we can write this conditim as

0 1 -1 0 I 0 - (-2.) 1 -2a 0 c 1 r n 1 -2a 0 0 = 0. rxpanding along the first raw.

a h g p

h b f q

g f c r

P Q r O

* (-24 (cm - 2a) - m(2ac) = 0

= 0 ...( 16)

a 3 c = -.

m

This is the same condi~ion that we derived in Sec. 2.3.2. Why don't yo11 try these exercises now?

Thus, (15) or the determinant condition (16) tell us if px + qy + r = 0 is a tangent to the general conic or not.

For example; the line y = mx + c will touch the parabola y2 = 4ax, i f

E14) Is x + 4y = 0 a tangent to the conic x 2 + 4xy + 3y2 - 5x - 6y + 3 = O? Find all the tangents to this conic that are parallel t o the given Iinc.

E15) a) Prove that the condition for ax .+ by + 1 = 0 to touch

x 2 + y 2 + 2gx + 2fy + c = Ois (ag i- bf - I ) ~ = (a2 + b2) (g2 + f 2 - c).

b) In partkular, under what conditions on C will y = Mx + C touch x 2 ,+ y2 = A'?

Conics In this section you saw that a line and a conic intersect in at most two points. Now'let us see what we get when two conics intersect.

3.6 INTERSECTION OF CONICS

Consider the intersection of an ellipse and a circle (Fig. 4(a)) or of an ellipse and a hyperbola (Fig. 4(b)).

Fig. 4: Intersecting conics.

You can see that these conics' intersect in four points. But, do any two conics 'intersect in four points? The foIlowing result answers this question.

Theorem 5 : In general, two conics intersect in four points.

Proof : Let the equations of the two conics be

ax2 + 2 (hy + g) x + by2 + 2fy + c = 0, and.

a l x2 + 2(hIy + g,) x + b ly2 + 2fly + cl = 0.

These equations can be considered as quadratic equations in x. If we eliminate x from them, we will get a fourth degree equation in y. This will have four roots. Corresponding to each of these roots, we will get a root of x. So there are,in general, four points of intersection for the two conics.

Since a fourth degree equation with real coefficients may have two or four complex roots (see MTE-04, Unit 3), two conics can intersect in i) four real points, ii) two real and two imaginary points, or iii) four imaginary points.

These points of intersection can be distinc't, or some may 'coincide, or all of them may coincide.

Let us consider an example.

Example 7: Find the points of intersection of the parabola y2 = 2x and the circle x Z + y2 = 1 (see Fig. 5).

Solution: I f (xi, y,) is a point of intersection, then x: + y: = 1 and y: = 2x1. Eliminating y, from these equations, we get

x: + 2xl = 1, that is, (x, + 112 = 2.

S o x , = -1 & a. Then y: = 2x, gives us

y, = & fi(a -1)'12 if x, = -1 + a, and y, = * ai ( a + 1)'12 i f xl = -1 - a. Thus, there are only two real points of intersection, namely,

( f i - 1, a ( f i - 1)'j2) and (a - 1, - fi (fi - I)"~). This.is why you see only two points of intersection in Fig. 4.

Here is an exercise for you now.

x y2 x y2 E16) Find the points of intersection of - + -- = 1 aiid - + - = 1. a 2 b2 b2 a 2

you have seen that two conics intersect in four real or imaginary points. Now we will firid the equation of any conic that passes through these points.

Let ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 and alx

2 + 2hl XY + bly2 + 2gl x + 2Fl y + c, = 0 be thg equations of two conics.

Let US briefly denote them by S = 0 and S , = 0, respectively.

Then, Tor each k f R, S + kS1 = 0 is a second degree equation in x and y . So it is a conic, for each value of k.

On the other hand, any point of intersection of the two conics satisfies both the equations S = 0 and SI = 0. Hence it satisfies S + kSI = 0. Thus, the conic S + k S , = 0 passes through all the points of intersectios of S = 0 and SI = 0.

So we have proved

Theorem 6: The equation of any conic passing through the intersection of two conics S = 0 and S l .= 0 is of the form S + kSI = 0, where k E R.

For different values o f k, we get different conics passing through the points of intersection of S = 0 and SI = 0. But, will all these conics be of the same type? If you do the following exercises, you may answer this question.

E17) 1 f : ~ = 0 and SI = 0 are rectangular hyperbolas, then show that S + kS, = 0 is a rectangular hyperbola, for all real k .

(Irint: Recall that ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is a rectangular hyperbola if a + b = 0.)

x2 y2 E18) Let S = - + - - 1 = 0 and SI = xy - 9 = 0.

9 4

Under what conditions on k will S + kS1 = 0 be a) an ellipse ? '

b) a parabola ? c) a hyperbola ?

Now we have come to the end of our discussion on conics. Let us see what we have covered in this unit.

3.7 SUMMARY.

In this unit we discussed the following points:

1) The general second degree equation

ax2 + 2hxy + by2 + 2gx + 2fy' + c = 0

Further, if the condition is satisfied, then the angle between the lines is

represents a conic. It is

2 . J K L b tan-' ( ) ;

a + b

ii) a parabola if ab - h 2 = 0, and the determinant condition in fi) is not satisfied ;

= 0. i) a pair of straight lines iff

iii) an ellipse if ab - h 2 > 0 ; iv) a hyperbola if ab - h2 < 0.

a h g h b f

g f c

General Theory of Conics ,, :

P Conics 2) An ellipse and a hyperbola are central conics; a parabola is a non-central conic.

3) A central conic with centre at the origih is of the form ax2 + 2hxy + by2

= 1, where a , h, b 6 R.

4) ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a central conic if ax + hy + g = 0 and,h%+ by + f - 0 intersect. And then, the centre of the conic is the point of intersection bf these lines. The slopes of rhe axes of this conic are the roots of the equation

5) ~ r a c i n i a conic.

6) The tangent to the conic ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 at the point ( X I , Y ~ ) is axx, + h(xyl + xly) + byy, + g(x + xl) + f(y + yl) + c = 0. Provided by, + hx, + f # 0

Further, a line px + qy + r = 0 is tangent to the given conic if

h b f q = 0

g - f c r

7) Two conics intersect in four points, which can be real or imaginary.

8) The equation of a conic passing through the four points of intersection of the conics S = 0 and S1 = 0 is S + kSI = 0, where k t R.

E l ) a) There are 3 types of non-degenerate conics: parabola, ellipse, hyperbola. There are 5 types of degenerate conics: point, pair of intersecting lines, pair of distinct parallel lines, pair of coincident lines, empty set.

b) A circle is a particular case of an ellipse. Thus, if (1) represents a circle then ab'- h2 > 0.

E2) a) x2 - 2xy + y2 + f i x - 2 = 0. Herea = 1 = b , h = -1. If we rotate the axes through ~ / 4 , then the new coordinates x ' and y' are given by

1 ' 1 x = fi - (x - y ' ) and y = fi - (xf + Y ' )

Thus, the given equation becomes

Now, if we shift the origin to (+, $), the equation becomes the . parabola

1 Y' = - - X, where X and'Y are the new coordinates.

2

b) 9x2 - 6xy + y2 - 40x - 20y + 75 = 0, 64 Herea = 9, b = 1, h = -3.

So, let us rotate the axes through 8, where

-

1 B = - tan-' (- $). :. tan 20 = - - 3 .

2 4' 3

I So we can take tan9 = 3, so that sin0 = - - I '

m1 coso = - - \ me

Then the equation in the X1Y&system.becomes I

Hence the given equation represents a pair of lines.

: E4) Here the discriminant concerned is

I 59 - y t 2 - 1 m i ' - lmyl + 75 = 0,.

I 5

I which can be transformed and seen to be,the equation of a parabola. I

I 5 7

I E3) In this case a = 3, b = 2, c = 2, f = - = g, h = - I 2 2 '

Thus, the givep equation represents a pair of lines. I

E5) a) The lines will be parallel if J G b = 0, that is, ab - h2 = 0. '

b) The lines will be perpendicular if a +' b = 0.

E6) Sirice P lies on the ellipse, so does P'.

Y - Y l x - X I I

The equation of PO is - = - that is, xl (y - yl) = y, (x - x,). -Y1 -XI

I I I

i

I . I

I

I

P' also lies on this since (-xl, -yl) satisfies this equation.

Hence, we have shown the result. I 927) In this case ab # h2. So the conic is central. Its centre is the intersection of .

=

a h g

h b f

g f c

7 5 7 . 5, 3 x t - y + - = Oand - x + 2y + - = 0,

2 2 2 2

3 ' I

that is, ( - --, - f ).

. 7 3 - -

2 2

- 5 7 2 -

2 2 5 5 - - 2 2 ' 2 .

3 ~ 8 ) ' Id this case a = 1 = b, h = - -

2 '

5 .

= o

So the given equation is central, andcan be a hyperbola or a pair of intersecting lines.

# O , Since

'

using Theorem 2 we can say t b t the equation represents a hjlperbola. Its centie is-the intersection of

3 3 x - - y + 5 = 0 and - - x + y - 5 = 0, that is, (-2, 2). 2 2 65

3 .- 1 - - 5 2

3 - - 1 - 5 2 .

' f

5 - 5 , 2 1

w -

Conics EQ) The central degenerate conics : point, pair of intersecting .lines.

The non-central degenerate conics : pair of distinct parallel lines, pair of coincident lines. The empty set is both central and non-central.

E10) The equation represents a hyperbola with centre (-2, 2). If we shift the origin to (-2, 2), the equation becomes - X I 2 + 3x1y' - y ' 2 = 1.

I 3

Herea = -1, b = -1, h = -. 2

7r 3* Thus, the axes of the conic are at an angle of - and - to the x-axis.

4 4

So, putting these values of 0 in (7), we get the lengths r l and f2, of the semi- 2

axes, on solving r: = 2 and rg = - -. 5

Thus, r, = \6 and r2 = ,I:. Note that over here, though r; is negative, we-only want its magnitude to compute the length of the axis.

Now, you know that if e is the eccentricity of the hyperbola then 7

r2 = r, m, that is, - = a li f

NOW let us also see where the hyperbola cuts the x and y axes. Putting y = 0 in the given equation, we get

x2 + lox + 21 = 0 * x = -3, -7.

So, the hyperbola intersects the x-axis in (-3, 0) and (-7, 0). Similarly, putting x = 0 in the given equation and solving for y, we see that the hyperbola intersects the y-axis in (0, 3) and (0, 7).

With all this information the curve is as given in Fig. 6 .

~ 1 1 ) It will be central if h 2 # 1. And then its centre will be the intersection of x + hy = 0 and hx + y + f = 0, which is

If we shift the origin to this point, the given equation is transformed to

! This is in the standard form A x 2 + 2HXY + BY^ = 1 of a central cora'ic. ! 1 - h 2

\ e ' HereA = B = - f 2 . Theretore, the,axes of the conic are at an angle of

1

i 45" and 135" to the x-axis. Since they pass through the centre, their I equations are

f - 1 - h

hf and y + - - x - - 1 - h 2

E12) The conic is a parabola since ab = h2, and the determinant condition for it to represent a pair of lines is not satisfied.

We can rewrite the equation ai

i2x- y12 = 8x + 6y - 5 .

We introduce a constant c to the equation, to get

(2x - y + c ) ~ = 8x + 6y - 5 + 4cx - 2cy + c2.

e, (2x - y + c12 = 4(2 + c) x + 2(3 - c) y. -1; c2

We choose c in such a way that

hen the eqdatioq of'the curve becomes

(2x.- y - 1)2 = 4(x + 2y - 1)

The vertex of this parabola is the intersection of 2x - y - 1 = 0 and

( , : ). The focus lies at x + 2y - 1 = 0, that is: - -

2 1 :. sine = - cose = - 6' x 6

:. The focus lies at

The curve intersects the y-axis in (0, 1) and (0, 5). 'It doesn't intersect the x-axis.

General Theory of Conics

Thus, the shape of the parabola is as given in Fig. 7.

Conics

Fig. 7

E13) The conic's equation is xa - 3xy + y2 + lox - 10y + 21 = 0.

From El0 you know it intersects the axes in (-3, O), (-7, O), (0, 3), (0, 7). The tangent at (-3, 0) is

Its slope is 4. 1

Thus, the slope of the normal at (-3, 0) is - -. Hence, its equation is 4

1 y , = - - ( x + 3).

4

You can similarly check that the tangents atF(-7, O), (0, 3) and (0, 7) are respectively, 4,

4 x - l l y + 28 = 0,

x - 4y + 12 = 0,

l l x - < 4 y + 28 = 0. '.

, The normals at these points. are respectively,

General Theory of Conics

# 40 = 0, which is false.

Thus, the given line is not a tangent to the given conic. . ,

Any line parallel to the given line is of the form x + 4y + c = 0. This will be a tangent to the given conic if (15) is satisfied, that is,

(5c + 2812 = 3(3c2 .+ 24c + 48) o c = -5 or -8.

Thus, the required tangents are , '

x + 4 y - 5 = O a n d x + 4 y - 8 = 0.

E15) a) Using (15), we see that the condition is

o b(f - bc) - (1 -bf) + g [af + b(bg-af)) + a [(ac-g) + f(bg-af)) = 0

e, b2g2 + a2f2 + 2bf - 2abfg - b2c'- 1 + 2ag - a2c = , 0 . . Adding a2g2 on both sides and simplifying, we get the given condition.

. . a 1 . b) In (a) we put g = 0 = f, c = - A2, - - = M , - - = C.

b b - SO the condition for y = Mx + , Z to touch .x2 .+ y2 = is

a2b2 ab Then x2 = a2 1 - '

a2 1 = a2 + b2 ( .. a 2 + b2 * X * J-F>-b=

Thus the.4 points of intersection are

f

and (6&,Js)- at

I i t

, We have drawn the situation in Fig. 8. i E17) Let S s ax2 +. 2hxy + by2 ; 2gx + 2fy + c = 0 1

and S , *.alx2 + 2hlxy + bjy2 .+ 2glx + 2fly + c, = 0 I 1

be rectangular hyperbolas. Then ' ,

a + b = O a n d a l + . b l = O . ' 69

L- -. ,. .

Conics :. (a + b) + k(a, + b,) = 0 v k E R ++ (a2 + ka,) + (t, + kbl) = 0 v k E R ++ S + kS1 = 0 is a rectangular hyperbola v k E R.

E18)S + kS1 = 0

a) This conic will be an ellipse if

k 1 (i) (+) - - > 0, that is, k2 < - - a

4 9

b) The conic will be a parabola if

But this can't be.

1 k 2 = - a n d

9

So the conic can't be a parabola.

1 But it will be a pair of lines if k = rt - - 3

k - - 1 - 2 4

0

1 c) The conic will be a hyperbola if k2 > -..

9

f 0, that is,

Genemi Tbccrry of Conlcs

MISCELLANEOUS EXERCISES

(Tbis section is optionall)

In this section we have gathered some problems related to the contents of this block. You may like to do them to get a better understanding of conics. Our solutions to the questions follow the list of problems, in case you'd like to counter-check your answers.

Find the equation of the path traced by a point P , the sum of the squares of the distances from (1, 0) and (-1, 0) of which is 8.

A path traced by a moving point is called its locus. Find the equation of the circle which passes through (1, O), (0, -6) and (1,4).

(Hint: The general equation of a circle is x2 + y2 + 2gx + 2fy + c = 0).

Prove the reflecting property for k parabola.

(Hint: Show that CY = /3 in Fig. 9, Unit 2.)

Prove Theorem 2 of Unit 3.

A circle cuts the parabola yZ = 4ax in the points (at:, 2ati) for i = 1, 2, 3, 4. Prove that tl + t2 + tg + ' t4 = 0.

(Hint: tl, t2, t$ t4 are the solutions of the quadric equation obtained by putting x = o.), y = 2at in the equation of a circle.)

Trace the curves xy = 0 and xy - 4x - 5y + 20 = 0.

What relations must hold between the coefficients of ax2 + by2 + cx + cy = 0 for it to represent a pair of straight lines? .

Find the angle through which the axes should be rotated so that the equation Ax + By + C = 0 is reduced to the form x = constant, and find the value of the constant.

Prove that y2 + 2Ax + 2By + C = 0 represents a parabola whose axis is parallel to the x-axis. Find its vertex and the equation of its latus rectum.

Prove that the set of midpoints of all chords of y2=4ax which are drawn through its vertex is the parabola y2 = 2ax.

Y! a) Prove that -!& + 7 - 1 is negative, zero or positive, according as the a Z b *

x y" point (xl, yl) lies inside, on or outside the ellipse - + , = 1.

a2 b

b) Is the point (4, -3) inside or outside the ellipse 5x2 + *2 = 11?

A line segment of fixed length a + b moves so that its ends are always on two fixed perpendicular lines (see Fig. 1). Prove that the path traced by a point which divides 'this segment in. the ratio a : b is an ellipse.

Find the equation of the common tangent to the hyperbolas Fig, 1

x2 y2 A normal to - - - = 1 meets the x and y axes in M and N, respectively.

a Z b2'

The lines through M and N drawn perpendicular to the x and y-axes, respectively, meet in the point P. Prove that the locus of P is the hyperbola azx2'- b S 2 = (a2 + b2)'.

Consider the hyperbola in Fig. 20 d Unit 2. Through A and A' draw parallels.to the conjugate axis, and through B and B' draw parallels to the transverse axis. Show that the diagonals of the rectangle so formed lie along the asymptotes of the hyperbola. A method for drawing

asymptotes of a hyperbola. Which conics are represented by the following equations?

a) ( x - ~ ) ~ + (x-a12 = 0,

Cpnlca b) r sin28 = 2a cos 8,

- 1 + Ease + fi sine. c) - - I. .

17) Trace the conics

18) Find the equation to t'he conic which passes through (1, 1) and the intersection of x2 + 2xy + 5y2 - 7x - 8y t 6 7 0 with the pair of straight lines 2x'- y - 9 = 0 and 3x+y-11 = 0.

Solutions ,

1) Let P be (x, y). Then ((x - 112 + y2) + ((x + 1)' + Y') = 8 e 2x2 + 2y2 = 6 e x2 + y2 = 3, which is a circle with centre (0, 0) and radius fi.

2) Let the equation be x 2 + y2 + 2gx. + 2fy +: c = 0. Since (1, O), (0, -6) and : (3,4)lieo,nit,'

Solving these three linear equations in g, ,f and c, we get

Thus the equation is

3) The parabola is y2' = 4ax. The tangent T at a point P(xl, y,) is yyl. = 2a(x + XI.).

Y1 Y 1 - Y l x - X l The line PF, where Fa,, 0) is the focus, is - = -

-Y1 a - X l . Y1 Its s l ~ p e is -.

x1 - a 2a Y1 - --

Thus, tan@ = 'I - a 'I , using (1 1) cif Unit 1. -

#a 1 + ---- X I - a

2a ='-, using the fact that ~f = 4ax1.

Y l

Thus tanm = tan@ and cr and /3 are both less than or equal t o 90".

P k 3,' 4) We want to show that

{" ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ...( 1)

ii chn be written as a product of two linear factors iff its discriminant is 0. i* ; $ I if a # 0, we multiply (1) throughout by a and arrange it in decreasing ,

powers of x. We get

a2x2 + 2ax (hy. + gL= - aby2 - 2afjl - ac.

completing the square on the left hand, we get , 72 . a2x2 + 2ax (hy + g) + (hy + g)2 = y2(hZ - ab) + 2y (gh - af) + g2 - ac

-- -

a ax + hy + g = * dy2(h2 - ab) + 2y (gh-af) + g2-ac ' General Theory of Conics

From this we can obtain x in terms of y, only involving the first degree iff the quantity under the square root sign is a perfect square, that is, iff (gh - ao2 = (h2 - ab) (g2 - ac),

s abc + 2fgh - afT - bg2 - ch2 = 0.

5) Let the circle's equation be

Substituting x = at2, y = 2at in this, we get

a2t4 + 4 a V + 2agi2 + 4aft + c = 0. '

We,know that it has 4 roots tl , t2, tJ, t4. SO, from MTE-04 you know that

1 the sum.of the roots will be - (coefficient of t3) = 0.

a

6) xy = 0 is the pair of lines x =O and y = 0. We. have traced it in Fig.. 2.

xy - 4x - 5y + 20 = 0 is a pair of lines since its discriminant is 0. In fact, we can easily factorise it as

!x - 5) (y - 4) = 0.

Thus, it represents the pair of lines x = 5 and y = 4, which we have traced in Fig. 3.

7). ax2 + by2 + cx + cy = 0 represents a pair of lines iff

Fig. 2

8) Let us rotate the axes through 8. Then the equation becomes

A(x1cost9 - y'sint9) i B(x1sint9 + yfcost9) + C = '0,

s xl(A cost9 + B sin@) + y'(B cose - A sine) + C = 0.

This will reduce to the form x' = constant iff B cost9 - A sing,

B that is, t9 = tan-' - .

h . .. -

~ n d - t h e n the equation becomes

-C Thus, the constant is JFXi=-.

9) We rewrite the given equation as

Y2 = -(2Ax + 2By + C)

e (y + kl2: = -2Ax - 2By - C + 2ky + k2, where k is a constant.

e (y + k)2 = -2Ax + 2 ( k - ~ ) y + k ~ - ~

We choose k so that

I

Fig. 3 .

Conics

Fig. 4

C-k2

Ax + (B - k)y + - = 0 is parallel to the y-axis, that is, 2

k = B. Then the equation becomes

Its axis is y + B = 0, vertex and the equatibn of

B~ - A~ - C its latus rectum is x =

2A

10) The midpoint of any chord through P(xl, yl) and O(0, 0) is '

Q ---, - Since y: = 4axl, (: t ) (+)' = 2a (+). Thus, the set of all such Q is y2 = 2ax.

11) a) Firstly, if (xl, y,) lies on the ellipse, then clearly

x : Y: - + 7 = 1 . a2 b

Now, if (x,, y,) lies outside the ellipse (see Fig. 4) then either Ixl 1 > a

or 1 ~ 1 1 > b. :. x; > a2 or y: > b2

Y: , x: , - . . + - > I . a2 b2

Similarly, you can show that if (xl, yl) lies inside the ellipse,

b) Since 5(16) + 7(9) = 143 > l l , the point lies outside the ellipse.

12) Let the perpendicular lines be the coordinate axes. Let the segment intersect 'the axes in (x,O) and (0,y). Then the coordinates of the point P are

Now, since x2 +- y2 = (a+b12

x i y2 Thus, the path traced by P is the ellipse - + = 1.

a 2 b

xZ y2 13) Any tangent to 7 - -? = 1 is y = rnx + 4- and any '

a b y 2 x2

tangent to 7 - - = 1 is x = rn, y + J a w 2 . a b 2

1 For these two lines to be the same, we must have - = rn and

ml

Thus the commbn tangents are y = x + Jz2 and

x2 y2 14) See Fig. 5 for a diagram of the situation. The normal to - - 7 = 1

at (XI, YI) is a2 b

Thus, M is , 0 ) and N is ( 0, ( a 2 + b p 2 ) ~ ~ )

Thus, the coordinates of P are

Now, since (x,, y,) lies on the hyperbola,

a2 + b2 . X, and Y =

a 2 + b2

a a2x2 - b2y2 = (a2 + b2)2, where X = a2 - b

Y I .

Now, as P varies, X and Y vary; but always satisfy the relationship a2x2 - b2y2 = (a2 + b?12. Thus, this is the locus of the point P.

15) The lines meet in (d, b), (a, -b), (-a, b) and (-a, -b). Thus the diagonals of b b

the rectangles lie along y = - x and y = - -x, which are the asymptotes of the hyperbola. a a

16) a) (x - y)2 + (x - a)2 = 0 * 2x2'- 2xy + y2 -2ax + a2 = 0:

2 Herea = 2, b = 1, h = -1 ,g = -a, f = 0 , c = a .

:. ab - h 2 > 0. Thus, the conic is an ellipse.

b) r sin2@ = &a cod. Changing to Cartesian coordinates, this equation is y2 = 2ax, a parabola.

o 1 = F + - y 2 + x + \/g since x = r cos 0 , y = r 'sin 8.

* 2y2'+ 2Gxy + 2x -t 2fly + I = 0

Here ab - h2 .< 0 and its discriminant is

Thus, the curve represents a hyperbola.

17) a) You can check that ab-h2 = 0 and the discrimtnant is non-zero. Thus, the eqyation represents a parabola. We can write it as

(3x - 4 ~ ) ~ = 18x + lOly -19.

* (3x - 4y + c12 = (6c + 18) x + y(101 - Bc) + c2 -19, where wc

General Theory of Conics

Fig. 5

Conics choose the constant c so that

3 ( 6 ~ + 18) -4 (101 - 8 ~ ) = 0

Fig. 6

1 This is a hyperbola. Its centre is the intersection of - - y = 0 and

2 1 - - x + y = 0, that is, (0, 0). Its axes are inclined to the coordinate 2

axes at an angle of 8, where tan 28 = 1. Thus, the slope of the

+ c = 7 .

Then ;he given equation becomes

(3x-4y + 7)' = 15(4x t 3y + 2)

Thus, the axis of the parabola is 4x + 3y t 2 = 0. The vertex is the intersection of 3x - 4y + 7 = 0 and 4x + 3y + 2 = 0;

29 22 that is, (- -, 25 -) 25

The length of its latus rectum is 3. It's focus F lies

29 3 22 3 4 at (- - + - cos 0 , - + - sin , where tan 0 = - -

25 4 25 4 3 '

:. F is (-0. 71, 0.28).

101 & 4 ( lo l l2 - 64 x 19 The curve intersects the y-axis in

'3 2 I

49 3 that is, approximately, - and -.

8 16

It doesn't intersect the x-axis.

We have traced it in Fig. 6.

s , 5s transverse axis is 0 , = -, and of the conjugate axis is O2 = -.

8 8 Since tan 8, = .41, the length of the transverse axis, rl, is given by

:. r, = 1.24, approximately. We similarly find r2 = .91. Thus its eccentricity is 1.24. It doesn't interesect the axes. With all this information, we have traced the curve in Fig. 7.

Fig. 7

c) The equation is a hyperbola whose centre is the intersection of

3x - 4y + 1 = 0 and 4x + 3y + 1 = 0, that is', m I

Here a = 12, b = - 12, h = - -. ,2

Thus, its axes are inclined at angles 0, and O2 to the coordinate axes, where tan 0, and tan O2 are roots of the equation

a-b tan28 + - tan e -i = 0

48 a tan2 8 - - tan 8 - 1 = 0

7 1

a tan = 7 and tan O2 = .- - * 8, = 81. 9' (approx.) and 7

O2 = 171.9' (approx.).

The length of its axes are rl and r2, where

The curve intersects the coordinate axes in (0, 0), (-&. 0).

General Theory of Conics

So, its curve is as given in Fig. 8.

Conics

18) Let SI = x2 + 2xy + 5y2 - 7x -8y % 6 = Oand

Sz " (2x - y -"5) (3x + y - 11) = 0.

. Then the required conic is SI + kS2 = 0, where we choose k so that (1, 1) lies on the curve. Thus, the curve is

' 1 Since (I, -1) lies on it, k = -. 28

Thus, the conic is

34x2 + 55xy + 139y2 - 233x - 218x - 218y + 223 = 0.

N O T E S