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  • 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.