The details of the derivation of the equations of conics are com-

Transcription

1 Part 6 Conic sections Introduction Consider the double cone shown in the diagram, joined at the verte. These cones are right circular cones in the sense that slicing the double cones with planes at right-angles to the ais gives cross section which are circular. Now consider starting with a horizontal slicing plane and tilting it slightl. We obtain an oval curve called an ellipse. Suppose now that the slicing plane has the same slope as the surface of the cone that does not pass thru the verte. In this case we obtain a parabola. If we tilt the slicing plane further so that it cuts both cones, we obtain a hperbola. There are other trivial (degenerate) cases which give point and straight line intersections.] These curves turn up in various ares of mathematics and phsics. e.g. the motion of planets and comets around the sun follow an elliptical orbit, the path turned out b a particle (eg a ball) moving under gravit (ignoring air resistance) is a parabola, the path of an alpha particle moving near to the nucleus of an atom is a hperbola. Halle s commet Sun ellipse motion under gravit The details of the derivation of the equations of conics are com- 1

2 plicated and not useful: we will simpl state the standard equations of these curves in a suitable (, ) coordinate sstem in the slicing plane. 1/2 =2 2 (1,2) alpha particle nucleus (0,0) 1-2 (1,-2) 1/2 =-2 The parabola The standard eqn for the parabola is 2 = 4a, parabola in standard position where a > is a constant. eg with a = 1, 2 = 4, (i.e. = ±2 ). In general the parabola with eqn 2 = 4a (i) meets the aes at (0, 0) (ii) is smmetric w.r.t. the -ais (iii) is unbounded (i.e. as, ± ). The point (0, 0) is the verte of the parabola. The focus-directri propert Consider the following diagram. We show that the eqn of the 2

3 parabola is the locus of points such that PF=PD. D(-a,) P(,) -a a -a O F(a,0) =-a Figure 1: 2 = 4a, a > 0 P F = ( a) 2 + 2, P D = + a Squaring ( a) = ( + a) 2 2 2a + a = 2 + 2a + a 2 2 = 4a The point F (a, 0) is the focus and the line = a the directri. Eample 1: Find the focus and directri of the parabola 2 = 12 Solution Form 2 = 4a with a = 3 4a = 12, a = 3 focus is ( 3, 0) Directri is = 3 3

4 There are four standard positions for the focus-directri combinations. The remaining three are as follows. a a O O -a -a Figure 2: 2 = 4a, a > 0 Figure 3: 2 = 4a, a > 0 Eample 2: Find the verte, focus, directri and ais of 2 = Solution Write as ie = ( + 3 ) 2 = ( ( = 4 2) 17 ) 16 which is of the form X 2 = 4AY with A = 1 and X = + 3 2, Y =

5 Verte is at (X, Y ) = (0, 0) i.e. ( (, ) = 3 2, 17 ) 16 Focus is on (X, Y ) = (0, A) (, ) = ( 3 2, 1 ) 16 Directri is on Y = A = 1 = =33/16 verte (-3/2,17/16) focus (-3/2,1/16) -1 1 Tangent at a point on the parabola Eample 3: Find the eqn of the tangent to 2 = 20 at the point (3, 2 15). 5

6 [Check: when = 3, 2 = 60, = ±2 15] Solution Differentiating 2 = 20 gives 2 d d = 20 d d = 10 At (3, 2 15), d d = 5 15 Hence = c, c = 15 =

7 Tangent to parabola from a point not on the parabola As in the case of circles there will be either two solutions or no solutions. Eample 4: Find the eqns of the tangents to the parabola = 0 which pass thru ( 5, 2). Solution or Slope of curve at an point is given b Line thru ( 5, 2) has eqn 2 d d d d 2 = 0 d d = = m( + 5), i.e.m = Hence =

8 or Eqn of parabola is Eliminating b (1)-(2): gives = 0, (1) = 0, (2) = 0 ( 6)( + 2) = 0, = 6, 2 When = 6, 2 = 2(36) 30 8, = 17 Slope Tangents are = 2, = = 2 11, = = 2 11 ( + 5), 2 = 2 ( + 5) 5 Reflection propert of the parabola Let P ( 0, 0 ) be an point on the parabola 2 = 4a, with focus F (a, 0). Then the angle between FP and the tangent at P is the same as the angle between the tangent at P and the -ais. d d = 2a, d d = 2a 0 at ( 0, 0 ) 8

9 Eqn of tangent at P is 0 = 2a 0 ( 0 ) This line meets the -ais at A ( = 0) where PF = = 2 0 2a + 0 = = 0 AF = 0 + a ( 0 a) = 2 0 2a 0 + a 2 + 4a 0 = 0 +a = AF AFP is an isosceles triangle and the stated result follows. Hence a ra of light originated from the focus of F is reflected parallel to the ais of parabola. If a parabola is revolved about its ais the surface generated is called a paraboloid. Such surfaces are used in headlights, optical and radio telescopes, radar etc. 9

10 Parametric form of the parabola The standard parametrisation is = at 2, = 2at (eliminating t gives 2 = 4a) Note that positive values of t give points (, ) on the upper arm of the parabola whereas negative values of t give points on the lower arm. 10

11 (0,b) (0,b) (-a,0) (0,0) (a,0) (-a,0) (0,0) (a,0) (0,-b) (0,-b) The Ellipse An ellipse is obtained when the slicing plane is less steep than the surface of the cone. Its standard equation is 2 a b 2 = 1, where a, b > 0. If a > b we have ( a, 0), (a, 0), (0, b) and (0, b) are vertices. (left: standard position) whereas if a < b (right: reflected standard position - b reflection in the line =.) (If a = b the eqn represents the circle = a 2.) The and aes are the aes of smmetr of the ellipse, for the ellipse in standard position the line segment from ( a, 0) to (a, 0) is the major ais and the line segment from (0, b) to (0, b) is the minor ais. (The minor ais is never longer than the major ais.) It turns out that an ellipse is the locus all points the sum of whose distances from two fied points is constant. The two fied points are called foci, F 1 (c, 0) and F 2 ( c, 0) which, here, we assume to be on the ais. 11

12 P(,) d2 d1 F2(-c,0) (0,0) F1(c,0) Thus, d 1 + d 2 = const = 2a, sa where 2a > 2c. Hence ( c)2 + ( 0) 2 + ( + c) 2 + ( 0) 2 = 2a or ( c) = 4a 2 4a ( + c) ( + c) c = 4a 2 4a ( + c) ( + c)2 + 2 = a + c a Thus which simplifies to ( a 2 c 2 ( + c) = a 2 + 2c + c2 a 22 a 2 ) = a 2 c 2 Hence 2 a a 2 c 2 = 1, 12

13 b letting b 2 = a 2 c 2 we have 2 a b 2 = 1. (This construction allows an ellipse to be drawn via pencil, string etc. - the geometric method.) The eccentricit of an ellipse is defined as e = c a so that the foci are at (±ae, 0) and since c 2 = a 2 b 2 e = 1 b2 a 2. Clearl 0 < e < 1. For a nearl circular ellipse e is close to 0, whereas for a ver squashed ellipse e is close to 1. Eample 5: Find the equations of the ellipse satisfing (i) Vertices at (±6, 0) eccentricit 2 (ii) Foci at (0, ±4), eccentricit 4/5 3 Solution (i) a = 6, 2/3 = 1 b 2 /36 Hence 4 9 = 1 b2 36, b2 = = 1. (ii) The foci (0, ±4) are on the -ais, hence major ais is along the ais. Here it is essential to retain a for the major and b for 13

14 the minor ais for our formula to be valid. Hence 4 = ae = 4 5 a, a = = 1 b2 25 b2 = 9, b = 3 Thus, the equation is 2 ( = 1, 2 ) In general b a = 1 2 Tangent at a point on the ellipse Eample 6: Find the eqn of the tangent at the point (2, 1) on the ellipse = 6. Solution d d = 0, d d = 2 At (2, 1), d d = 1. Hence tangent is ( 1) = 1( 2) or 3 = 0 Tangent from a point outside the ellipse 14

15 Eample 7: Find the equations of the lines thru (1, 4) which are tangents to the ellipse = 6. Solution The eqn of the line thru (1, 4) with gradient m is 4 = m( 1). (1,4) The line = m (m 4) meets the ellipse when 2 + 2[m (m 4)] 2 = 6 or (1 + 2m 2 ) 2 4m(m 4) + 2m 2 16m + 26 = 0 (1) As each tangent meets the ellipse in eactl one point we look for repeated roots of this eqn. Hence we require [4m(m 4)] 2 4(1 + 2m 2 )(2m 2 16m + 26) = 0 or 5m 2 + 8m 13 = 0 (5m + 13)(m 1) = 0 15

16 m = 13 5, or 1 Hence tangents are 4 = 13 ( 1) ie = 33 5 and 4 = 1 ie = 3. The points of tangenc From (1) with m = 1 we find With m = 13/ = 0 ( + 2) 2 = 0, = 2 and = 3 + ( 2) = = 0 ie = 0 (11 26) 2 = 0, = = 5 11 The focus-directri propert Here we show that P F 1 = ep D 1 where D 1 is on the line = a. The eqn of the ellipse is e 2 a a 2 (1 e 2 ) = 1, ( b2 = a 2 (1 e 2 )) 2 (1 e 2 ) + 2 = a 2 (1 e 2 ) 2 + a 2 e = a 2 + e

17 Substituting 2ae from both sides we have ( ae) = (a e) 2 = e 2 ( a e ) 2 Now P F 1 = ( ae) and P D 1 = a e if D 1 is on the line = a/e. P F 1 = ep D 1 SimilarlP F 2 = ep D 2 17

18 The reflection propert beta P(,) alpha tangent at P F2 F1 It can be shown that α = β (proof omitted). This means that if the ellipse were a mirrored surface then an light emitted at F 1 will reflect onto F 2 and vice versa. The same idea is eploited in the construction of whispering galleries. Parametric equations The standard parametrisation of the ellipse is 2 a b 2 = 1 = a cos t, = b sin t (0 t 2π) As t increases from 0 to 2π the point (a cos t, b sin t) travels once round the ellipse (anticlockwise) starting and finishing at (a, 0). 18

19 d2 P(,) d2 d1 d1 F2(-c,0) F1(c,0) The hperbola A hperbola is obtained when the slicing plane is steeper than the surface of the double cone. The standard equation of the hperbola is 2 a 2 2 b = 1, 2 where a, b > 0. There is no requirement that a > b. Writing this in the form ( ) 2 = b 2 2 a 1 = b2 2 ) (1 a2 2 a 2 2 we find = ± b a As ± this tends to the lines ) 1/2 (1 a2. 2 = ± b a which are the asmptotes of the hperbola. When = 0, = ±a, the vertices. 19

20 It turns out that a hperbola is the locus of all points the difference of whose coordinates from two fied points is a positive constant. The two fied points are called foci, F 1 (c, 0) and F 2 ( c, 0). Thus either d 1 d 2 or d 2 d 1 is a positive constant which we call 2a, i.e. d 1 d 2 = ±2a. Thus ( c) ( + c) = ±2a Transferring second to right and squaring ( c) = ( + c) ± 4a ( + c) a 2 which reduces to Dividing b 4a and squaring or ( c 2 a 2 ±4a ( + c) = 4a 2 + 4c 2 + 2c + c = a 2 + 2c + c2 a 2 ) 2 2 = c 2 a 2 a 22 Hence 2 a 2 2 c 2 a = 1. 2 Recall that the sum of the lengths of an two sides of a triangle must be greater than the length of the third side. Hence 2c+d 2 > d 1 and 2c + d 1 > d 2 and thus 2c > d 1 d 2. But d 1 d 2 = 2a, c > a. We write b = c 2 a 2 ( where b > 0) and hence 2 a 2 2 b 2 = 1. 20

21 The eccentricit of a hperbola is defined as e = c so that the a foci are at (±ae, 0) and since c 2 = b 2 + a 2 e = 1 + b2 a2.( Note e > 1) For a ver thin hperbola b/a is small and e is close to 1. For a fat hperbola b/a is ver large and hence e is ver large. Eample 8: Find the equation of the hperbola satisfing (i) vertices at (±5, 0), foci at (±7, 0) (ii) vertices at (0, ±7), eccentricit 4/3. Solution (i) a = 5, ae = 7, 7/5 7 5 = 1 + b2 25, = 1 + b2 25, b2 = 24 Hence = 1 (ii) Here the foci are on the ais and the eqn takes the form 2 a 2 2 b 2 = 1 with the vertices (0, ±a), foci (0, ±ae) and e as before. Hence a = 7, ae = 28 ( 3 foci 0, ± 28 ) = 1 + b2 49, b2 =

22 = 1 Tangent at a point on a hperbola Eample 9: Find the eqn of the tangent to the hperbola = 1 at the point (5, 16/3) on the hperbola. Solution d d = 0, 16 d d = 16 9 at (5, 16/3), d d = = 5 3 ( 16 ) 3 3 = = 5 ( 5) 3 22

23 Tangents from a point not on the hperbola It turn out that we can onl construct tangents to a hperbola f(,)<1 P(,) f(,)>1 e < -a f(,)>1 e>a (-a,0) (a,0) from points as P where 2 a 2 2 b < 1. 2 Here, f(, ) = 2 a 2 2 b 2. Eample 10: Find the eqns of the tangents from P (1, 3) to the hperbola = 1. Solution 23

24 Line thru (1, 3) with slope m is 3 = m( 1) This line meets the hperbola at or [m (m 3)]2 = 1 (3 4m 2 ) 2 + 8m(m 3) 4m m 48 = 0 (1) For repeated roots 64m 2 (m 3) 2 + 4(3 4m 2 )(4m 2 24m + 48) = 0 which reduces to Tangents are m 2 + 2m 4 = 0 m = 1 ± 5 3 = ( 1 + 5)( 1) ie = ( 1 + 5) = ( 1 5)( 1) ie = ( 1 5) Points of tangenc can be found b solving (1) for each m Ans; ( 4 ± 12 ) 5, 12 ±

25 The focus-directri propert The relationship P F 1,2 = ep D 1,2 holds as for the ellipse and the directri are = ± a e. e.g. (e > 1) D1 P(,) a F1(ae,0) =a/e Reflection propert The ra is reflected as if it came from the other focus F 2 ( ae, 0). F2(-ae,0) F1(ae,0) (This idea is used in some kind of telescope.) 25

26 Parametric equations The standard parametrisation of the hperbola is = a sec t, = b tan t ( 12 π < t < 12 π, 12 π < t < 32 ) π As t increases from 1 2 π to 1 2π the point (a sec t, b tan t) moves upwards along the right-hand branch of the hperbola. The lefthand branch is obtained as t increases from 1 2 π to 3 2 π. The rectangular hperbola If b = a then the asmptotes are = ± (intersect at right angles). It turns out that if we make a change of coordinates so that = becomes the new -ais and = the new -ais the equation has a simpler form. 2 2 = a 2 = Y P (X,Y) X cos P(,) =- Consider the following : X = cos θ sin θ Y = sin θ + cos θ 26

27 or ( ) X = Y ( cos θ sin θ sin θ cos θ ) ( ) In our case we need to choose θ = π/4 hence =c 2 O Thus X = 2, Y = = ( )( + ) = 2XY = a 2 Changing the notation we can write = c 2 where c = a/ 2 The standard parametrisation of = c 2 is = ct, = c t (t 0) 27

28 (2,3) slope -1/2 slope 1/2 (2,1) (5,1) (1,-5) (5,-5) (3,-5) Quadratic curves Consider the general quadratics curve A 2 + B + c 2 + D + E + F = 0 with A, B, C not all zero. Here we focus on the case B = 0. B completing the square (if necessar) we can show that such eqns often represent non-degenerate conics which are shifted or translated from the standard position. (Alread seen a shifted parabolas.) Eample 11: Identif the following conics (i) = 0 (ii) = 0 Solution 28

29 (i) 4( 2 4) + 9( 2 2) 11 = 0 4[( 2) 2 4] + 9[( 1) 2 1] 11 = 0 4( 2) 2 + 9( 1) 2 = 36 ( 2) 2 ( 1)2 + = Ellipse with center (2, 1) major ais 3, minor ais 2 (ii) 2 6 4( ) 95 = 0 ( 3) 2 9 4( + 5) = 0 ( 3) 2 4( + 5) 2 = 4 ( 3) 2 ( + 5) 2 = 1 4 Shifted hperbola (centre (3, 5)) Note: general quadratic could represent nothing eg. ( 2) 2 + ( 3) = 0. Can also represent two parallel lines eg 2 2 = 0 (A = 1, C = 0, D = 2, E = F = 0) or just one line eg = 0 or = = 0, ie ( 1) 2 = 0. The case B 0 involves rotating the coordinate sstem in such a wa that in the new sstem (, ) there is no term. (see Maple e.) 29