Senior Math Circles February 18, 2009 Conics III
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1 University of Waterloo Faculty of Mathematics Senior Math Circles February 18, 2009 Conics III Centre for Education in Mathematics and Computing Eccentricity of Conics Fix a point F called the focus, a line d called the directrix, and a real number e > 0 (the eccentricity). Claim: The locus {P : P F = e P d } is an ellipse for e < 1, parabola for e = 1, and a hyperbola for e > 1. Proof: Use a similarity transformation to take d to the y-axis and F to (1, 0). Then P = (x, y) is on the locus if and only if (x 1)2 + y 2 = e x (x 1) 2 + y 2 = e 2 x 2 if e = 1 this gives 2x y 2 = 0 or x = 1+y2, a parabola. 2 Otherwise we can rewrite as (1 e 2 )(x 1 ) 2 + y 2 = e 2 1 e 2 which is an ellipse when 1 e 2 > 0, a hyperbola when 1 e 2 < 0. Reflective Properties When light reflects off a curve, the angles of incidence and angles of reflection are equal (with respect to the tangent): This definition of reflection can be used to prove reflective properties of conics. Proof of reflective property of ellipse: (rays leaving one focus will pass through the other focus after reflecting): 1
2 Let F 1, be the foci and X any point on the ellipse. If F 1 X is indeed the light s path, then the tangent must be the extended angle bisector l of F 1 X and vice-versa. X F 1 l Suppose, for the sake of contradiction, that l is not tangent to the ellipse. Then it hits the ellipse twice. Let X be a point on l but strictly inside the ellipse; then F 1 X + X < 2a where a is the focal radius. On the other hand, construct the image of after reflecting through l; call is F 2. Since F 1 XF 2 is a straight line and F 1 X F 2 is not, the triangle inequality gives: F 1 X + X = F 1 X + F 2X > F 1 X + F 2X = F 1 X + X = 2a. Since F 1 X + X cannot be both larger than and smaller than 2a, we obtain the desired contradiction. More generally, for conics, A B C, with B on the conic, is a valid reflection if B is a local minimizer or maximizer of AB + BC. This approach can be used to prove the reflective property of ellipses and hyperbolas. Envelope Envelope problems are similar to locus problems, except we are given a moving line instead of a moving point... A family of lines is said to envelop a curve if each line in the family is tangent to the curves; the curve is called the envelope of the family. Duality between ellipse and its set of tangents: set of tangents envelope 2
3 Despite the apparent grossness, envelope problems are often solvable by simple algebraic methods. Key observation: any point on the envelope lies in exactly one member of the family of lines. Sample Problem: Consider the family of lines whose x- and y-intercept add up to 1. Find the envelope of this family. Solution: Let t denote a parameter which we will use to describe the family of lines; specifically, let any line in this family have x-intercept (t, 0) and y-intercept (0, 1 t). This line has equation = 1. (*) Now we use the key observation : for any point x, y on the curve, there is only one t for which (*) holds. Solve (*) for t: (1 t)x + ty = t(1 t) t 2 + t(y x 1) + x = 0, a quadratic in t. This has one root if and only if the discriminant B 2 4AC vanishes, i.e. (y x 1) 2 4x = 0. x + y t 1 t Rule. Let αx 2 + βxy + δx + ɛy + η = 0 be a conic. It is a ellipse β 2 4αδ < 0 parabola if β 2 4αδ = 0 hyperbola β 2 4αδ > 0 3
4 Ellipse Parabola Hyperbola F 1 F F 1 Picture d Typical Equations Typical Definition Reflective Property x 2 + y 2 = 1, x 2 /a + y 2 /b = c, (x x 0 ) 2 /a+(y y 0 ) 2 /b = c y x 2 = 0, y = ax 2 + bx + c, x = ay 2 + by + c x 2 y 2 = 1, x 2 /a y 2 /b = c, (x x 0 ) 2 /a (y y 0 ) 2 /b = c, xy = c, (x x 0 )(y y 0 ) = c {P : P F 1 + P = 2a} {P : P F = P d } {P : P F 1 P = ±2a} Rays leaving F Rays leaving F 1, after reflecting off the ellipse, hit flecting off the parabola, Rays leaving F, after re- 1, after reflecting off the hyperbola, travel directly away from. are perpendicular to d.. Eccentricity* 0 < e < 1 e = 1 e > 1 *: This gives an alternate way of defining ellipses and hyperbolas. Namely, for a line d and point F, these curves can be defined to be the locus {P : P F = e P d }. Obtaining as a Conic Section 4 1
5 Exercises 1. Find an equation of the envelope of the family of lines whose x- and y-intercepts have a constant product k. 2. Find an equation of the envelope of the family of circles whose centers lie on the parabola {y = x 2 } and who are tangent to the x-axis. 3. The midpoint of a chord of the circle x 2 + y 2 = r 2 is on a fixed straight line x = a, 0 < a < r. Find an equation of the envelope of the family of such chords. (Hint: let the midpoint be (b, a) & consider the chord s slope). 4. Let P be any fixed point inside a circle C with centre O and radius r. Let A be a variable point on the circle. Let l be the perpendicular to AP through A. In this problem we find the envelope of l as A varies. (note: you can use a set square or any rectangular object to sketch the envelope.) A P O l (a) Let P be the image of P after reflecting P through O, and let P be the image of P after reflecting P through l. Show P P = 2r (Hint: inscribe a rectangle in C such that one side is l). (b) Show that exactly one point X on l satisfies P X + P X = 2r, and that the rest satisfy P X + P X > 2r. (c) Determine, with proof, the envelope of l as A varies. 5
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