Conic Sections Session 3: Hyperbola Toh Pee Choon NIE Oct 2017 Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 1 / 16
Problem 3.1 1 Recall that an ellipse is defined as the locus of points P such that PF 1 + PF 2 = k. What happens if k F 1 F 2? 2 What about the locus of P where PF 1 PF 2 = k? (Compare the value of k with F 1 F 2.) Q1) The locus is the segment F 1 F 2 when k = F 1 F 2. This can be considered a degenerate ellipse. There are no solutions for smaller values of k. (Triangle inequality.) Q2) If k = F 1 F 2, the locus is a ray from F 2 in the direction of F 1 F 2. If k = 0, it is the line equidistant from F 1 and F 2. If k > F 1 F 2, no solutions since PF 1 PF 2 F 1 F 2. In GeoGebra, create F 1, F 2 such that F 1 F 2 = 2 and slider 5 k 5. The equations PF 2 = r and PF 1 = r + k independently define a pair of circles whose intersections give feasible points for P. To visualize the locus, create slider 0 r 20 and use the locus tool. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 2 / 16
Problem 3.2 (Geometric = Cartesian) Derive the Cartesian form of PF 1 PF 2 = 2a where 2a < F 1 F 2. Let F 1 = (c, 0) and F 2 = ( c, 0) with c > 0, c < a < c. (x c) 2 + y 2 (x + c) 2 + y 2 = 2a ( ) 2 = (x c) 2 + y 2 = 2a + (x + c) 2 + y 2 = 4a 2 + 4a (x + c) 2 + y 2 + (x + c) 2 + y 2 = 4xc 4a 2 = 4a (x + c) 2 + y 2 = ( cx + a 2) 2 = a 2 (x + c) 2 + a 2 y 2 = (c 2 a 2 )x 2 a 2 y 2 = a 2 (c 2 a 2 ) = x 2 a 2 y 2 c 2 a 2 = 1. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 3 / 16
Definition 3.3 A hyperbola in standard position with transverse axis on x-axis is defined by x 2 a 2 y 2 b 2 = 1. Vertices: A = (a, 0), B = ( a, 0). F2 B A F1 AB is the transverse or major axis. Foci are at (±c, 0) with c = a 2 + b 2. Asymptotes: y = ± b a x. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 4 / 16
Definition 3.4 A hyperbola in standard position with transverse axis on y-axis is defined by y 2 b 2 x 2 a 2 = 1. F1 C D F2 Vertices: C = (0, b), D = (0, b). CD is the transverse or major axis. Foci are at (0, ±c) with c = a 2 + b 2. Asymptotes: y = ± b a x. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 5 / 16
Definition 3.5 (Parametric Form) The parametric form of the hyperbola x2 y 2 a 2 b 2 = 1 is given by x = ±a cosh t and y = b sinh t, t R. cosh t = et + e t 1 and so a cosh t a. 2 sinh t = et e t and takes on all real values. 2 cosh 2 t sinh 2 t = 1. The gradient of the tangent line to the hyperbola at (x, y) where x > 0 is given by dy dx = b cosh t a sinh t = b2 x a 2 y. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 6 / 16
Definition 3.6 (Alternative Parametric Form) The parametric form of the hyperbola x2 y 2 a 2 b 2 = 1 is given by x = a sec θ and y = b tan θ, 0 θ < 2π, θ π 2, 3π 2. Use GeoGebra to plot the locus of intersections of the lines given by the above parametrization and note the discontinuities. Do the same for the parametrization by hyperbolic functions. The gradient of the tangent line to the hyperbola at (x, y) is given by dy dx = b sec t a tan t = b2 x a 2 y. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 7 / 16
Problem 3.7 (Cartesian = Geometric ) Let x2 = 1 be a hyperbola and let F b 2 1 be the right focus. Using parametrization or otherwise, show that for an arbitrary point P = (x 0, y 0 ), with x 0 > 0, PF 1 = cx 0 a 2. a PF 1 = (a cosh t a 2 + b 2 ) 2 + b 2 sinh 2 t = a 2 cosh 2 t 2a(cosh t) a 2 + b 2 + a 2 + b 2 + b 2 (cosh 2 t 1) a 2 y 2 = ( a 2 + b 2 cosh t a = c cosh t a = cx 0 a 2 a. ) 2 Similarly PF 2 = c cosh t + a = cx 0+a 2 a. Consequently, PF 2 PF 1 = 2a, if P has positive x-coordinate. For general P, PF1 PF 2 = 2a, a > 0. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 8 / 16
Exploring eccentricity Problem 3.8 Let F = (0, 0) and directrix d be the line x = k, with k > 0. If P satisfies PF = e Pd for e 0, find the possible solutions for P that lie on the x-axis. If P = (x, 0), the equation is simply x = e x k. Case: x > k > 0, i.e. x = e(x k) = x = ek e 1. In this case, we see that e > 1. Case: 0 < x < k, i.e. x = e(k x) = x = ek e+1. In this case, there are no additional constraints on e. Case: x = 0, e = 0 (degenerate case). Case: x < 0, i.e. x = e(k x) = x = ek e 1. In this case, we require e < 1. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 9 / 16
Definition of Hyperbola Definition 3.9 (Directrix-Eccentricity-Focus) A hyperbola is the locus of points whose distance from the focus F (focus) and { the directrix d are } in a constant ratio greater than 1, i.e. P : PF Pd = e, e > 1. d D P A F Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 10 / 16
Varying eccentricity and directrix: Repeat Exercise Exercise 3.10 In GeoGebra, create sliders with variables 0 e 5, 5 k 20 and 0 m 100. Create the directrix d : x = k and the focus F at the origin. Since PF = e Pd, we shall let m be the value of Pd, so P lies on one of the lines x = k ± m. On the other hand, P lies on a circle of radius em about F. (Use circle tool.) Plot the (four possible) intersections to find positions of P. Use the locus tool to draw the conic. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 11 / 16
Linking the definitions Definition 3.11 (D-E-F = Cartesian) Let F = (c, 0), c > 0 and directrix d be the line x = c e 2. The equation PF Pd = e for e > 1 is equivalent to e 2 x 2 e 2 y 2 c 2 c 2 (e 2 1) = 1 Thus a = c e and b = (e 2 1)a. Directrix becomes x = a e. e = 1 + b2 a 2 c = ae = a 2 + b 2. (So the focus defined in the G and D-E-F definitions coincide.) By symmetry F 2 = ( c, 0) and x = a e also qualify as focus and directrix. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 12 / 16
Linking the definitions Problem 3.12 (D-E-F = Cartesian) Let F = (0, c), c > 0 and directrix d be the line y = k. Show that equation PF Pd = e for e > 1 defines a hyperbola and determine the value of e, c and k for a hyperbola in standard position with transverse axis on y-axis. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 13 / 16
Polar form with respect to the focus Problem 3.13 Let F = (0, 0) be a focus and directrix d given by x = k, k > 0. Show that the curve with locus defined by {P : PF = e Pd } has the polar form. r = ek 1 + e cos θ. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 14 / 16
Reflection Property Problem 3.14 Show that light coming from F 1 is reflected in a hyperbolic mirror in such a way that it appears to have come from F 2. P F 2 F 1 Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 15 / 16
Problem 3.15 (Apollonius) Let T be a point on a hyperbola in standard position with transverse axis on x-axis. The tangent line to the hyperbola at T cuts the asymptotes at P and Q. Show that T is the midpoint of PQ. If O is the origin, show that OQ OP is a constant and hence the triangle OPQ has constant area. Toh Pee Choon (NIE) Session 3: Hyperbola Oct 2017 16 / 16