Solutions

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: d the tangent lines are x = ±a which also follows from x0 x/a2 + y0 y/b2 = 1. 64. By implicit differentiation, dy dx = (x0 ,y0 ) b2 x0 b2 x0 if y0 = 0, the tangent line is y − y0 = 2 (x − x0 ), a2 y0 a y0 = a2 b2 , x0 x/a2 − y0 y/b2 = 1. If y0 = 0 then x0 = ±a and the tangent b x0 x − a y0 y = lines are x = ±a which also follow from x0 x/a2 − y0 y/b2 = 1. 2 2 2 b2 x2 − a2 y0 0 y2 x2 y2 x2 + 2 = 1 and 2 − 2 = 1 as the equations of the ellipse and hyperbola. If (x0 , y0 ) is a2 b A B 2 x2 y0 x2 y2 1 1 1 1 0 2 a point of intersection then 2 + 2 = 1 = 0 − 0 , so x2 − 2 = y0 + 2 and 0 a b A2 B2 A2 a B2 b 65. Use 2 a2 A2 y0 (b2 + B 2 ) = b2 B 2 x2 (a2 − A) . Since the conics have the same foci, a2 − b2 = c2 = A2 + B 2 , 0 2 2 so a − A2 = b2 + B 2 . Hence a2 A2 y0 = b2 B 2 x2 . From Exercises 63 and 64, the slopes of the 0 tangent lines are − perpendicular. B 2 x0 b2 B 2 x2 b2 x0 0 and 2 , whose product is − 2 2 2 = −1. Hence the tangent lines are a2 y0 A y0 a A y0 dy x0 =− where (x0 , y0 ) is the point dx (x0 ,y0 ) 4y0 of tangency, but −x0 /(4y0 ) = −1/2 because the slope of the line is −1/2 so x0 = 2y0 . (x0 , y0 ) is 2 on the ellipse so x2 + 4y0 = 8 which when solved with x0 = 2y0 yields the points of tangency (2, 1) 0 and (−2, −1). Substitute these into the equation of the line to get k = ±4. 66. Use implicit differentiation on x2 + 4y 2 = 8 to get √ √ 67. Let (x0 , y0 ) be such a point. The foci are at (− 5, 0) and ( 5, 0), the lines are perpendicular if y0 y0 2 2 √· √ = −1, y0 = 5 − x2 and 4x2 − y0 = 4. Solve the product of their slopes is −1 so 0 0 x0 + 5 x0 − 5 √ √ √ √ √ √ to get x0 = ±3/ 5, y0 = ±4/ 5. The coordinates are (±3/ 5, 4/ 5), (±3/ 5, −4/ 5). 68. Let (x0 , y0 ) be one of the points; then dy/dx (x0 ,y0 ) = 4x0 /y0 , the tangent line is y = (4x0 /y0 )x +4, 2 2 but (x0 , y0 ) is on both the line and the curve which leads to 4x2 − y0 + 4y0 = 0 and 4x2 − y0 = 36, 0 0 √ solve to get x0 = ±3 13/2, y0 = −9. 69. Let d1 and d2 be the distances of the first and second observers, respectively, from the point where the gun was fired. Then t = (time for sound to reach the second observer) − (time for sound to reach the first observer) = d2 /v − d1 /v so d2 − d1 = vt. For constant v and t the difference of distances, d2 and d1 is constant so the gun was fired somewhere on a branch of a vt v 2 t2 , and hyperbola whose foci are where the observers are. Since d2 − d1 = 2a, a = , b2 = c2 − 2 4 2 2 y x −2 = 1. 2 t2 /4 v c − (v 2 t2 /4) 70. As in Exercise 69, d2 − d1 = 2a = vt = 299,792,458 × 10−7 , a2 = (vt/2)2 ≈ 224.6888; c2 = (50)2 y2 x2 − = 1. But y = 200 km, so x ≈ 64.6 km. = 2500, b2 = c2 − a2 ≈ 2275.3112, 224.6888 2275.3112 433 Chapter 12 71. (a) Use y2 3 x2 + = 1, x = 9 4 2 −2+h V= −2 = 54 4 − y2 , (2)(3/2) 4 − y 2 (18)dy = 54 y 2 4 − y 2 + 2 sin−1 y 2 −2+h −2+h 4 − y 2 dy −2 = 27 4 sin−1 −2 h−2 + (h − 2) 2 4h − h2 + 2π ft3 (b) When h = 4 ft, Vfull = 108 sin−1 1 + 54π = 108π ft3 , so solve for h when V = (k/4)Vfull , k = 1, 2, 3, to get h = 1.19205, 2, 2.80795 ft or 14.30465, 24, 33.69535 in. 72. We may assume A > 0, since if A < 0 then one can multiply the equation by −1, and if A = 0 then one can exchange A with C (C cannot be zero simultaneously with A). Then 2 2 E2 D E D2 − = 0. +C y+ +F − Ax2 + Cy 2 + Dx + Ey + F = A x + 2A 2C 4A 4C D2 E2 + the equation represents an ellipse (a circle if A = C ); 4A 4C D2 E 2 D2 E 2 if F = + , the point x = −D/(2A), y = −E/(2C ); and if F > + then there is 4A 4C 4A 4C no graph. (a) Let AC > 0. If F < D2 E2 + , then 4A 4C √ √ D E + −C y + A x+ 2A 2C otherwise a hyperbola (b) If AC < 0 and F = √ A x+ D 2A − √ −C y + E 2C = 0, a pair of lines; (c) Assume C = 0, so Ax2 +Dx+Ey +F = 0. If E = 0, parabola; if E = 0 then Ax2 +Dx+F = 0. If this polynomial has roots x = x1 , x2 with x1 = x2 then a pair of parallel lines; if x1 = x2 then one line; if no roots, then no graph. If A = 0, C = 0 then a similar argument applies. 73. (a) (x − 1)2 − 5(y + 1)2 = 5, hyperbola √ (b) x2 − 3(y + 1)2 = 0, x = ± 3(y + 1), two lines (c) 4(x + 2)2 + 8(y + 1)2 = 4, ellipse (d) 3(x + 2)2 + (y + 1)2 = 0, the point (−2, −1) (degenerate case) (e) (x + 4)2 + 2y = 2, parabola (f ) 5(x + 4)2 + 2y = −14, parabola 74. distance from the point (x, y ) to the focus (0, p) = distance to the directrix y = −p, so x2 + (y − p)2 = (y + p)2 , x2 = 4py 75. distance from the point (x, y ) to the focus (0, −c) plus distance to the focus (0, c) = const = 2a, x2 + (y − c)2 = 2a, x2 + (y + c)2 = 4a2 + x2 + (y − c)2 − 4a x2 + (y − c)2 , c x2 y2 x2 + (y − c)2 = a − y , and since a2 − c2 = b2 , 2 + 2 = 1 a b a x2 + (y + c)2 + 76. distance from the point (x, y ) to the focus (−c, 0) less distance to the focus (c, 0) is equal to 2a, (x + c)2 + y 2 − (x − c)2 + y 2 = ± (x − c)2 + y 2...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online