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

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