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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.
 Spring '14
 The Land

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