Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 10 REVIEW ¤ 451
45.
x 2
9 + y2
=1is an ellipse with center (0, 0).
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a =3, b =2 √ 2, c =1
⇒
foci (±1, 0), vertices (±3, 0).
47. 6y 2 + x − 36y +55=0 ⇔
6(y 2 − 6y +9)=−(x +1) ⇔
(y − 3) 2 = − 1 (x +1), a parabola with vertex (−1, 3),
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opening to the left, p = − 1 24
⇒ focus − 25
24 , 3 and
directrix x = − 23
24 .
49. The ellipse with foci (±4, 0) and vertices (±5, 0) has center (0, 0) and a horizontal major axis, with a =5and c =4,
so b 2 = a 2 − c 2 =5 2 − 4 2 =9. An equation is x2
25 + y2
9 =1.
51. The center of a hyperbola with foci (0, ±4) is (0, 0),soc =4andanequationis y2
a 2 − x2
b 2 =1.
The asymptote y =3x has slope 3,so a b = 3 1
⇒ a =3b and a 2 + b 2 = c 2 ⇒ (3b) 2 + b 2 =4 2 ⇒
10b 2 =16 ⇒ b 2 = 8 and so 5 a2 =16− 8 = 72 . Thus, an equation is y 2
5 5
72/5 − x2 5y2
=1,or
8/5 72 − 5x2
8 =1.
53. x 2 = −(y − 100) has its vertex at (0, 100), so one of the vertices of the ellipse is (0, 100). Another form of the equation of a
parabola is x 2 =4p(y − 100) so 4p(y − 100) = −(y − 100) ⇒ 4p = −1 ⇒ p = − 1 . Therefore the shared focus is
4
found at
0, 399
4 so 2c =
399
− 0 ⇒ c = 399 and the center of the ellipse is
0, 399
4 8 8 .Soa = 100 −
399
= 401 and
8 8
b 2 = a 2 − c 2 = 4012 − 399 2
=25. So the equation of the ellipse is x2 y −
399 2
8 2 b + 8
=1 ⇒ x2 y −
399 2
2 a 2 25 + 8
401
2
=1,
or x2 (8y − 399)2
+ =1.
25 160,801
55. Directrix x =4 ⇒ d =4,soe = 1 3
⇒ r =
ed
1+e cos θ = 4
3+cosθ .
57. In polar coordinates, an equation for the circle is r =2a sin θ. Thus, the coordinates of Q are x = r cos θ =2a sin θ cos θ
and y = r sin θ =2a sin 2 θ. The coordinates of R are x =2a cot θ and y =2a. SinceP is the midpoint of QR,weusethe
midpoint formula to get x = a(sin θ cos θ +cotθ) and y = a(1 + sin 2 θ).
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