Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 10.5 CONIC SECTIONS 0 29
35. A parabola with vertical axis and vertex (2, 3) has equation y- 3 = a(x- 2?. Since it passes through (I; 5), we have
5-3=a(I-2) 2
'* a =2,soanequationisy -3 =2(x-2?.
37. The ellipse with foci (±2, 0) and vertices (± 5, 0) has center (0, 0) and a horizontal major axis, with a = 5 and c = 2,
x2 1 2
so b 2 = a 2 - c 2 = 25- 4 = 21. An equation is 25
+ ; 1
= 1.
39. Since the vertices are (0, 0) and (0, 8), the ellipse has center (0, 4) with a vertical axis and a = 4. The foci at (0, 2) and (0, 6)
are 2 units from the center, soc= 2 and b = .,Ja 2 - c 2 = .,j4 2 - 2 2 = v'f2. An equation is (x - 2
0) 2 + (y- 4 ) 2 = 1 '*
b a 2
_41. An equation of an ellipse with center (-1,4) and vertex (- 1,0) is (x ! 2
1 )
2
+ (y ~2 4 ) 2 = 1. The focus (-1,6) is 2 units
from the center, soc= 2. Thus, b 2 + 2 2 = 4 2 '* b 2 = 12 and the equation is (x + 1 ? + (y - 4 ) 2 = 1
. , 12 16 .
2 2
43. An equation of a hyperbola with vertices (±3, 0) is ....:. ; ~2 2
= 1. Foci (±5, 0) '* c = 5 and 3 2 + b 2 =5 '*
2
. 2 2
b 2 = 25 - 9 = 16, so the equation is ~ - r6 = 1.
45. The center of a hyperbola with vertices ( -3, - 4) and ( -3, 6) is ( -3, 1), so a = 5 and a.n equation is
(x + 3) 2 . · ·
b 2
= l.Focl(- 3, - 7)and(- 3,9)-* c =8,so5 2 +b 2 =8 2 '* b 2 =64-25=39andthe
( - 1} 2 {x + 3) 2
equation is y - = I
25 39 .
. . 2 • 2
47. The center of a hyperbola with vertices (±3, 0) is (0, 0), so a = 3 and an equation is ~ 2
- t 2
= 1.
b . 2 2
Asymptotes y = ±2x '* - = 2 '* b = 2(3) = 6 and the equation is ~ - JL = 1.
a 9 36
49. In Figure 8, we see that the point on the ellipse closest to a focus is the closer vertex (which is a distance
a - c from it) while the farthest point is the other vertex (at a distance of a+ c). So for this lunar orbit,
(a- c)+ (a+ c) = 2a = (1728 + 110) + (1728 + 314), or a = 1940; and (a+ c)- .(a - c) = 2c = 314 - 110,
or c = 102. Thus, b 2 = a 2 -
x2 y2
c 2 = 3,753,196, and the equation is , , 3 763 600
+ , , =I.
3 753 196
51. (a) Set up the coordinate system so that A is ( -200, 0) and B is (200, 0) .
IPAI-\PBI = (1200)(980) = 1,176,000 ft = 2 1~ mi = 2a -* a= 1 ~ 5 , and c ~ 200 so
b 2 - 2 - 2 - 3,339,375
- c a - 121
121x 2
1,500,625
121y 2 - 1
3,339,375 - .
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