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

CHAPTER 10 REVIEW 0 41
2 2 .
45. ~ + 'lL = 1 is an ellipse with center (0, 0).
9 8
a = 3, b = 2 J2, c = 1 =?
foci (±1, 0), vertices (±3, 0).
y J-
2 2
47. 6y 2 + X - 36y + 55 = 0 ~
6(y 2 - 6y + 9) = - (x + 1) ~
(y - 3) 2 = --[;(x + 1), aparabola with vertex (-1, 3),
opening to the left, p = - f.t =? focus (- ;~, 3) and
directrix x = -~.
y
- 3 3 X
(- 1,3)
0 X
49. The ellipse with foci (±4, 0) and vertices (±5, 0) has center (0, 0) and a horizontal major axis, with a = 5 and c = 4,
2 2
so b 2 = a 2 - c 2 ·= 5 2 - 4 2 = 9. An equation is ~ 5
+ Y 9
= 1.
2 ?
51 . The center of a hyperbola with foci {0, ±4) is (0, 0), so c= 4 and an equation is y 2
- x: = 1.
. a b
a 3 ·
The asymptote y = 3x has slope 3, sob = l - ::::} a= 3b and a 2 + b 2 = c 2 =? {3b? + b 2 = 4 2 =?
10b 2 = 16 ::::}
b 2 = * and so a 2 = 16 - ~ = L 5
2 • Thus, an equation is -
" " 72 5 8 5
y2 x2 5y2 5x2
1 - - - 1 = 1, or - 72 - - 8
= 1.
53. x 2 = -(y - 100) has its vertex at (0,100), so one of the vertices ofthe 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 = -~.The refo re the shared focus is
found at .(0, 3 ~ 9 ) so 2c = 3 ~ 9 -0 ::::} c = 3 ~ 9 and the center of the ellipse is (0, 3 ~ 9 ). So a= 100- 3 ~ 9 = 4 ~ 1 and
2 2 . 2 ( 39!))2
b 2 __ a 2 _ c 2 __ 401 - 399 x y - 8
82
----,=--- = 25. So the equation of th e ell ipse is /1 + a 2
= 1 =?
x 2 (8y - 399? = 1
or 25 + 160,801 ·
55. Directrix x = 4 =? d = 4, so e = ~
ed 4
=? r - - ..,..--- -.,
- 1 +ecosfJ- 3+cosfJ'
57. (a) Jf (a, b) lies on the curve, then there is some parameter value t1 such that~ = a and
1 + tl 1 + t l
3 tr ~ =b. lft 1 = 0,
the point is (0, 0), which lies on the line y = x. lft1 =I 0, then the point corresponding tot= ..!. is given by
tl
3(1/tl ) 3ti 3(1/td 3tl
x = 1
+ ( 1
/t
1
) 3 = t~ + 1
= b, Y = 1
+ ( 1
/h) 3 = tt + 1
= a. So (b, a) also lies on the curve. [Another way to see
2
3
this is to do part (e) first; the result is immediate.] The curve intersects the line y = x when ~3
= t
1 + t 1+t
t = t 2 ::::} t = 0 or. 1, so the points are (0, 0) and ( ~, ~).
3
::::}