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

SECTION 10.5 CONIC SECTIONS 0 27
5. (x + 2) 2 = 8 (y- 3). 4p = 8, sop= 2. The vertex is
( - 2, 3), the focus is ( -2, 5), and the directrix is y = 1.
y
7. y 2 + 2y + 12x + 25 = 0 =l>
y2 + 2y + 1 = - 12x- 24 =;..
(y + 1) 2 = - 12(x + 2). 4p = - 12, sop= -3.
The vertex is (- 2, -1), the focus is (-5, - 1), and the
directrix is x = 1.
-------------------------- __ ,y_~_!_
X
0 :
~ X
I
jx=l
9. The equation has the form y 2 = 4px, where p < 0. Since the parabola passes through (- 1, 1), we have 1 2 = 4p( - 1), so
4p = - 1 and an equation is y 2 = - x or x = - y 2 • 4p = - 1, sop = -i and the focus is ( -;i, 0) while the directrix
. 1
ISX = 4·
ellipse is centered at {0, 0), with vertices at (0, ±2). The foci are (0, ±v'2).
y
2
-..fi
x2
Y2
13. x 2 + 9y 2 = 9 9 + l = 1 =l> a = v'9 = 3, 15. 9x 2 - 18x + 4y 2 = 27
b = Vi.= 1, c = ~ = v'9=I = J8 = 2v'2.
The ellipse is centered at {0, 0), with vertices (± 3, 0).
The foci are (±2v'2, 0).
y
I
-2
9(x 2 - 2x + 1) + 4y 2 = 27 + 9
9(x- 1) 2 + 4y 2 = 36
a = 3, b = 2, c = J5 =:- center (1, 0),
vertices (1 , ± 3}, foci (1, ±J5)
(x- 1)2 y2
4 + g- = 1 =}
- 3
0
3 X
y (1, 3)
- I
3 X
2 2
17. The center is (0, 0), a= 3, and b = 2, so an equation is ~ + Y 9
= 1. c = ~ = J5, so the foci are (0, =;I=J5).
(1,- 3)
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