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

SECTION 10.6 CONIC SECTIONS IN POLAR COORDINATES 0 33
13. r = : 6 9 · 1 6 3 2
1 = (
+ cos 1
6 1+ 3 cos0
(a) Eccentricity= e = t
(b) Since e = t < 1, the conic is an ellipse.
, where e = ~ and ed = ~ :::} d = ~ -
(c) Since "+ e cos e" appears in the denominator, the directrix is to the right of
the focus at the origin. d = !Fll = ~.so an equation of the directrix is
x = ~-
(d) The vertices are ( ~, 0) and ( t, 1r), so the center is midway between them~
that is, ( fn, 1r) .
(f,1T)
I
9 :
x=21
:
!
3 1/ 4 3/ 4 3
15. r = 4 - 8 cos 9 . 1 I 4 = 1 - 2 cos 9' where e = 2 and ed = 4
d - 1 - s ·
(a) Eccentricity = e = 2
(b) Since e = 2 > 1, the conic is a hyperbola.
(c) Since "- e cos 0" appears in the denominator, the directrix is to the left of
the focus at the origin. d'= IFll = ~.so an equation of the directrix is
. X - - s·
3
(d) The vertices are ( - ~. 0) and (i, 1r), so the center is midway between them,
that is, (~ , 1r).
17. (a) r = ; .
1 - sm 9 , where e = 2 and ed = 1 :::} d = 1 . The eccentricity
2
e = 2 > 1, so the conic is a hyperbola. Since" -e sin B" appears in the
denominator, the directrix is below the focus at the origin. d = IFll = ~.
so an equation of the directrix is y = - ~ . The vertices are (- 1, ~) and
( 1 3 '~~" ) h . "d b th h . ( 2 3 " )
3 , 2 , sot e center IS m1 way etween l'!m, t at 1s, 3 , 2 .
(b) By the discussion that precedes Example 4, the equation
. 1
IS T = ( 3 ) •
1 - 2sin e- ;
-3
-2
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