Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 10.6 CONIC SECTIONS IN POLAR COORDINATES ¤ 443
13. r =
9
6+2cosθ · 1/6
1/6 = 3/2
1+ 1 cos θ ,wheree = 1 and ed = 3 ⇒ d = 9 .
3 2 2
3
(a) Eccentricity = e = 1 3
(b) Since e = 1 3
< 1, the conic is an ellipse.
(c) Since “+e cos θ ” appears in the denominator, the directrix is to the right of
the focus at the origin. d = |Fl| = 9 2 ,soanequationofthedirectrixis
x = 9 2 .
(d) The vertices are 9
8 , 0 and 9
4 ,π , so the center is midway between them,
15. r =
that is, 9
16 ,π .
3
4 − 8cosθ · 1/4
1/4 = 3/4
1 − 2cosθ ,wheree =2and ed = 3 4
⇒ d = 3 . 8
(a) Eccentricity = e =2
(b) Since e =2> 1, the conic is a hyperbola.
(c) Since “−e cos θ ” appears in the denominator, the directrix is to the left of
the focus at the origin. d = |Fl| = 3 8 ,soanequationofthedirectrixis
x = − 3 8 .
(d) The vertices are − 3 4 , 0 and 1
4 ,π , so the center is midway between them,
17. (a) r =
that is, 1
2 ,π .
1
1 − 2sinθ ,wheree =2and ed =1 ⇒ d = 1 . The eccentricity
2
e =2> 1, so the conic is a hyperbola. Since “−e sin θ ” appears in the
denominator, the directrix is below the focus at the origin. d = |Fl| = 1 2 ,
so an equation of the directrix is y = − 1 . The vertices are
−1, π 2 2 and
1 , 3π
3 2 , so the center is midway between them, that is, 2 , 3π
3 2 .
(b) By the discussion that precedes Example 4, the equation
1
is r =
1 − 2sin .
θ − 3π 4