Solução_Calculo_Stewart_6e

F.
442 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10
10.6 Conic Sections in Polar Coordinates
1. The directrix y =6is above the focus at the origin, so we use the form with “ + e sin θ” in the denominator. [See Theorem 6
7
ed
and Figure 2(c).] r =
1+e sin θ = · 6 4
1+ 7 sin θ = 42
4+7sinθ
4
3. The directrix x = −5 is to the left of the focus at the origin, so we use the form with “ − e cos θ” in the denominator.
3
ed
r =
1 − e cos θ = · 5 4
1 − 3 cos θ = 15
4 − 3cosθ
4
5. The vertex (4, 3π/2) is 4 units below the focus at the origin, so the directrix is 8 units below the focus (d =8), and we use the
form with “−e sin θ ” in the denominator. e =1for a parabola, so an equation is r =
ed
1 − e sin θ = 1(8)
1 − 1sinθ = 8
1 − sin θ .
7. The directrix r =4secθ (equivalent to r cos θ =4or x =4) is to the right of the focus at the origin, so we will use the form
with “+e cos θ” in the denominator. The distance from the focus to the directrix is d =4,soanequationis
1
ed
r =
1+e cos θ = (4) 2
1+ 1 cos θ · 2
2 = 4
2+cosθ .
2
9. r =
1
1+sinθ = ed
,whered = e =1.
1+e sin θ
(a) Eccentricity = e =1
(b) Since e =1, the conic is a parabola.
(c) Since “+e sin θ ” appears in the denominator, the directrix is above the
focus at the origin. d = |Fl| =1, so an equation of the directrix is y =1.
(d) The vertex is at 1
2 , π 2
, midway between the focus and the directrix.
11. r =
12
4 − sin θ · 1/4
1/4 = 3
1 − 1 sin θ ,wheree = 1 4
and ed =3 ⇒ d =12.
4
(a) Eccentricity = e = 1 4
(b) Since e = 1 < 1, the conic is an ellipse.
4
(c) Since “−e sin θ ” appears in the denominator, the directrix is below the focus
at the origin. d = |Fl| =12,soanequationofthedirectrixisy = −12.
(d) The vertices are
4, π 2 and 12 , 3π
5 2 , so the center is midway between them,
that is, 4
, π
5 2 .