Solução_Calculo_Stewart_6e

F.
438 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10
13. 4x 2 + y 2 =16 ⇒ x2
4 + y2
16 =1 ⇒
a = √ 16 = 4, b = √ 4=2,
c = √ a 2 − b 2 = √ 16 − 4=2 √ 3. The ellipse is
centered at (0, 0), with vertices at (0, ±4). Thefoci
are 0, ±2 √ 3 .
15. 9x 2 − 18x +4y 2 =27 ⇔
9(x 2 − 2x +1)+4y 2 =27+9 ⇔
9(x − 1) 2 +4y 2 =36 ⇔
(x − 1)2
4
a =3, b =2, c = √ 5 ⇒ center (1, 0),
vertices (1, ±3),foci 1, ± √ 5
+ y2
9 =1 ⇒
17. The center is (0, 0), a =3,andb =2, so an equation is x2
4 + y2
9 =1. c = √ a 2 − b 2 = √ 5, so the foci are 0, ± √ 5 .
19.
x 2
144 − y2
25 =1 ⇒ a =12, b =5, c = √ 144 + 25 = 13 ⇒
center (0, 0),vertices(±12, 0),foci(±13, 0),asymptotesy = ± 5 12 x.
Note: It is helpful to draw a 2a-by-2b rectangle whose center is the center
of the hyperbola. The asymptotes are the extended diagonals of the
rectangle.
21. y 2 − x 2 =4 ⇔ y2
4 − x2
4 =1 ⇒ a = √ 4=2=b,
c = √ 4+4=2 √ 2 ⇒ center (0, 0),vertices(0, ±2),
foci 0, ±2 √ 2 ,asymptotesy = ±x
23. 4x 2 − y 2 − 24x − 4y +28=0 ⇔
4(x 2 − 6x +9)− (y 2 +4y +4)=−28 + 36 − 4 ⇔
4(x − 3) 2 − (y +2) 2 =4 ⇔
(x − 3)2
1
a = √ 1=1, b = √ 4=2, c = √ 1+4= √ 5
−
(y +2)2
4
⇒
=1 ⇒
center (3, −2), vertices (4, −2) and (2, −2),foci 3 ± √ 5, −2 ,
asymptotes y +2=±2(x − 3).