Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 10.5 CONIC SECTIONS ¤ 439
25. x 2 = y +1 ⇔ x 2 =1(y +1). This is an equation of a parabola with 4p =1,sop = 1 4
.Thevertexis(0, −1) and the
focus is 0, − 3 4
.
27. x 2 =4y − 2y 2 ⇔ x 2 +2y 2 − 4y =0 ⇔ x 2 +2(y 2 − 2y +1)=2 ⇔ x 2 +2(y − 1) 2 =2 ⇔
x 2
2
+
(y − 1)2
1
=1.Thisisanequationofanellipse with vertices at ± √ 2, 1 . The foci are at ± √ 2 − 1, 1 =(±1, 1).
29. y 2 +2y =4x 2 +3 ⇔ y 2 +2y +1=4x 2 +4 ⇔ (y +1) 2 − 4x 2 =4 ⇔
(y +1)2
4
− x 2 =1.Thisisanequation
of a hyperbola with vertices (0, −1 ± 2) = (0, 1) and (0, −3). Thefociareat 0, −1 ± √ 4+1 = 0, −1 ± √ 5 .
31. The parabola with vertex (0, 0) and focus (0, −2) opens downward and has p = −2, so its equation is x 2 =4py = −8y.
33. Thedistancefromthefocus(−4, 0) to the directrix x =2is 2 − (−4) = 6, so the distance from the focus to the vertex is
1
(6) = 3 and the vertex is (−1, 0). Since the focus is to the left of the vertex, p = −3. An equation is 2 y2 =4p(x +1) ⇒
y 2 = −12(x +1).
35. A parabola with vertical axis and vertex (2, 3) has equation y − 3=a(x − 2) 2 . Since it passes through (1, 5),wehave
5 − 3=a(1 − 2) 2 ⇒ a =2,soanequationisy − 3=2(x − 2) 2 .
37. The ellipse with foci (±2, 0) and vertices (±5, 0) has center (0, 0) and a horizontal major axis, with a =5and c =2,
so b 2 = a 2 − c 2 =25− 4=21. An equation is x2
25 + y2
21 =1.
39. Since the vertices are (0, 0) and (0, 8), the ellipse has center (0, 4) with a vertical axis and a =4. The foci at (0, 2) and (0, 6)
are 2 units from the center, so c =2and b = √ a 2 − c 2 = √ 4 2 − 2 2 = √ 12.Anequationis
x 2 (y − 4)2
+ =1.
12 16
41. An equation of an ellipse with center (−1, 4) and vertex (−1, 0) is
(x − 0)2 (y − 4)2
+ =1 ⇒
b 2 a 2
(x +1)2 (y − 4)2
+ =1.Thefocus(−1, 6) is 2 units
b 2 4 2
from the center, so c =2.Thus,b 2 +2 2 =4 2 ⇒ b 2 =12, and the equation is
(x +1)2
12
+
(y − 4)2
16
=1.
43. An equation of a hyperbola with vertices (±3, 0) is x2
3 2 − y2
b 2 =1.Foci(±5, 0) ⇒ c =5and 32 + b 2 =5 2 ⇒
b 2 =25− 9=16, so the equation is x2
9 − y2
16 =1.
45. The center of a hyperbola with vertices (−3, −4) and (−3, 6) is (−3, 1), soa =5and an equation is
(y − 1) 2 (x +3)2
− =1.Foci(−3, −7) and (−3, 9) ⇒ c =8,so5 2 + b 2 =8 2 ⇒ b 2 =64− 25 = 39 and the
5 2 b 2
equation is
(y − 1)2
25
−
(x +3)2
39
=1.