Conic Sections - 10.5 - Solutions
Conic Sections - 10.5 - Solutions
Conic Sections - 10.5 - Solutions
56. (a) Rotation around =
2 is the same as rotation around the -axis, that is, =
2 where
= ()2 + ()2 for a parametric equation, and for the special case of a polar equation, = cos and
= ()2 + ()2 = 2 + ()2 [see the derivation of Equation 10.4.5]. Therefore, for a polar
equation rotated around = 2,
=
2 cos 2 + ()2 .
(b) As in the solution for Exercise 55(b), we can double the surface area generated by rotating the curve from = 0 to =
4
LE
0 0 cos 2
4 √
4 √ 1 4 2 √
= 4 cos 2 cos √ = 4 cos = 4 sin 0 = 4 −0 = 2 2
0 cos 2 0 2
SA
10.5 Conic Sections
1. 2 = 6 and 2 = 4 ⇒ 4 = 6 ⇒ = 32 . 2. 2 2 = 5 ⇒ 2 = 52 . 4 = 52 ⇒ = 58 .
The vertex is (0 0), the focus is 0 32 , and the directrix The vertex is (0 0), the focus is 58 0 , and the directrix
is = − 32 . is = − 58 .
R
FO
T
directrix is = 12 . is 0 − 23 , and the directrix is = 23 .
N
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°
2
5. ( + 2) = 8 ( − 3). 4 = 8, so = 2. The vertex is 6. ( − 2)2 = 2 + 1 = 2 + 1
2
. 4 = 2, so = 12 . The
(−2 3), the focus is (−2 5), and the directrix is
vertex is − 12 2 , the focus is (0 2), and the directrix is
= 1.
= −1.
LE
7. 2 + 6 + 2 + 1 = 0 ⇔ 2 + 6 = −2 − 1 8. 22 − 16 − 3 + 38 = 0 ⇔ 22 − 16 = 3 − 38
⇔ 2 + 6 + 9 = −2 + 8 ⇔ ⇔ 2(2 − 8 + 16) = 3 − 38 + 32 ⇔
( + 3)2 = −2( − 4). 4 = −2, so = − 12 . 2( − 4)2 = 3 − 6 ⇔ ( − 4)2 = 32 ( − 2).
The vertex is (4 −3), the focus is 72 −3 , and the 4 = 32 , so = 38 . The vertex is (4 2), the focus is 4 19 ,
SA
8
R
FO
9. The equation has the form 2 = 4, where 0. Since the parabola passes through (−1 1), we have 12 = 4(−1), so
4 = −1 and an equation is 2 = − or = − 2 . 4 = −1, so = − 14 and the focus is − 14 0 while the directrix
is = 14 .
10. The vertex is (2 −2), so the equation is of the form ( − 2)2 = 4( + 2), where 0. The point (0 0) is on the parabola,
T
directrix is = − 52 .
O
2 2 √ √ 2 2 √ √
11. + = 1 ⇒ = 4 = 2, = 2, 12. + = 1 ⇒ = 36 = 6, = 8,
2 4 36 8
√ √ √ √ √ √ √
N
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°
2 2 √ 2 2
13. 2 + 9 2 = 9 +
⇔ = 1 ⇒ = 9 = 3, 14. 1002 + 36 2 = 225 ⇔ 225 + 225 =1 ⇔
9 1 100 36
√ √ √ √ √
2 2
= 1 = 1, = − = 9 − 1 = 8 = 2 2.
2 2
9 + 25 =1 ⇒ = 25
4
= 52 , = 9
4
= 32 ,
The ellipse is centered at (0 0), with vertices (±3 0). 4 4
√ √
The foci are (±2 2 0). = 2 − 2 = 25 − 9
= 2. The ellipse is centered
4 4
at (0 0), with vertices 0 ± 52 . The foci are (0 ±2).
LE
15. 92 − 18 + 4 2 = 27
9(2 − 2 + 1) + 4 2 = 27 + 9 ⇔
⇔
SA
16. 2 + 3 2 + 2 − 12 + 10 = 0
2 + 2 + 1 + 3( 2 − 4 + 4) = −10 + 1 + 12 ⇔
⇔
R
( − 1)2 2 ( + 1)2 + 3( − 2)2 = 3 ⇔
9( − 1)2 + 4 2 = 36 ⇔ + =1 ⇒
4 9
( + 1)2 ( − 2)2 √
FO
√ + = 1 ⇒ = 3, = 1,
= 3, = 2, = 5 ⇒ center (1 0), 3 1
√ √ √
vertices (1 ±3), foci 1 ± 5 = 2 ⇒ center (−1 2), vertices −1 ± 3 2 ,
√
foci −1 ± 2 2
T
O
N
2 2 √ √ √
17. The center is (0 0), = 3, and = 2, so an equation is + = 1. = 2 − 2 = 5, so the foci are 0 ± 5 .
4 9
( − 2)2 ( − 1)2 √ √
18. The ellipse is centered at (2 1), with = 3 and = 2. An equation is + = 1. = 2 − 2 = 5, so
9 4
√
the foci are 2 ± 5 1 .
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°
2 2 √ √
19. − = 1 ⇒ = 5, = 3, = 25 + 9 = 34 ⇒
25 9
√
center (0 0), vertices (0 ±5), foci 0 ± 34 , asymptotes = ± 53 .
2 2 √
LE
20. − = 1 ⇒ = 6, = 8, = 36 + 64 = 10 ⇒
36 64
center (0 0), vertices (±6 0), foci (±10 0), asymptotes = ± 86 = ± 43
21. 2 − 2 = 100
2
⇔
100
−
2
100
= 1 ⇒ = = 10,
SA
R
√ √
= 100 + 100 = 10 2 ⇒ center (0 0), vertices (±10 0),
√
foci ±10 2 0 , asymptotes = ± 10
10 = ±
FO
2 2
22. 2 − 162 = 16 ⇔ − = 1 ⇒ = 4, = 1,
16 1
√
T
√
= 16 + 1 = 17 ⇒ center (0 0), vertices (0 ±4),
√
foci 0 ± 17 , asymptotes = ± 41 = ±4
O
N
23. 2 − 2 + 2 = 2 ⇔ 2 − ( 2 − 2 + 1) = 2 − 1 ⇔
2 ( − 1)2 √ √
− = 1 ⇒ = = 1, = 1 + 1 = 2 ⇒
1 1
√
center (0 1), vertices (±1 1), foci ± 2 1 ,
asymptotes − 1 = ± 11 = ±.
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°
2 2
25. 42 = 2 + 4 ⇔ 42 − 2 = 4 ⇔ − = 1. This is an equation of a hyperbola with vertices (±1 0).
1 4
√ √
LE
The foci are at ± 1 + 4 0 = ± 5 0 .
SA
27. 2 = 4 − 2 2 ⇔ 2 + 2 2 − 4 = 0 ⇔ 2 + 2( 2 − 2 + 1) = 2 ⇔ 2 + 2( − 1)2 = 2 ⇔
2 ( − 1)2 √ √
+ = 1. This is an equation of an ellipse with vertices at ± 2 1 . The foci are at ± 2 − 1 1 = (±1 1).
2 1
2 ( − 1)2
28. 2 − 2 = 2 − 2 ⇔ 2 − 2 + 2 = 2 ⇔ 2 − (2 − 2 + 1) = 2 − 1 ⇔ − = 1. This is an
1 1
R
√ √
equation of a hyperbola with vertices (1 ±1). The foci are at 1 ± 1 + 1 = 1 ± 2 .
( − 1)2 = 23 ( + 2). This is an equation of a parabola with 4 = 23 , so = 16 . The vertex is (1 −2) and the focus is
1 −2 + 16 = 1 − 11 6
.
( − 1)2 ( − 2)2 √
= 1. This is an equation of an ellipse with vertices at 1 ± 2 2 . The foci are at
T
+
2 1
√
1 ± 2 − 1 2 = (1 ± 1 2).
O
31. The parabola with vertex (0 0) and focus (1 0) opens to the right and has = 1, so its equation is 2 = 4, or 2 = 4.
32. The parabola with focus (0 0) and directrix = 6 has vertex (0 3) and opens downward, so = −3 and its equation is
N
33. The distance from the focus (−4 0) to the directrix = 2 is 2 − (−4) = 6, so the distance from the focus to the vertex is
1
2
(6) = 3 and the vertex is (−1 0). Since the focus is to the left of the vertex, = −3. An equation is 2 = 4( + 1) ⇒
2 = −12( + 1).
34. The parabola with vertex (2 3) and focus (2 −1) opens downward and has = −1 − 3 = −4, so its equation is
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°
35. The parabola with vertex (3 −1) having a horizontal axis has equation [ − (−1)]2 = 4( − 3). Since it passes through
36. The parabola with vertical axis and passing through (0 4) has equation = 2 + + 4. It also passes through (1 3) and
(−2 −6), so
3=++4 −1 = + −1 = +
⇒ ⇒
−6 = 4 − 2 + 4 −10 = 4 − 2 −5 = 2 −
Adding the last two equations gives us 3 = −6, or = −2. Since + = −1, we have = 1, and an equation is
= −22 + + 4.
LE
37. The ellipse with foci (±2 0) and vertices (±5 0) has center (0 0) and a horizontal major axis, with = 5 and = 2,
2 2
so 2 = 2 − 2 = 25 − 4 = 21. An equation is + = 1.
25 21
SA
√ √
38. The ellipse with foci 0 ± 2 and vertices (0 ±2) has center (0 0) and a vertical major axis, with = 2 and = 2,
2 2
so 2 = 2 − 2 = 4 − 2 = 2. An equation is + = 1.
2 4
39. Since the vertices are (0 0) and (0 8), the ellipse has center (0 4) with a vertical axis and = 4. The foci at (0 2) and (0 6)
√ √ √ ( − 0)2 ( − 4)2
R
are 2 units from the center, so = 2 and = 2 − 2 = 42 − 22 = 12. An equation is 2
+ =1 ⇒
2
2 ( − 4)2
+ = 1.
FO
12 16
40. Since the foci are (0 −1) and (8 −1), the ellipse has center (4 −1) with a horizontal axis and = 4.
√ √ √
The vertex (9 −1) is 5 units from the center, so = 5 and = 2 − 2 = 52 − 42 = 9. An equation is
( + 1)2 ( − 4)2
41. An equation of an ellipse with center (−1 4) and vertex (−1 0) is 2
+ = 1. The focus (−1 6) is 2 units
42
O
( + 1)2 ( − 4)2
from the center, so = 2. Thus, 2 + 22 = 42 ⇒ 2 = 12, and the equation is + = 1.
12 16
N
2 2
42. Foci 1 (−4 0) and 2 (4 0) ⇒ = 4 and an equation is + = 1. The ellipse passes through (−4 18), so
2 2
2 = | 1 | + | 2 | ⇒ 2 = 18 + 82 + (18)2 ⇒ 2 = 18 + 82 ⇒ = 5.
2 2
2 = 2 − 2 = 25 − 16 = 9 and the equation is + = 1.
25 9
2 2
43. An equation of a hyperbola with vertices (±3 0) is − = 1. Foci (±5 0) ⇒ = 5 and 32 + 2 = 52 ⇒
32 2
2 2
2 = 25 − 9 = 16, so the equation is − = 1.
9 16
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°
2 2
44. An equation of a hyperbola with vertices (0 ±2) is 2
− 2 = 1. Foci (0 ±5) ⇒ = 5 and 22 + 2 = 52 ⇒
2
2 2
2 = 25 − 4 = 21, so the equation is − = 1.
4 21
45. The center of a hyperbola with vertices (−3 −4) and (−3 6) is (−3 1), so = 5 and an equation is
( − 1)2 ( + 3)2
− = 1. Foci (−3 −7) and (−3 9) ⇒ = 8, so 52 + 2 = 82 ⇒ 2 = 64 − 25 = 39 and the
52 2
( − 1)2 ( + 3)2
equation is − = 1.
25 39
LE
( − 3)2 ( − 2)2
46. The center of a hyperbola with vertices (−1 2) and (7 2) is (3 2), so = 4 and an equation is − = 1.
42 2
Foci (−2 2) and (8 2) ⇒ = 5, so 42 + 2 = 52 ⇒ 2 = 25 − 16 = 9 and the equation is
( − 3)2 ( − 2)2
− = 1.
SA
16 9
2 2
47. The center of a hyperbola with vertices (±3 0) is (0 0), so = 3 and an equation is − = 1.
32 2
2 2
Asymptotes = ±2 ⇒ = 2 ⇒ = 2(3) = 6 and the equation is − = 1.
9 36
R
( − 4)2 ( − 2)2
48. The center of a hyperbola with foci (2 0) and (2 8) is (2 4), so = 4 and an equation is − = 1.
2 2
FO
1
The asymptote = 3 + 12 has slope 12 , so = ⇒ = 2 and 2 + 2 = 2 ⇒ 2 + (2)2 = 42 ⇒
2
( − 4)2 ( − 2)2
52 = 16 ⇒ 2 = 16
5
and so 2 = 16 − 16
5
= 64
5
. Thus, an equation is − = 1.
165 645
49. In Figure 8, we see that the point on the ellipse closest to a focus is the closer vertex (which is a distance
− from it) while the farthest point is the other vertex (at a distance of + ). So for this lunar orbit,
T
2 2
or = 102. Thus, 2 = 2 − 2 = 3,753,196, and the equation is + = 1.
3,763,600 3,753,196
50. (a) Choose to be the origin, with -axis through and . Then is ( 0), is ( 5), so substituting into the
N
51. (a) Set up the coordinate system so that is (−200 0) and is (200 0).
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°
2 2
(2 − 2 )2 − 2 2 = 2 (2 − 2 ) ⇔ 2 2 − 2 2 = 2 2 [where 2 = 2 − 2 ] ⇔ − 2 =1
2
53. The function whose graph is the upper branch of this hyperbola is concave upward. The function is
LE
2 √ 2
= () = 1+ = + 2 , so 0 = (2 + 2 )−12 and
2
00 = (2 + 2 )−12 − 2 (2 + 2 )−32 = (2 + 2 )−32 0 for all , and so is concave upward.
SA
54. We can follow exactly the same sequence of steps as in the derivation of Formula 4, except we use the points (1 1) and
(−1 −1) in the distance formula (first equation of that derivation) so ( − 1)2 + ( − 1)2 + ( + 1)2 + ( + 1)2 = 4
will lead (after moving the second term to the right, squaring, and simplifying) to 2 ( + 1)2 + ( + 1)2 = + + 4,
2 2
(b) If 0 16, then − 16 0, and + = 1 is a hyperbola since it is the difference of two squares on the
− 16
left side.
(c) If 0, then − 16 0, and there is no curve since the left side is the sum of two negative terms, which cannot equal 1.
(d) In case (a), 2 = , 2 = − 16, and 2 = 2 − 2 = 16, so the foci are at (±4 0). In case (b), − 16 0, so 2 = ,
T
2 0
57. 2 = 4 ⇒ 2 = 4 0 ⇒ 0 = , so the tangent line at (0 0 ) is − 0 = ( − 0 ). This line passes
2 4 2
20 0
through the point ( −) on the directrix, so − − = ( − 0 ) ⇒ −42 − 20 = 20 − 220 ⇔
4 2
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°
58. Without a loss of generality, let the ellipse, hyperbola, and foci be as shown in the figure.
LE
The curves intersect (eliminate 2 ) ⇒
2 2
2 2 2
2 − + + = 2 + 2 ⇒
2 2 2 2
2
SA
2 2 2 2 2 2 2 2
+ = + ⇒ + = 2 + 2 ⇒
2 2 2 2
2 + 2 2 2 ( 2 + 2 )
2 = = .
2 2 + 22 2 2 + 22
2 2
2 2 (2 − 2 )
Similarly, 2 = .
R
22 + 2 2
2 2 2 2 0 0
Next we find the slopes of the tangent lines of the curves: + =1 ⇒ + =0 ⇒ =− 2 ⇒
2 2 2 2 2
FO
2 2 2 2 20 0 2
0
=− and − =1 ⇒ − = 0 ⇒ = ⇒
0
= . The product of the slopes
2 2 2 2 2 2 2 2
2 2 2 2
2 2 ( + )
2 2 2 2 2 + 22 2 + 2
at (0 0 ) is
0 0
= − 2 2 02 = − 2 2 2 2 = − 2 2
. Since 2 − 2 = 2 and 2 + 2 = 2 ,
0 ( − ) −
22
22 + 2 2
T
we have 2 − 2 = 2 + 2 ⇒ 2 − 2 = 2 + 2 , so the product of the slopes is −1, and hence, the tangent lines at
each point of intersection are perpendicular.
O
2 2
59. 92 + 4 2 = 36 ⇔ + = 1. We use the parametrization = 2 cos , = 3 sin , 0 ≤ ≤ 2. The circumference
4 9
N
is given by
2 2
= 0
()2 + ()2 = 0 (−2 sin )2 + (3 cos )2
2 2 √
= 0 4 sin2 + 9 cos2 = 0 4 + 5 cos2
2 − 0 √
Now use Simpson’s Rule with = 8, ∆ = = , and () = 4 + 5 cos2 to get
8 4
4 3
≈ 8 = 3 (0) + 4 4 + 2 2 + 4 4 + 2 () + 4 5 4
+ 2 3
2
+ 4 7 4
+ (2) ≈ 159.
2 2
= 12 (114 × 1010 ) = 57 × 109 . An equation of the ellipse is + 2 = 1, or converting into parametric equations,
2
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°
2 − 0
Using Simpson’s Rule with = 10, ∆ = 10 = 20 ,
and () = 2 sin2 + 2 cos2 , we get
≈ 4 · 10 = 4 ·
20 · 3
(0) + 4 20 + 2 2
20
+ · · · + 2 8
20
+ 4 9
20
+ 2 ≈ 364 × 1010 km
2 2 2 2 − 2 √ 2
61. 2
− 2 =1 ⇒ 2
= 2
⇒ =± − 2 .
39 2 2
= 2 2 − 2 = 2 − 2 − ln + 2 − 2
2 2
LE
√ 2 √
= − 2 − 2 ln + 2 − 2 + 2 ln ||
√
Since 2 + 2 = 2 2 − 2 = 2 , and 2 − 2 = .
SA
= − 2 ln( + ) + 2 ln = + 2 (ln − ln( + ))
= 2 + ln[( + )], where 2 = 2 + 2 .
2 2 2 2 − 2 √ 2
62. (a) 2
+ 2 =1 ⇒ 2
= 2
⇒ =± − 2 .
2
2 2
( − 2 )
R
= 2 − 2 = 2 2
− 0
22 22 23 4
= 2 2 − 13 3 = 2 = 2
FO
0 3 3
2 2 2 2 − 2 2
(b) 2
+ 2 =1 ⇒ 2
= 2
⇒ =± − 2 .
2
2 2
= 2 − 2 = 2 2 ( − 2 )
− 0
22
T
22 23 4
= 2 2 − 13 3 = 2 = 2
0 3 3
O
2 2
63. 92 + 4 2 = 36 ⇔ + = 1 ⇒ = 3, = 2. By symmetry, = 0. By Example 2 in Section 7.3, the area of the
4 9
N
top half of the ellipse is 12 () = 3. Solve 92 + 4 2 = 36 for to get an equation for the top half of the ellipse:
√
92 + 4 2 = 36 ⇔ 4 2 = 36 − 92 ⇔ 2 = 94 (4 − 2 ) ⇒ = 3
2
4 − 2 . Now
2 2 2
1 1 1 1 3 3
= [ ()]2 = 4 − 2 = (4 − 2 )
2 3 −2 2 2 8 −2
2 2
3 2 3 1 3 3 16 4
= ·2 (4 − ) = 4 − = =
8 0 4 3 0 4 3
so the centroid is (0 4).
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°
2 2
64. (a) Consider the ellipse 2
+ 2 = 1 with , so that the major axis is the -axis. Let the ellipse be parametrized by
= cos , = sin , 0 ≤ ≤ 2. Then
2 2
+ = 2 sin2 + 2 cos2 = 2 (1 − cos2 ) + 2 cos2 = 2 + (2 − 2 ) cos2 = 2 − 2 cos2 ,
where 2 = 2 − 2 . Using symmetry and rotating the ellipse about the major axis gives us surface area
2 0
1 = cos
= 2 = 2 2( sin ) 2 − 2 cos2 = 4 2 − 2 −
= − sin
0
4 30 4 2 2 2
= 2 − 2 = 2 − 2 + sin−1 = − 2 + 2 sin−1
0 2 2 0
LE
2
= + 2 sin−1
(b) As in part (a),
2 2
SA
+ = 2 sin2 + 2 cos2 = 2 sin2 + 2 (1 − sin2 ) = 2 + (2 − 2 ) sin2 = 2 + 2 sin2 .
Rotating about the minor axis gives us
2
1 = sin
= 2 = 2 2( cos ) 2 + 2 sin2 = 4 2
+2
= cos
0 0
21 4 √ 2 2 √ 2 √ 2 √
+ 2 + 2 ln + 2 + 2 − 2 ln
R
= + 2 + ln + 2 + 2 =
2 2 0
2 +
= + 2 ln
FO
2 2 2 20 2
65. Differentiating implicitly, 2
+ 2 =1 ⇒ 2
+ 2 = 0 ⇒ 0 = − 2 [ 6= 0]. Thus, the slope of the tangent
2 1 1 1
line at is − . The slope of 1 is and of 2 is . By the formula in Problem 21 on text page 273,
2 1 1 + 1 −
we have
2 1
T
1
+ 2
1 + 1 2 2 + 2 1 (1 + ) 2 2 + 2 1 using 2 21 + 2 12 = 2 2 ,
tan = 2 = 2 1 = 2 2 2
1 1 1 (1 + ) − 2 1 1 2 1 1 + 2 1 and − =
1− 2
O
1 (1 + )
2 1 + 2 2
= 2
=
1 (1 + ) 1
N
2 1 1
−−
2 1 1 − −2 12 − 2 1 (1 − ) −2 2 + 2 1 2 1 − 2 2
and tan = 2 = 2 2
= 2 2
= 2
=
1 1 1 (1 − ) − 1 1 1 1 − 1 1 (1 − ) 1
1− 2
1 (1 − )
Thus, = .
1 1
66. The slopes of the line segments 1 and 2 are and , where is (1 1 ). Differentiating implicitly,
1 + 1 −
2 20 2 2 1
2
− 2 = 0 ⇒ 0 = 2 ⇒ the slope of the tangent at is , so by the formula in Problem 21 on text
2 1
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°
page 273,
2 1 1
−
2 1 1 + 2 1 (1 + ) − 2 12 2 (1 + 2 ) using 21 2 − 12 2 = 1, 2
tan = 2 = 2 = 2 2 2
=
1 1 1 (1 + ) + 2 1 1 1 (1 + 2 ) and + = 1
1+ 2
1 (1 + )
2 1 1
− +
and 2 1 1 − −2 1 (1 − ) + 2 12 2 (1 − 2 ) 2
tan = 2 = 2 = =
1 1 1 (1 − ) + 2 1 1 1 (1 − 2 ) 1
1+ 2
1 (1 − )
So = .
LE
10.6 Conic Sections in Polar Coordinates
1. The directrix = 4 is to the right of the focus at the origin, so we use the form with “+ cos ” in the denominator.
SA
1
2 ·4 4
(See Theorem 6 and Figure 2.) An equation of the ellipse is = = = .
1 + cos 1 + 12 cos 2 + cos
2. The directrix = −3 is to the left of the focus at the origin, so we use the form with “− cos ” in the denominator.
1·3 3
= 1 for a parabola, so an equation is = = = .
1 − cos 1 − 1 cos 1 − cos
R
3. The directrix = 2 is above the focus at the origin, so we use the form with “+ sin ” in the denominator. An equation of
15(2) 6
FO
the hyperbola is = = = .
1 + sin 1 + 15 sin 2 + 3 sin
4. The directrix = 3 is to the right of the focus at the origin, so we use the form with “+ cos ” in the denominator. An
3·3 9
equation of the hyperbola is = = = .
1 + cos 1 + 3 cos 1 + 3 cos
5. The vertex (2 ) is to the left of the focus at the origin, so we use the form with “− cos ” in the denominator. An equation
T
2
2
of the ellipse is = . Using eccentricity = with = and = 2, we get 2 = 3
2 ⇒
1 − cos 3 1 − 3 (−1)
O
2
2 (5) 10
2= ⇒ = 5, so we have = 3
= .
5 1 − 23 cos 3 − 2 cos
N
6. The directrix = 4 csc (equivalent to sin = 4 or = 4) is above the focus at the origin, so we will use the form with
“+ sin ” in the denominator. The distance from the focus to the directrix is = 4, so an equation of the ellipse is
(06)(4) 5 12
= = · = .
1 + sin 1 + 06 sin 5 5 + 3 sin
7. The vertex 3
2
is 3 units above the focus at the origin, so the directrix is 6 units above the focus ( = 6), and we use the
1(6) 6
form “+ sin ” in the denominator. = 1 for a parabola, so an equation is = = = .
1 + sin 1 + 1 sin 1 + sin
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°