26. Since the vertex is 2 units to the right of the focus, the directrix must be 4 units to the right
of the focus and perpendicular to the
x
axis. Therefore,
r
=
4
1 + cos
θ
.
27. Since the vertex is
1
4
units below the focus, the directrix must be
1
2
units below the focus and
parallel to the
x

axis. Therefore,
r
=
1
/
2
1

sin
θ
.
28. Since the vertex is
3
2
units above the focus, the directrix must be 3 units above the focus and
parallel to the
x
axis. Therefore,
r
=
3
1 + sin
θ
.
29. This is the parabola
r
=
4
1 + cos
θ
rotated counterclockwise by
π/
4
.
The original parabola
had its focus at the origin and its directrix at
x
= 4
.
The original vertex therefore had polar
coordinates (2
,
0). After rotation, the vertex is located at (2
, π/
4)
.
728
CHAPTER 10.
CONICS AND POLAR COORDINATES
30. This is the ellipse
r
=
5
3 + 2 cos
θ
=
5
/
3
1 + 2
/
3 cos
θ
rotated counterclockwise by
π/
3.
The
original ellipse had vertices at
θ
= 0 and
θ
=
π.
The polar coordinates of the vertices were
(1
,
0) and (5
, π
). After rotation, the vertices are located at (1
, π/
3) and (5
,
4
π/
3)
.
31. This is the ellipse
r
=
10
2

sin
θ
=
5
2

1
2
sin
θ
rotated clockwise by
π/
6
.
The original ellipse
had vertices at
θ
=
π/
2 and
θ
= 3
π/
2
.
The polar coorinates of the vertices were (10
, π/
2) and
(10
.
3
,
3
π/
2). After rotation, the vertices are located at (10
.π/
3) and (10
/
3
,
4
π/
3)
.
32. This is the hyperbola
r
=
6
1 + 2 sin
θ
rotated clockwise by
π/
3
.
The original hyperbola had
vertices at
θ
=
π/
2 and
θ
= 3
π/
2. The polar coordinates of the vertices were (2
, π/
2) and
(

6
,
3
π/
2). After rotation, the vertices are located at (2
, π/
6) and (

6
,
7
π/
6)
.
33. Identifying
r
a
= 12
,
000 and
e
= 0
.
2
,
we have from (7) in the text 0
.
2 =
12
,
000

r
p
12
,
000 +
r
p
.
Solving
for
r
p
,
we obtain
r
p
= 8
,
000 km.
34. The equation of the orbit is
r
=
0
.
2
p
(1

0
.
2 cos
θ
)
.
When
θ
= 0
, r
= 12
,
000 so 12
,
000 =
0
.
2
p
(1

0
.
2)
=
p
4
. Thus,
p
= 48
,
000 and the equation of the orbit is
r
=
9
,
600
(1

0
.
2 cos
θ
)
.
35. The equation of the orbit is
r
=
ep
(1

e
cos
θ
)
.
From (7) in the text,
e
=
1
.
5
×
10
8

1
.
47
×
10
8
1
.
52
×
10
8
+ 1
.
47
×
10
8
=
5
299
≈
1
.
67
×
10

2
.
When
θ
= 0
, r
=
r
a
= 1
.
52
×
10
8
=
ep
(1

1
.
67
×
10

2
)
.
Thus
ep
≈
1
.
52
×
10
8

2
.
52
×
10
6
≈
1
.
49
×
10
8
and the equation of the orbit is
r
=
(1
.
49
×
10
8
)
(1

1
.
67
×
10

2
cos
θ
)
.
36.
(a) The equation of the orbit is
r
=
0
.
97
p
(1

0
.
97 cos
θ
)
.
The length of the major axis is the sum
of
r
a
=
r
(0) =
0
.
97
p
0
.
03
and
r
p
=
r
(
π
) =
0
.
97
p
1 + 0
.
97
=
0
.
97
p
1
.
97
.
That is
r
a
+
r
p
= 0
.
97
p
1
0
.
03
+
1
1
.
97
= 3
.
34
×
10
9
.
Solving for
p
we obtain
p
= 1
.
02
×
10
8
.
The equation of the orbit is
r
=
0
.
97(1
.
02
×
10
8
)
(1

0
.
97 cos
θ
)
.
(b) From part (a)
r
a
=
r
(0) =
9
.
87
×
10
7
1

0
.
97
=
9
.
87
×
10
7
0
.
03
≈
3
.
29
×
10
9
miles
10.7.
CONIC SECTIONS IN POLAR COORDINATES
729
and
r
p
=
r
(
π
) =
9
.
87
×
10
7
1 + 0
.
97
=
9
.
87
×
10
7
1
.
97
≈
5
.
01
×
10
7
miles
.
37.
polar
axis
38.
polar
axis
39.
polar
axis
40.
polar
axis
730
CHAPTER 10.
CONICS AND POLAR COORDINATES
41.
polar
axis
42.
polar
axis
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 Fall '13
 Cartesian Coordinate System, Polar Coordinates, Polar coordinate system, Conic section