Chapter 7: Conic Sections
SSM: College Algebra
Section 7.2
3.
Check Point Exercises
1. a.
a 2 = 25, a = 5
vertices: (5, 0) and (–5, 0)
b 2 = 16
c 2 = a 2 + b 2 = 25 + 16 = 41
c = 41
The foci are at
c = 45 = 3 5
The foci are at ( −3 5 , 0) and (3 5 , 0) .
( 41, 0) and ( − 41, 0) .
b.
x2 y2
−
=1
36 9
a 2 = 36, a = 6
The vertices are (6, 0) and (-6, 0).
b 2 = 9, b = 3
b
3
1
asymptotes: y = ± x = ± x = ± x
a
6
2
c 2 = a 2 + b 2 = 36 + 9 = 45
y
10
a 2 = 25, a = 5
vertices: (0, 5) and (0, –5)
b 2 = 16
F1
c 2 = a 2 + b 2 = 25 + 16 = 41
F2
10 x
c = 41
The foci are at
( 0,
41) and ( 0, − 41) .
4.
2. a = 3, c = 5
b 2 = c 2 − a 2 = 25 − 9 = 16
y2 x2
−
=1
9 16
y2 − 4 x2 = 4
y2 4 x2 4
−
=
4
4
4
2
y
− x2 = 1
4
a 2 = 4, a = 2
The vertices are (0, 2) and (0, –2).
b 2 = 1, b = 1
a
asymptotes: y = ± x = ±2 x
b
2
2
2
c = a + b = 4 +1 = 5
c= 5
395
Chapter 7: Conic Sections
SSM: College Algebra
Exercise Set 7.2
The foci are at ( 0, 5 ) and ( 0, − 5 ) .
y
5
1. a2 = 4, a = 2
The vertices are (2, 0) and (–2, 0).
b2 = 1
c2 = a2 + b2 = 4 + 1 = 5
F1
5 x
c= 5
F2
The foci are located at ( 5 , 0) and
( − 5 , 0).
graph (b)
( x − 3)2 ( y − 1)2
5.
−
=1
4
1
center at (3, 1)
a 2 = 4, a = 2
3. a2 = 4, a = 2
The vertices are (0, 2) and (0, –2).
b2 = 1
c2 = a2 + b2 = 4 + 1 = 5
b 2 = 1, b = 1
The vertices are (1, 1) and (5, 1).
1
asymptotes: y − 1 = ± ( x − 3)
2
c2 = a2 + b2 = 4 + 1 = 5
c= 5
The foci are located at
(0, 5) and ( 0, − 5 ) .
graph (a)
c= 5
The foci are at (3 − 5 , 1) and (3 + 5 , 1) .
5.
y
5
a = 1, c = 3
b2 = c2 – a2 = 9 – 1 = 8
y2 −
F1
F2
8 x
7.
x2
=1
8
a = 3, c = 4
b2 = c2 – a2 = 16 – 9 = 7
x2 y2
−
=1
9
7
6.
c = 5280
2 a = 3300, a = 1650
b 2 = c 2 − a 2 = 5280 2 − 1650 2 = 25, 155, 900
The explosion occurred somewhere at the
right branch of the hyperbola given by
x2
y2
−
= 1.
2, 722, 500 25, 155, 900
396
Chapter 7: Conic Sections
SSM: College Algebra
9. 2a = 6 – (–6)
2a = 12
a=6
a
=2
b
6
=2
b
6 = 2b
3=b
Transverse axis is vertical.
y2 x2
−
=1
36
9
15.
x2
y2
−
=1
100 64
a2 = 100, a = 10
b2 = 64, b = 8
vertices: (10, 0) and (–10, 0)
b
8
x
asymptotes: y = ± x = ±
a
10
4
or y = ± x
5
c2 = a2 + b2 = 100 + 64 = 164
c = 164 = 2 41 on x-axis
The foci are at (2 41, 0) and ( −2 41, 0) .
y
20
11. a = 2, c = 7 – 4 = 3
2 2 + b 2 = 32
4 + b2 = 9
b2 = 5
Transverse axis is horizontal.
( x − 4 )2 ( y + 2 )2
−
=1
4
5
13.
F1
F2
20 x
17.
x2 y2
−
=1
9 25
a2 = 9, a = 3
b2 = 25, b = 5
vertices: (3, 0) and (–3, 0)
b
5
asymptotes: y = ± x = ± x
a
3
c2 = a2 + b2 = 9 + 25 = 34
c = 34 on x-axis
y2 x2
−
=1
16 36
a2 = 16, a = 4
b2 = 36, b = 6
vertices: (0, 4) and (0, –4)
a
4
asymptotes: y = ± x = ± x
b
6
2
or y = ± x
3
c2 = a2 + b2 = 16 + 36 = 52
c = 52 = 2 13 on y-axis
The foci are at ( 34 , 0) and ( − 34 , 0) .
The foci are at ( 0, 2 13 ) and ( 0, − 2 13 ) .
y
10
y
10 F
1
F1
F2
10 x
10 x
F2
397
Chapter 7: Conic Sections
19.
SSM: College Algebra
y2
− x2 = 1
1
4
1
1
a2 = , a =
4
2
2
b = 1, b = 1
3
b
x=± x
2
a
c2 = a2 + b2 = 4 + 9 = 13
asymptotes: y = ±
c = 13 on x-axis
The foci are at ( 13 , 0) and ( − 13 , 0) .
y
5
c2 = a2 + b2
1
c2 = + 1
4
5
c2 =
4
F1
F2
5 x
5
2
c ≈ ±1.1
The foci are located at


5
5
 0, 2  and  0, − 2  .




c=±
23.
9y 2 − 25x 2 = 225
9y 2 25x 2 225
−
=
225 225 225
y2 x2
−
=1
25 9
a2 = 25, a = 5
b2 = 9, b = 3
vertices: (0, 5) and (0, –5)
a
5
asymptotes: y = ± x = ± x
b
3
c2 = a2 + b2 = 25 + 9 = 34
1
2
y=± x
asymptotes:
1
1
y=± x
2
y
5
c = 34 on y-axis
The foci are at ( 0, 34 ) and ( 0, − 34 ) .
F¡
y
10
5 x
F1
F™
10 x
21.
9x 2 − 4 y 2 = 36
F2
9x 2 4 y 2 36
−
=
36
36
36
2
2
x
y
−
=1
4
9
a2 = 4, a = 2
b2 = 9, b = 3
vertices: (2, 0) and (–2, 0)
25. y 2 = x 2 − 2
398
Chapter 7: Conic Sections
SSM: College Algebra
2 = x2 − y2
x2 y2
1=
−
2
2
2
a = 2, a = 2
b 2 = 2, b = 2
c2 = 2 + 2
4
( x + 4)
3
c2 = a2 + b2 = 9 + 16 = 25
c = ±5 parallel to x-axis
The foci are at (–9, –3) and (1, –3).
asymptotes: y + 3 = ±
y
2
1 x
c2 = 4
c =2
The foci are located at (2,0) and (–2, 0).
asymptotes:
y=±
2
F1
x
2
y = ±x
35.
y
5
F¡
F™
( x + 3)2 y 2
−
=1
25
16
center: (–3, 0)
a 2 = 25, a = 5
b 2 = 16, b = 4
vertices: (2, 0) and (–8, 0)
4
asymptotes: y = ± ( x + 3)
5
2
2
2
c = a + b = 25 + 16 = 41
5 x
27. a = 3, b = 5
c = 41
x2 y2
−
=1
9 25
The foci are at ( −3 + 41, 0) and
( −3 − 41, 0) .
y
10
29. a = 2, b = 3
2
F2
2
y
x
−
=1
4
9
31. Center (2, –3), a = 2, b = 3
( x − 2)2 ( y + 3)2
−
=1
4
9
F1
2
2
33. ( x + 4) − ( y + 3) = 1
9
16
center: (–4, –3)
a2 = 9, a = 3
b2 = 16, b = 4
vertices: (–7, –3) and (–1, –3)
37.
399
F2 8 x
( y + 2)2 ( x − 1)2
−
=1
4
16
center: (1, –2)
a2 = 4, a = 2
b2 = 16, b = 4
vertices: (1, 0) and (1, –4)
Chapter 7: Conic Sections
SSM: College Algebra
41. (x – 1)2 – (y – 2)2 = 3
( x − 1)2 ( y − 2)2
−
=1
3
3
center: (1, 2)
1
( x − 1)
2
c2 = a2 + b2 = 4 + 16 = 20
asymptotes: y + 2 = ±
c=
20 = 2 5 parallel to y-axis
The foci are at (1, − 2 + 2 5 ) and
a2 = 3, a =
(1, − 2 − 2 5 ) .
b2 =
y
3
3
,b= 3
vertices: (–1, 2) and (3, 2)
asymptotes: y – 2 = ± (x – 1)
c2 = a2 + b2 = 3 + 3 = 6
F1
6 x
c=
6 parallel to y-axis
The foci are at (1 + 6 , 2) and (1 − 6 , 2).
y
7
F2
39.
( x − 3)2 − 4( y + 3)2 = 4
F1
( x − 3)2 4( y + 3)2 4
−
=
4
4
4
2
( x − 3)
− ( y + 3)2 = 1
4
center: (3, –3)
a 2 = 4, a = 2
6 x
43.
b 2 = 1, b = 1
vertices: (1, –3) and (5, –3)
1
asymptotes: y + 3 = ± ( x − 3)
2
c2 = a2 + b2 = 4 + 1 = 5
c= 5
The foci are at (3 + 5 , − 3) and
x2 – y2 – 2x – 4y – 4 = 0
(x2 – 2x) – (y2 + 4y) = 4
(x2 – 2x + 1) – (y2 + 4y + 4) = 4 + 1 – 4
(x – 1)2 – (y + 2)2 = 1
center: (1, –2)
a2 = 1, a = 1
b2 = 1, b = 1
c2 = a2 + b2 = 1 + 1 = 2
c= 2
asymptotes: y + 2 = ±( x − 1)
(3 − 5 , − 3).
The foci are at (1 + 2 , − 2) and
y
2
(1 − 2 , − 2) .
y
3
8 x
F1
F2
F2
6 x
F1
400
F2
Chapter 7: Conic Sections
SSM: College Algebra
45. 16x2 – y2 + 64x – 2y + 67 = 0
(16x2 + 64x) – (y2 + 2y) = –67
16(x2 + 4x + 4) – (y2 + 2y + 1)
= –67 + 64 – 1
2
2
16( x + 2) − ( y + 1) = −4
a2 = 9, a = 3
b2 = 4, b = 2
c2 = a2 + b2 = 9 + 4 = 13
c = 13
2
( x − 2)
3
The foci are at (2 + 13 , 3) and
asymptotes: y − 3 = ±
−4
16( x + 2)
( y + 1)
−
=
−4
−4
−4
2
2
( y + 1)
( x + 2)
−
=1
1
4
2
2
(2 − 13 , 3) .
y
8
4
center: (–2, –1)
a2 = 4, a = 2
1
1
b2 = , b =
4
2
c2 = a2 + b2 = 4 +
c=
17
4
1
17
=
4
4
7 x
= 4.25
49. 4x2 – 25y2 – 32x + 164 = 0
(4x2 – 32x) – 25y2 = –164
4(x2 – 8x + 16) – 25y2 = –164 + 64
4(x – 4)2 – 25y2 = –100
2
( y + 1) = ± ( x + 2)
1
asymptotes:
2
y + 1 = ±4( x + 2)
(
4( x − 4)2 25y 2 −100
−
=
−100
−100 −100
y 2 ( x − 4 )2
−
=1
4
25
center: (4, 0)
a2 = 4, a = 2
b2 = 25, b = 5
c2 = a2 + b2 = 4 + 25 = 29
)
The foci are at −2, − 1 + 4.25 and
(−2, − 1 −
)
4.25 .
y
4
F2
F1
F1
3 x
c = 29
2
( x − 4)
5
The foci are at ( 4, 29 ) and ( 4, − 29 ) .
F2
asymptotes: y = ±
47. 4x2 – 9y2 – 16x + 54y – 101 = 0
(4x2 – 16x) – (9y2 – 54y) = 101
4(x2 – 4x + 4) – 9(y2 – 6y + 9)
= 101 + 16 – 81
4(x – 2)2 – 9(y – 3)2= 36
y
10
F1
14 x
( x − 2)2 ( y − 3)2
−
=1
9
4
center: (2, 3)
F2
401
Chapter 7: Conic Sections
SSM: College Algebra
67. 4x2 – 6xy + 2y2 – 3x + 10y – 6 = 0
2y2 + (10 – 6x)y + (4x2 – 3x – 6) = 0
51. | d2 – d1 | = 2a = (2 s)(1100 ft / s) = 2200 ft
a = 1100 ft
2c = 5280 ft, c = 2640 ft
b2 = c2 – a2 = (2640)2 – (1100)2
= 5,759,600
x2
y2
−
=1
(1100)2 5, 759, 600
x2
y2
−
=1
1, 210, 000 5, 759, 600
If M1 is located 2640 feet to the right of the
origin on the x-axis, the explosion is located
on the right branch of the hyperbola given
by the equation above.
y=
6x − 10 ± (10 − 6x )2 − 8( 4 x 2 − 3x − 6)
4
y=
6x − 10 ± 4( x 2 − 24 x + 37)
4
y=
3x − 5 ± x 2 − 24 x + 37
2
50
–50
70
–30
625y 2 − 400 x 2 = 250, 000
53.
The xy-term rotates the hyperbola.
Separation of terms into ones containing
only x or only y would be impossible.
625y 2
400 x 2
250, 000
−
=
250, 000 250, 000 250, 000
y2
x2
−
=1
400 625
a2 = 400, a = 400 = 20
2a = 40
The houses are 40 yards apart at their closest
point.
69. a.
x2
x2 y2
−
=0
4
9
9
y2 = x2
4
3
y=± x
2
c.
2
x is one of the asymptotes
3
of the hyperbola and they will not
intersect.
True; y = −
d. False; for example,
4
⫺6
−
y2
= 1 will not pass the vertical
a2 b2
line test, so will not define y as a
function of x.
b. False; none of the points on the
asymptotes satisfy the hyperbola’s
equation, since the hyperbola never
touches its asymptotes.
55.–61. Answers may vary.
65.
False; one branch of the hyperbola
x2 y2
−
= 1 and
4
4
y2 x2
−
= 1 each have asymptotes
4
4
y = ± x, but are different hyperbolas.
(c) is true.
6
⫺4
No; in general, the graph is two intersecting
lines.
71. The center is at the midpoint of the line
segment joining the vertices, so it is located
at (5, 0). The standard form is:
402
Chapter 7: Conic Sections
SSM: College Algebra
( y − k )2
2
−
( x − h )2
2
directrix: y = 3
=1
y
10
a
b
(h, k) = (5, 0), and a = 6, so a2 = 36.
Directrix:
y=3
y 2 ( x − 5)2
−
= 1.
36
b2
Substitute x = 0 and y = 9:
(–6, –3)
92 ( 0 − 5)2
−
=1
36
b2
25
5
− 2 =−
4
b
−100 = −5b 2
3.
p=8
y 2 = 4 ⋅ 8x
y 2 = 32 x
b 2 = 20
y 2 ( x − 5)2
Standard form:
−
=1
36
20
4. ( x − 2)2 = 4( y + 1)
4 p = 4, p = 1
vertex: (2, –1)
focus: (2, 0)
directrix: y = –2
Section 7.3
Check Point Exercises
1.
10 x
(6, –3)
Focus
(0, –3)
y
5
y 2 = 8x
4 p = 8, p = 2
foci: (2, 0)
directrix: x = –2
F
7 x
y
5
5. y 2 + 2 y + 4 x − 7 = 0
F
5 x
y 2 + 2 y = −4 x + 7
y 2 + 2 y + 1 = −4 x + 7 + 1
2.
( y + 1)2 = −4( x − 2)
4 p = −4, p = −1
vertex: (2, –1)
focus: (1, –1)
x 2 = −12 y
4 p = −12, p = 3
focus: (0, –3)
403