Solution manual calculus 8th edition varberg, purcell, rigdon ch10

CHAPTER

10

10.1 Concepts Review
1. e < 1; e = 1; e > 1

Conics and Polar
Coordinates
3. x 2 = –4(3) y
Focus at (0, –3)
Directrix: y = 3

2. y 2 = 4 px
3. (0, 1); y = –1
4. parallel to the axis

Problem Set 10.1
1. y 2 = 4(1) x
Focus at (1, 0)
Directrix: x = –1

4. x 2 = –4(4) y
Focus at (0, –4)
Directrix: y = 4

2. y 2 = –4(3) x
Focus at (–3, 0)
Directrix: x = 3

⎛1⎞

5. y 2 = 4 ⎜ ⎟ x
⎝4⎠
⎛1 ⎞
Focus at ⎜ , 0 ⎟
⎝4 ⎠

Directrix: x = –

590

Section 10.1

1
4

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by any means, without permission in writing from the publisher.


6. y 2 = –3 x

8. 3x 2 = 9 y

⎛3⎞
y 2 = –4 ⎜ ⎟ x
⎝4⎠
⎛ 3 ⎞
Focus at ⎜ – , 0 ⎟

⎝ 4 ⎠
3
Directrix: x =
4

7. 2 x 2 = 6 y

Directrix: y = –

3
4

9. The parabola opens to the right, and p = 2.

⎛3⎞
x = 4⎜ ⎟ y
⎝4⎠
⎛ 3⎞
Focus at ⎜ 0, ⎟
⎝ 4⎠

y2 = 8x

2

Directrix: y = –

⎛3⎞
x2 = 4 ⎜ ⎟ y
⎝4⎠

⎛ 3⎞
Focus at ⎜ 0, ⎟
⎝ 4⎠

10. The parabola opens to the left, and p = 3.
y 2 = –12 x
3
4

11. The parabola opens downward, and p = 2.
x 2 = –8 y

1
12. The parabola opens downward, and p = .
9
4
x2 = – y
9
13. The parabola opens to the left, and p = 4.
y 2 = –16 x
7
14. The parabola opens downward, and p = .
2
x 2 = –14 y

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Section 10.1

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15. The equation has the form y 2 = cx, so
(–1)2 = 3c.

c=

1
3

y2 =

18. The equation has the form x 2 = cy, so
(–3) 2 = 5c .

c=

9
5

x2 =

⇒

9
y
5

1
x
3

19. y 2 = 16 x
2 yy ′ = 16
16. The equation has the form y 2 = cx, so
(4) 2 = –2c.
c = –8
y 2 = –8 x

17. The equation has the form x 2 = cy, so
(6)2 = –5c.
36
5
36
x2 = –
y
5

c=–

y′ =

16
2y

At (1, –4), y ′ = –2.
Tangent: y + 4 = –2(x – 1) or y = –2x – 2

1
1
9
Normal: y + 4 = ( x –1) or y = x –
2
2
2

20. x 2 = –10 y
2 x = –10 y ′

y′ = –

(

x
5

)

At 2 5, – 2 , y ′ = –
Tangent: y + 2 = –
y=–

Section 10.1

(

)

2 5
x – 2 5 or
5

2 5
x+2
5

Normal: y + 2 =

592

2 5
.
5

(

)

5
5
x – 2 5 or y =
x–7
2
2

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21. x 2 = 2 y
2 x = 2 y′
y′ = x
At (4, 8), y ′ = 4 .
Tangent: y – 8 = 4(x – 4) or y = 4x – 8
1
1
Normal: y – 8 = – ( x – 4) or y = – x + 9
4
4

y=–

2 5
21 5
x–
5
5

24. x 2 = 4 y
2 x = 4 y′
x
2
At (4, 4), y ′ = 2.
Tangent: y – 4 = 2(x – 4) or y = 2x – 4
1
1

Normal: y – 4 = – ( x – 4) or y = – x + 6
2
2
y′ =

22. y 2 = –9 x
2 yy ′ = –9
y′ = –

9
2y

3
2
3
3
3
Tangent: y + 3 = ( x + 1) or y = x –
2
2
2
2
2
11
Normal: y + 3 = – ( x + 1) or y = – x –
3
3
3

At (–1, –3), y ′ =

25. x 2 = –6 y
2 x = –6 y ′
y′ = –

(

x
3

)

At 3 2, – 3 , y ′ = – 2.

y′ = –

(

(

)

Tangent: y + 3 = – 2 x – 3 2 or y = – 2 x + 3

23. y 2 = –15 x
2 yy ′ = –15

Normal: y + 3 =

15

2y

(

)

2
2
x – 3 2 or y =
x–6
2
2

)

5
.
2
5
( x + 3) or
Tangent: y + 3 5 =
2
5
3 5
y=
x–
2
2
2 5
( x + 3) or

Normal: y + 3 5 = –
5

At –3, – 3 5 , y ′ =

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by any means, without permission in writing from the publisher.


26. y 2 = 20 x
2 yy ′ = 20

29. The slope of the line is
y 2 = −18 x; 2 yy ′ = −18

10
y′ =
y

(

)

10

.
2
10
( x – 2) or
Tangent: y + 2 10 = –
2
10
y=–
x – 10
2
10
( x – 2) or
Normal: y + 2 10 =
5

At 2, – 2 10 , y ′ = –

y=

10
12 10
x–
5
5

y′ =

⎛3⎞
2 y ⎜ ⎟ = −18; y = −6
⎝2⎠

(−6)2 = −18 x; x = −2
The equation of the tangent line is
3
3
y + 6 = ( x + 2) or y = x – 3 .
2
2

30. Place the x-axis along the axis of the parabola

such that the equation y 2 = 4 px describes the
⎛y 2
⎞
parabola. Let ⎜ 0 , y0 ⎟ be one of the
⎜ 4p
⎟
⎝
⎠

⎛y2
⎞
extremities and ⎜ 1 , y1 ⎟ be the other.
⎜ 4p
⎟
⎝
⎠
First solve for y1 in terms of y0 and p. Since the
focal chord passes through the focus (p, 0), we
have the following relation.
y

y1
= 20
2
y1
y
– p 40p – p
4p
y1 ( y02 – 4 p 2 ) = y0 ( y12 – 4 p 2 )

27. y 2 = 5 x
2 yy ′ = 5
y′ =

3
.
2

y0 y12 – ( y02 – 4 p 2 ) y1 – 4 p 2 y0 = 0
( y1 – y0 )( y0 y1 + 4 p 2 ) = 0

5
2y
5
when y = 2 5, so x = 4.
4

(

)

The point is 4, 2 5 .

y1 = y0 or y1 = –

4 p2
y0

⎛ 4 p3
4 p2
Thus, the other extremity is ⎜
,–
⎜ y 2
y0
⎝ 0

⎞
⎟.
⎟
⎠

Implicitly differentiate y 2 = 4 px to get
2 yy ′ = 4 p, so y ′ =

2p
.
y

⎛y 2
⎞
2p

. The equation of the
At ⎜ 0 , y0 ⎟ , y ′ =
⎜ 4p
⎟
y0
⎝
⎠

tangent line is y – y0 =
28. x 2 = –14 y
2 x = –14 y ′

x = –p, y = –

y′ = –

x
7

y′ = –

2 7
when x = 2 7, so y = –2.
7

(

)

The point is 2 7, – 2 .

594

Section 10.1

y 2⎞
2p ⎛
⎜ x – 0 ⎟ . When
y0 ⎜⎝
4 p ⎟⎠

2 p 2 y0
+ .
2
y0

⎛ 4 p3
4 p2
At ⎜
,–
⎜ y 2
y0
⎝ 0

⎞
y
⎟ , y ′ = – 0 . The equation of
⎟
2p
⎠

the tangent line is y +

y ⎛
4 p2
4 p3 ⎞
= – 0 ⎜x–
⎟.
2 p ⎜⎝
y0
y0 2 ⎟⎠

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by any means, without permission in writing from the publisher.


35. Let the y-axis be the axis of the parabola, so the
Earth’s coordinates are (0, p) and the equation of

y0 2 p 2
.
–
2
y0
Thus, the two tangent lines intersect on the
⎛
y
2 p2 ⎞

directrix at ⎜ – p, 0 –
⎟.
⎜
2
y0 ⎟⎠
⎝

When x = –p, y =

the path is x 2 = 4 py , where the coordinates are
in millions of miles. When the line from Earth to
the asteroid makes an angle of 75D with the axis
of the parabola, the asteroid is at

31. From Problem 30, if the parabola is described by

( 40sin 75D , p + 40 cos 75D ). (See figure.)

the equation y 2 = 4 px, the slopes of the tangent
y
2p
and – 0 . Since they are negative
y0
2p
reciprocals, the tangent lines are perpendicular.

lines are

32. Place the x-axis along the axis of the parabola

such that the equation y 2 = 4 px describes the
⎛ 1⎞
parabola. The endpoints of the chord are ⎜1, ⎟
⎝ 2⎠
2

1
1⎞
⎛
⎛1⎞
and ⎜1, – ⎟ , so ⎜ ⎟ = 4(1) p or p = . The
16
2
2
⎝
⎠
⎝ ⎠
1
.
distance from the vertex to the focus is
16

33. Assume that the x- and y-axes are positioned such
that the axis of the parabola is the y-axis with the
vertex at the origin and opening upward. Then

the equation of the parabola is x 2 = 4 py and
(0, p) is the focus. Let D be the distance from a
point on the parabola to the focus.
⎛ x2

D = ( x − 0)2 + ( y − p) 2 = x 2 + ⎜
−
⎜ 4p
⎝

⎞
p⎟
⎟
⎠

2

x4

x2
x2
=
+
+ p2 =
+p
4p
16 p 2 2
D′ =

x x
;
= 0, x = 0
2p 2p

D′′ > 0 so at x = 0, D is minimum. y = 0

Therefore, the vertex (0, 0) is the point closest to
the focus.

(40sin 75°)2 = 4 p( p + 40 cos 75°)
p 2 + 40 p cos 75° − 400sin 2 75° = 0
−40 cos 75° ± 1600 cos 2 75° + 1600sin 2 75°
2
= −20 cos 75° ± 20
p=

p = 20 − 20 cos 75° ≈ 14.8 ( p > 0)
The closest point to Earth is (0, 0), so the asteroid
will come within 14.8 million miles of Earth.

36. Let the equation x 2 = 4 py describe the cables.
The cables are attached to the towers at
(±400, 400).
(400)2 = 4 p (400), p = 100
The vertical struts are at x =±300.
(300)2 = 4(100) y, y = 225
The struts must be 225 m long.

37. Let RL be the distance from R to the directrix.

Observe that the distance from the latus rectum to
the directrix is 2p so RG = 2 p – RL . From the
definition of a parabola, RL = FR . Thus,
FR + RG = RL + 2 p – RL = 2 p.

34. Let the y-axis be the axis of the parabola, so

Earth’s coordinates are (0, p) and the equation of

the path is x 2 = 4 py , where the coordinates are
in millions of miles. When the line from Earth to
the asteroid makes an angle of 90D with the axis
of the parabola, the asteroid is at (40, p).
(40)2 = 4 p( p ), p = 20
The closest point to Earth is (0, 0), so the asteroid
will come within 20 million miles of Earth.

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Section 10.1

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by any means, without permission in writing from the publisher.


40. Let C denote the center of the circle and r the
radius. Observe that the distance from a point P
to the circle is PC – r. Let l be the line and

38. Let the coordinates of P be ( x0 , y0 ).
2 yy ′ = 4 p, so y ′ =

2p
. Thus the slope of the
y

PL the distance from the point to the line. Thus,

y0
.
2p
The equation of the normal line is
y
y – y0 = – 0 ( x – x0 ). When y = 0,
2p

normal line at P is –

PC – r = PL . Let l ' be the line parallel to l, r

units away and on the side opposite from the
circle. Then PL′ , the distance from P to l ' , is
PL + r ; so PL = PL′ – r. Therefore,

x = 2 p + x0 , so B is at (2 p + x0 , 0). A is at

PC – r = PL′ – r or PC = PL′ . The set of

( x0 , 0). Thus, AB = 2 p + x0 – x0 = 2 p.

points is a parabola by definition.

39. Let P1 and P2 denote ( x1 , y1 ) and ( x2 , y2 ),

41.

respectively. P1P2 = P1F + P2 F since the focal

dy δ x
=
dx H

δ x2

chord passes through the focus. By definition of a
parabola, P1F = p + x1 and P2 F = p + x2 .

y=

Thus, the length of the chord is
P1P2 = x1 + x2 + 2 p.
L = p + p + 2p = 4p

y(0) = 0 implies that C = 0. y =

42. a.

2H

+C

δ x2

2H
This is an equation for a parabola.

A(T1 ) is the area of the trapezoid formed by

(a, 0), P, Q, (b, 0) minus the area the two trapezoids formed by (a, 0), P, (c, c 2 ) , (c, 0) and by (c, 0), (c, c 2 ) ,
a+b b−c c−a b−a
=
=
,
.
2
2
2
4
b−a 2
c−a 2 2 b−c 2
b−a 2
b−a 2
A(T1 ) =
[a + b 2 ] −
[a + c ] −
[c + b 2 ] =
[a + b2 ] −
[a + 2c 2 + b 2 ]
2
2
2
2
4
2
2

⎤ b–a⎡ 2
b–a 2
b–a⎡ 2
b2 ⎤
⎛ a+b⎞
2
2 a
=
+
[a + b2 ] –
b
a
b
–
–
ab
–
=
⎢a + 2 ⎜
+
⎥
⎢
⎥
⎟
4 ⎢⎣
2
2 ⎥⎦
2
4 ⎢
⎝ 2 ⎠

⎥⎦
⎣

Q, (b, 0). Observe that since c =

=

( b –2a ) + ( b –2a )
3

b.

(c – a )3 (b – c)3
A(T2 ) =
+
=
8
8

c.

Using reasoning similar to part b, A(Tn ) =

d.

Area =

8

8

3

=

(b – a)3 A(T1 )
=
32
4

A(Tn –1 )
A(T1 )
.
, so A(Tn ) =
4
4n –1
∞
⎛ 1 ⎞ 4
A(T1 )
⎟ = A(T1 )
A( S ) = A(T1 ) + A(T2 ) + A(T3 ) +… = ∑
= A(T1 ) ⎜
n –1
⎜1– 1 ⎟ 3
n =1 4
4⎠
⎝

=

596

b – a ⎛ a2
b 2 ⎞ (b – a)3
– ab + ⎟ =
⎜
4 ⎜⎝ 2
2 ⎟⎠
8

b–a 2
(b – a)(a 2 + b 2 ) (b – a )3
[a + b 2 ] – A( S ) =
–
2
2
6

b–a
(b – a)
1
b3 a 3
[3a 2 + 3b 2 – (b 2 – 2ab + a 2 )] =
[2a 2 + 2ab + 2b 2 ] = (b3 – a3 ) =
–
6
6
3
3
3

Section 10.1

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by any means, without permission in writing from the publisher.


43.

Using a CAS, we find that the largest vertical
distance between the catenary and the
parabola is roughly 67 feet.

44. a.

Since the vertex is on the positive y-axis and
the parabola crosses the x-axis, its equation is
of the form: x 2 = 4 p( y − k ) , where k is the
y-coordinate of the vertex; that is
x 2 = 4 p ( y − 630) . Since the point (315, 0)
is on the parabola, we have
−315
(315) 2 = 4 p(0 − 630) or 4 p =
= −157.5
2
Thus the parabola has the equation
x 2 = −157.5( y − 630)

1.

2.

x2
a2

+

y2
b2

=1

x2 y 2
+
=1
9 16

3. foci
4. to the other focus; directly away from the other
focus

b. Solving for y, we get
1
y=−
x 2 + 630
157.5
The catenary for the Gateway Arch is
x

y = 758 − 128cosh
.
128
y = 758 − 128cosh

650

10.2 Concepts Review

Problem Set 10.2
1. Horizontal ellipse
2. Horizontal hyperbola
x
.
128

1
y=−
x 2 + 630
157.5

3. Vertical hyperbola
4. Horizontal hyperbola
5. Vertical parabola
6. Vertical parabola
7. Vertical ellipse

−350

c.

0

350

Because of symmetry, we can focus on the
largest vertical distance between the graphs
for positive x.
Let f ( x ) = yArch − yparabola . That is,
x ⎞
⎛
f ( x ) = ⎜ 758 − 128cosh
⎟
128 ⎠
⎝
1
⎛
⎞
x 2 + 630 ⎟
−⎜−
157.5
⎝
⎠
1
x
x 2 + 128
= −128cosh
+
128 157.5

Instructor’s Resource Manual

8. Horizontal hyperbola
9.

x2 y 2
+
= 1; horizontal ellipse
16 4

a = 4, b = 2, c = 2 3
Vertices: (±4, 0)

(

Foci: ±2 3, 0

)

Section 10.2

597

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by any means, without permission in writing from the publisher.


10.

x2 y 2

–
= 1; horizontal hyperbola
16 4

13.

a = 2 2, b = 2, c = 6

a = 4, b = 2, c = 2 5
Vertices: (±4, 0)

(

Foci: ±2 5, 0

Vertices: (0, ± 2 2)

)

(

Foci: 0, ± 6

1
Asymptotes: y = ± x
2

11.

y 2 x2

–
= 1; vertical hyperbola
4
9

14.

(

)

x2 y 2
+
= 1; horizontal ellipse
25 4

a = 5, b = 2, c = 21
Vertices: (±5, 0)

a = 2, b = 3, c = 13
Vertices: (0, ±2)
Foci: 0, ± 13

x2 y 2
+
= 1; vertical ellipse
2
8

(

Foci: ± 21, 0

)

)

2
Asymptotes: y = ± x
3

15.

x2 y 2
–
= 1; horizontal hyperbola
10 4
a = 10, b = 2, c = 14

12.

x2 y 2
+
= 1; horizontal ellipse
7
4

a = 7, b = 2, c = 3

(

Vertices: ± 7, 0

(

Foci: ± 3, 0

598

)

Section 10.2

)

(

Vertices: ± 10, 0

(

Foci: ± 14, 0

)

)

Asymptotes: y = ±

2

10

x

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by any means, without permission in writing from the publisher.


16.

x2 y 2
–
= 1; horizontal hyperbola
8
2

a = 2 2, b = 2, c = 10

(

Vertices: ±2 2, 0

(

Foci: ± 10, 0

)

)

1
Asymptotes: y = ± x
2

22. This is a horizontal hyperbola with a = 4 and
c = 5.
b = 25 –16 = 3
x2 y 2
–
=1
16 9

23. This is a vertical hyperbola with a = 4 and c = 5.
b = 25 –16 = 3
y 2 x2
–
=1
16 9

24. This is a vertical hyperbola with a = 3.
81
45
⎛3⎞ 9
c = ae = 3 ⎜ ⎟ = , b =
−9 =
4
2
⎝2⎠ 2

y 2 x2
−
=1
45
9
4

17. This is a horizontal ellipse with a = 6 and c = 3.
b = 36 − 9 = 27
x2 y 2
+
=1
36 27

18. This is a horizontal ellipse with c = 6.
c 6
a = = = 9, b = 81 − 36 = 45
e 2
3

2

2

x
y
+
=1
81 45

19. This is a vertical ellipse with c = 5.
c 5
a = = = 15, b = 225 – 25 = 200
e 1
3
2

x2
y
+
=1
200 225

20. This is a vertical ellipse with b = 4 and c = 3.
a = 16 + 9 = 5
x2 y 2
+
=1
16 25

21. This is a horizontal ellipse with a = 5.
x2 y 2
+
=1
25 b 2
4
9
+
=1
25 b 2

225
b2 =
21
2
x
y2
+
=1
25 225

25. This is a horizontal hyperbola with a = 8.
b 1
1
The asymptotes are y = ± x, so = or b = 4.
8 2
2
x2 y 2
–
=1
64 16

26. c = ae =
y2
a2

16
a

2

−

−

x2
1 a2
2

4
1 a2
2

6
3
1
a, b 2 = c 2 – a 2 = a 2 – a 2 = a 2
2
2
2
=1

=1

a2 = 8
y 2 x2
–
=1
8
4

27. This is a horizontal ellipse with c = 2.
a
a
8 = , 8 = , so a 2 = 8c = 16 .
c
e
a

b = 16 − 4 = 12
x2 y 2
+
=1
16 12

28. This is a horizontal hyperbola with c = 4.
a
a
1 = , 1 = , so a 2 = c = 4.
c
e
a

b = 16 − 4 = 12
x2 y 2
−
=1
4 12

21

Instructor’s Resource Manual

Section 10.2

599

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by any means, without permission in writing from the publisher.


1
29. The asymptotes are y = ± x. If the hyperbola is
2
b 1
horizontal = or a = 2b. If the hyperbola is
a 2
a 1
vertical, = or b = 2a.
b 2
Suppose the hyperbola is horizontal.
x2
y2
–
=1
4b 2 b 2
16
9
–
=1
2

4b
b2

b 2 = –5
This is not possible.
Suppose the hyperbola is vertical.
y 2 x2
–
=1
a 2 4a 2
9
16
–
=1
2
a
4a 2
a2 = 5
y 2 x2
–
=1
5 20

30.

x2
a2
25

+

y2
b2
1

+

equation is

x2 y 2
+
=1
49 33

33. This is an hyperbola whose foci are (7, 0) and
(- 7, 0) and whose axis is the x-axis. So the
x2 y 2
equation has the form
−
= 1 . Since
a 2 b2
2a = 12, a 2 = 36 and b 2 = c 2 − a 2 = (7 2 ) − 36 = 13

Thus the equation is

x2 y 2
−
=1
36 13

34. This is an hyperbola whose foci are (0, 6) and
(0, - 6) and whose axis is the y-axis. So the
y 2 x2
−
= 1 . Since
equation has the form
a 2 b2

=1

Thus the equation is

2

a 2 = 28
28
b2 =
3

35. Use implicit differentiation to find the slope:
2
2
x + y y ′ = 0 . At the point
27
9
2 2 6
6
+
y ′ = 0, or y ′ = −
so the

9
9
6
equation of the tangent line is
6
( y − 6) = −
( x − 3) or x + 6 y = 9 .
6

3

31. This is an ellipse whose foci are (0, 9) and (0, - 9)
and whose major diameter has length 2a = 26.
Since the foci are on the y-axis, it is the major axis
of the ellipse so the equation has the form
y 2 x2
+
= 1 . Since 2a = 26, a 2 = 169 and since
a 2 b2
a 2 = b 2 + c 2 , b 2 = 169 − (9)2 = 88 . Thus the

equation is

y 2 x2
−
=1
25 11

(3, 6),

x2 y 2
+
=1
28 28

600

a 2 = b 2 + c 2 , b 2 = 49 − (4)2 = 33 . Thus the

2a = 10, a 2 = 25 and b 2 = c 2 − a 2 = (62 ) − 25 = 11

⎫
= 1⎪
⎪ 84
a
b
⎬ – 2 = –3
16 4
+
= 1⎪ a
⎪⎭
a 2 b2
2

32. This is an ellipse whose foci are (4, 0) and (- 4, 0)
and whose major diameter has length 2a = 14.
Since the foci are on the x-axis, it is the major axis
of the ellipse so the equation has the form
x2 y2
+

= 1 . Since 2a = 14, a 2 = 49 and since
a 2 b2

36. Use implicit differentiation to find the slope:
1
y
x + y ′ = 0 . At the point
12
8
2 1
− y ′ = 0, or y ′ = 2 so the
4 4
equation of the tangent line is
( y + 2) = 2( x − 3 2) or 2 x − y = 8 .
(3 2, −2),

y 2 x2
+
=1
169 88

Section 10.2

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by any means, without permission in writing from the publisher.


37. Use implicit differentiation to find the slope:

2
2
x + y y ′ = 0 . At the point
27
9
2 2 6
6
−
y ′ = 0, or y ′ =
so the
9
9
6
equation of the tangent line is
6
( y + 6) = −
( x − 3) or x − 6 y = 9 .
6
(3, − 6),

38. Use implicit differentiation to find the slope:
1
x − y y ′ = 0 . At the point ( 3, 2)
2
2
3−
y ′ = 0, or y ′ = 6 so the equation of the
2
tangent line is ( y − 2) = 6( x − 3) or
6x − 6 y = 4 3 .

39. Use implicit differentiation to find the slope:
2 x + 2 y y ′ = 0 . At the point (5,12)
5
so the equation of the
12
5
tangent line is ( y − 12) = − ( x − 5) or
12
5 x + 12 y = 169 .
10 + 24 y ′ = 0, or y ′ = −

40. Use implicit differentiation to find the slope:
2 x − 2 y y ′ = 0 . At the point ( 2, 3)
2 2 − 2 3 y ′ = 0, or y ′ =

6
so the equation of
3

6
the tangent line is ( y − 3) =
( x − 2) or
3
3y − 6 x = 3 .

41. Use implicit differentiation to find the slope:
1
2
x+

y y ′ = 0 . At the point
44
169
2
(0,13),
y ′ = 0, or y ′ = 0 . The tangent line is
13
horizontal and thus has equation y = 13 .
42. Use implicit differentiation to find the slope:
2
2
x+
y y ′ = 0 . At the point (7, 0)
49
33
2
+ 0 y ′ = 0, or y ′ is undefined. . The tangent line
7
is vertical and thus has equation x = 7 .

Instructor’s Resource Manual

43. Let the y-axis run through the center of the arch
and the x-axis lie on the floor. Thus a = 5 and
x2 y 2
b = 4 and the equation of the arch is
+
= 1.
25 16
x 2 (2)2

5 3
+
= 1, so x = ±
.
25 16
2
The width of the box can at most be
5 3 ≈ 8.66 ft.

When y = 2,

44. Let the y-axis run through the center of the arch
and the x-axis lie on the floor.
x2 y 2
The equation of the arch is
+
= 1.
25 16
(2) 2 y 2
4 21
+
= 1, so y = ±
.
25 16
5
The height at a distance of 2 feet to the right of the
4 21
center is
≈ 3.67 ft.
5

When x = 2,

45. The foci are at (±c, 0).
c = a 2 – b2

a 2 – b2
a2
y2 =

b4
a2

+

y2
b2

=1

,y=±

b2
a

Thus, the length of the latus rectum is

2b 2
.
a

46. The foci are at (±c, 0)
c = a 2 + b2
a 2 + b2
a2
y2 =

b4
a2

–

y2
b2

=1

,y=±

b2
a

Thus, the length of the latus rectum is

2b 2
.
a

47. a = 18.09, b = 4.56,
c = (18.09) 2 − (4.56) 2 ≈ 17.51

The comet’s minimum distance from the sun is
18.09 – 17.51 ≈ 0.58 AU.

48. a − c = 0.13, c = ae, a (1 − e) = 0.13,
0.13
≈ 1733
1 − 0.999925
a + c = a (1 + e) ≈ 1733(1 + 0.999925) ≈ 3466 AU
a=

Section 10.2

601

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by any means, without permission in writing from the publisher.


49. a – c = 4132; a + c = 4583
2a = 8715; a = 4357.5
c = 4357.5 – 4132 = 225.5
225.5
c
e= =
≈ 0.05175
a 4357.5

54. The slope of the line is

y′ = −

b = a 2 − c 2 = 1462.0415 ≈ 38.24 . So the
major diameter = 2a = 78.96 and the minor
diameter = 2b = 76.48 .
x2 y 2
+
=1
4
9
Equation of tangent line at ( x0 , y0 ) :

3
At (0, 6), y0 = .
2
3 x2 1
When y = ,
+ = 1, x = ± 3.
2 4 4
3⎞
⎛
The points of tangency are ⎜ 3, ⎟ and
2⎠
⎝
3
⎛
⎞
⎜ − 3, ⎟
2⎠
⎝
2

x
y
52.
–
=1
4 36
Equation of tangent line at ( x0 , y0 ) :

Substitute x = − 2 y into the equation of the
ellipse.
1
2 y 2 + 2 y 2 − 2 = 0; y = ±
2
1 ⎞
⎛
The tangent lines are tangent at ⎜ −1,
⎟ and
2⎠
⎝
1 ⎞
⎛
⎜1, −
⎟ . The equations of the tangent lines are
2⎠
⎝
1
1
1
1

y−
( x + 1) and y +
( x − 1) or
=
=
2
2
2
2

A = 4b ∫

a

A = 4b ∫

a

a2
1−

x2

dx
a2
Let x = a sin t then dx = a cos t d t. Then the limits
π
are 0 and .
2
0

1−

x2
a

π/ 2

0

2

dx = 4ab ∫

π/2

0

cos 2 t dt
π/2

⎡ sin 2t ⎤
(1 + cos 2t )dt = 2ab ⎢t +
2 ⎥⎦ 0
⎣

56. x = ± a 1 −

(

)

The points of tangency are 2 2, – 6 and

V = 2 ⋅ π∫

=

53. 2 x 2 − 7 y 2 − 35 = 0; 4 x − 14 yy ′ = 0

b

0

−2 2, – 6 .

y′ =

x2

= πab

36
x
–
= 1, x = ±2 2.
4 36

)

55. y = ±b 1 −

= 2ab ∫

2

(

x 1
x
;
= − ; x = − 2y
2y 2
2y

0

xx0 yy0
–
=1
4
36
At (0, 6), y0 = –6 .
When y = –6,

.

x − 2 y + 2 = 0 and x − 2 y − 2 = 0 .

xx0 yy0

+
=1
4
9

2

2

x 2 + 2 y 2 − 2 = 0; 2 x + 4 yy ′ = 0

50. (See Example 5) Since a + c = 49.31 and
a − c = 29.65 , we conclude that
2a = 78.96, 2c = 19.66 and so
a = 39.48, c = 9.83 . Thus

51.

1

y2
b2

⎛
y2
a ⎜1 −
⎜ b2
⎝
2

b

⎞
⎡
y3 ⎤
⎟ dy = 2πa 2 ⎢ y − 2 ⎥
⎟
3b ⎦⎥ 0
⎠
⎣⎢

4πa 2 b
3

2x 2 2x
7y
;− =
;x=−
7y 3 7y
3

Substitute x = −

7y
into the equation of the
3

hyperbola.
98 2
y − 7 y 2 − 35 = 0, y = ±3

9
The coordinates of the points of tangency are
( −7, 3) and ( 7, − 3) .
602

Section 10.2

Instructor’s Resource Manual

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by any means, without permission in writing from the publisher.


60.

x2

57. y = ±b

−1

a2

The vertical line at one focus is x = a 2 + b 2 .
2

V = π∫

a 2 +b 2

a

⎛ x2
⎞
⎜b
– 1 ⎟ dx
⎜ a2
⎟
⎝
⎠

⎛ x2
⎞
⎡ x3
⎤
= b π∫
⎜ 2 – 1⎟ dx = b 2 π ⎢ 2 − x ⎥
⎜a
⎟
a
⎥⎦ a
⎝
⎠
⎣⎢ 3a
2
2
3
/
2
⎡ (a + b )

2 ⎤
– a 2 + b2 + a ⎥
= b2 π ⎢
2
3 ⎦⎥
3a
⎣⎢
a 2 +b 2

2

=

a 2 +b2

πb 2 ⎡ 2
(a + b 2 )3 / 2 – 3a 2 a 2 + b 2 + 2a3 ⎤
2 ⎢⎣
⎥⎦
3a

58. y = ±b 1 –

⎛ bx – ay0 ⎞
x⎜ 0
⎟ =1
⎝ a 2b ⎠

a2
2

x 2 ⎞⎟
V = 2 ⋅ π∫ ⎜ b 1 –
dx
2 ⎟
0⎜
a
⎝
⎠
a

⎡
a⎛
4
x2 ⎞
x3 ⎤
2
dx = 2πb 2 ⎢ x −
= 2πb 2 ∫ ⎜ 1 –
⎟
⎥ = πab
2⎟
2
0⎜
3
⎝ a ⎠
⎣⎢ 3a ⎦⎥ 0

y2
b2

y

2

b

2

= 4a y 2 −

y

4

b 2 x02 – a 2 y0 2

= x0

⎞
a 2 b 2 y0
= y0
⎟= 2 2
⎟ b x – a2 y 2
⎠
0
0
Thus, the point of contact is midway between the
two points of intersection.

b

b2

a 2b 2 x0

⎞
⎟
⎟
⎠

1 ⎛ ab 2
ab 2
–
⎜
2 ⎜⎝ bx0 – ay0 bx0 + ay0

b2

4 y3
2a ⎛⎜ 2 y − 2 ⎞⎟
b ⎠ dA
dA
;
= ⎝
= 0 when
dy
dy
y4
y2 − 2

2y

Observe that b 2 x0 – a 2 y0 = a 2 b 2 .

=

A = 4 xy = 4 ya 1 −

y−

a 2b
a 2b
x=
bx0 – ay0
bx0 + ay0
Thus the tangent line intersects the asymptotes at
⎛ a 2b
ab 2 ⎞
,
⎜
⎟ and
⎜ bx0 – ay0 bx0 – ay0 ⎟
⎝
⎠
2
⎛ a 2b
⎞
ab
,–
⎜

⎟.
⎜ bx0 + ay0
bx0 + ay0 ⎟⎠
⎝
1 ⎛ a 2b
a 2b
+
⎜
2 ⎜⎝ bx0 – ay0 bx0 + ay0

59. If one corner of the rectangle is at (x, y) the sides
have length 2x and 2y.

3

⎛ bx + ay0 ⎞
x⎜ 0
⎟ =1
⎝ a 2b ⎠

x=

x2

a⎛

x = ±a 1 −

Position the x-axis on the axis of the hyperbola
x2 y2

–
= 1 describes the
such that the equation
a 2 b2
hyperbola. The equation of the tangent line at
x x y y
( x0 , y0 ) is 0 – 0 = 1. The equations of the
a2
b2
b
asymptotes are y = ± x .
a
b
b
Substitute y = x and y = – x into the
a
a
equation of the tangent line.
x0 x y0 x
x0 x y0 x
–
=1
+
=1
2
ab
ab
a
a2

61. Add the two equations to get 9 y 2 = 675.

=0

y = ±5 3

Substitute y = 5 3 into either of the two
equations and solve for x ⇒ x = ±6

⎛ 2 y2
y ⎜1 −
⎜
b2
⎝

⎞
⎟=0
⎟
⎠
b
y = 0 or y = ±
2

(

)

The point in the first quadrant is 6, 5 3 .

The Second Derivative Test shows that y =

b
2

is

a maximum.
x = a 1−

( )
b
2
2

b

2

=

a
2

Therefore, the rectangle is a 2 by b 2 .
Instructor’s Resource Manual

Section 10.2

603

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by any means, without permission in writing from the publisher.


62. Substitute x = 6 – 2y into x 2 + 4 y 2 = 20.

68.

(6 – 2 y )2 + 4 y 2 = 20
8 y 2 – 24 y + 16 = 0
y2 – 3y + 2 = 0
(y – 1)(y – 2) = 0
y = 1 or y = 2
x = 4 or x = 2
The points of intersection are (4, 1) and (2, 2).

AP
u

AB

=

v

BP

+

u

2uc
AP – BP =
v
Thus the curve is the right branch of the horizontal

hyperbola with a =

uc
, so b =
v

⎛ u2
⎜1 – 2
⎜ v
⎝

⎞
⎟c.
⎟
⎠

The equation of the curve is
x2
y2
uc ⎞
⎛
–
= 1⎜ x ≥ ⎟.
2
2

u 2c 2
v ⎠
⎝
1– u c

63.

(

v2

v2

)

69. Let P(x, y) be the location of the explosion.
3 AP = 3 BP + 12
AP – BP = 4

Thus, P lies on the right branch of the horizontal
hyperbola with a = 2 and c = 8, so b = 2 15.
x2 y 2
–
=1
4 60
Since BP = CP , the y-coordinate of P is 5.

64. If the original path is not along the major axis, the
ultimate path will approach the major axis.

x 2 25
17
–
= 1, x = ±
4 60
3
⎛ 17 ⎞
P is at ⎜⎜
, 5 ⎟⎟ .
⎝ 3 ⎠

70.
65. Written response. Possible answer: the ball will
follow a path that does not go between the foci.
66. Consider the following figure.

lim ⎜⎛ x 2 − a 2 − x ⎟⎞
⎠
⎡⎛ 2
2
2
⎞ ⎛ 2
⎞⎤
⎢ ⎜⎝ x − a − x ⎟⎠ ⎜⎝ x − a + x ⎟⎠ ⎥
= lim ⎢
⋅
⎥
1
x →∞ ⎢
⎛ x2 − a2 + x ⎞ ⎥

⎜
⎟
⎢⎣
⎝
⎠ ⎥⎦
x →∞ ⎝

−a 2

= lim

x →∞

x2 − a2 + x

=0

71. 2a = p + q, 2c = p – q
b2 = a 2 – c2 =
b=

Observe that 2(α + β) = 180°, so α + β = 90°. The
ellipse and hyperbola meet at right angles.
67. Possible answer: Attach one end of a string to F
and attach one end of another string to F ′ . Place a
spool at a vertex. Tightly wrap both strings in the
same direction around the spool. Insert a pencil
through the spool. Then trace out a branch of the
hyperbola by unspooling the strings while keeping
both strings taut.

604

Section 10.2

( p + q)2 ( p – q)2
–
= pq
4
4

pq

72. x = a cos t , y = a sin t − b sin t = (a − b) sin t

cos t =
x2

+

x
y
, sin t =
a
a −b
y2

=1
a 2 ( a − b) 2
Thus the coordinates of R at time t lie on an

ellipse.

Instructor’s Resource Manual

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by any means, without permission in writing from the publisher.


73. Let (x, y) be the coordinates of P as the ladder
slides. Using a property of similar triangles,

75. The equations of the hyperbolas are

b2 − y 2
x
.
=
a
b

and

y2
b2

–

x2
a2

Square both sides to get
x2 b2 − y 2
=
or b 2 x 2 + a 2 y 2 = a 2b 2 or
a2
b2
x2 y2
+
= 1.
a 2 b2

x=

y2
b2

=1

c
=
a

a 2 + b2
a

E=

c
=
b

a 2 + b2
b
a2
a 2 + b2

+

b2
a 2 + b2

=1

76. Position the x-axis on the plane so that it makes
the angle φ with the axis of the cylinder and the
y-axis is perpendicular to the axis of the cylinder.
(See the figure.)

74. Place the x-axis on the axis of the hyperbola such
x2 y2
= 1. One focus is at
that the equation is
–
a 2 b2
b
(c , 0) and the asymptotes are y = ± x. The
a
equations of the lines through the focus,
perpendicular to the asymptotes, are
a

y = ± ( x – c). Then solve for x in
b
b
a
x = – ( x – c).
a
b
a 2 + b2
ac
x=
ab
b

a2

–

= 1.

e=

e –2 + E –2 =

x2

If P(x, y) is a point on C, ( x sin φ )2 + y 2 = r 2
where r is the radius of the cylinder. Then
x2
y2
+

= 1.
2
r2
r
2
sin φ

77.

a2c
a 2 + b2

Since c 2 = a 2 + b 2 , x =

a2
. The equation of the
c

a2
, so the line
directrix nearest the focus is x =
c
through a focus and perpendicular to an asymptote
intersects that asymptote on the directrix nearest
the focus.

Instructor’s Resource Manual

When a < 0, the conic is an ellipse. When a > 0,
the conic is a hyperbola. When a = 0, the graph is

two parallel lines.

Section 10.2

605

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by any means, without permission in writing from the publisher.


7. y 2 – 5 x – 4 y – 6 = 0

10.3 Concepts Review
1.

( y 2 – 4 y + 4) = 5 x + 6 + 4
( y – 2) 2 = 5( x + 2)
This is a parabola.

a2
4

8. 4 x 2 + 4 y 2 + 8 x – 28 y –11 = 0

2. 14; ellipse

49 ⎞
⎛
4( x 2 + 2 x + 1) + 4 ⎜ y 2 – 7 y + ⎟ = 11 + 4 + 49
4 ⎠

⎝

A−C
3.
B

2

7⎞
⎛
4( x + 1)2 + 4 ⎜ y – ⎟ = 64
2⎠
⎝
This is a circle.

4. rotation, translation

9. 3x 2 + 3 y 2 – 6 x + 12 y + 60 = 0

Problem Set 10.3
2

2

1. x + y – 2 x + 2 y + 1 = 0
( x 2 – 2 x + 1) + ( y 2 + 2 y + 1) = –1 + 1 + 1
( x – 1)2 + ( y + 1)2 = 1
This is a circle.

( x 2 + 6 x + 9) + ( y 2 – 2 y + 1) = –6 + 9 + 1

( x + 3) 2 + ( y – 1) 2 = 4
This is a circle.

3. 9 x 2 + 4 y 2 + 72 x –16 y + 124 = 0
9( x 2 + 8 x + 16) + 4( y 2 – 4 y + 4) = –124 + 144 + 16
9( x + 4) 2 + 4( y – 2)2 = 36
This is an ellipse.

4. 16 x 2 − 9 y 2 + 192 x + 90 y − 495 = 0
16( x 2 + 12 x + 36) − 9( y 2 − 10 y + 25)
= 495 + 576 − 225
16( x + 6) 2 − 9( y − 5) 2 = 846
This is a hyperbola.
2

5. 9 x + 4 y + 72 x –16 y + 160 = 0
9( x 2 + 8 x + 16) + 4( y 2 – 4 y + 4) = –160 + 144 + 16
9( x + 4)2 + 4( y – 2)2 = 0
This is a point.

6. 16 x 2 + 9 y 2 + 192 x + 90 y + 1000 = 0
2

2

16( x + 12 x + 36) + 9( y + 10 y + 25)
= –1000 + 576 + 225
16( x + 6) 2 + 9( y + 5)2 = –199
This is the empty set.

606

Section 10.3

3( x – 1) 2 + 3( y + 2) 2 = –45
This is the empty set.

10. 4 x 2 – 4 y 2 – 2 x + 2 y + 1 = 0

2. x 2 + y 2 + 6 x – 2 y + 6 = 0

2

3( x 2 – 2 x + 1) + 3( y 2 + 4 y + 4) = –60 + 3 + 12

1
1⎞
1
1⎞
1 1
⎛
⎛
4 ⎜ x 2 – x + ⎟ – 4 ⎜ y 2 – y + ⎟ = –1 + –
2
16 ⎠
2
16 ⎠
4 4
⎝
⎝

2

2

2

2

1⎞
1⎞
⎛
⎛
4 ⎜ x – ⎟ – 4 ⎜ y – ⎟ = −1
4⎠
4⎠
⎝
⎝
1⎞
1⎞
⎛
⎛
4⎜ y – ⎟ – 4⎜ x – ⎟ = 1
4⎠
4⎠
⎝
⎝
This is a hyperbola.

11. 4 x 2 – 4 y 2 + 8 x + 12 y – 5 = 0
9⎞

⎛
4( x 2 + 2 x + 1) – 4 ⎜ y 2 – 3 y + ⎟ = 5 + 4 – 9
4⎠
⎝
2

3⎞
⎛
4( x + 1)2 – 4 ⎜ y – ⎟ = 0
2⎠
⎝
This is two intersecting lines.

12. 4 x 2 – 4 y 2 + 8 x + 12 y – 6 = 0
9⎞
⎛
4( x 2 + 2 x + 1) – 4 ⎜ y 2 – 3 y + ⎟ = 6 + 4 – 9
4⎠
⎝
2

3⎞
⎛
4( x + 1) 2 – 4 ⎜ y – ⎟ = 1
2⎠
⎝
This is a hyperbola.

13. 4 x 2 – 24 x + 36 = 0
4( x 2 – 6 x + 9) = –36 + 36

4( x – 3)2 = 0
This is a line.

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14. 4 x 2 – 24 x + 35 = 0

19. ( x + 2)2 = 8( y − 1)

4( x 2 – 6 x + 9) = –35 + 36
4( x – 3) 2 = 1
This is two parallel lines.

15.

( x + 3) 2 ( y + 2)2
+
=1
4
16

20. ( x + 2)2 = 4
x + 2 = ±2
x = –4, x = 0

16. ( x + 3) 2 + ( y − 4) 2 = 25

21. ( y − 1) 2 = 16
y – 1 = ±4
y = 5, y = –3

17.

( x + 3) 2 ( y + 2)2
−
=1
4
16

22.
18. 4( x + 3) = ( y + 2) 2

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( x + 3)2 ( y − 2)2
+
=0
4
8
(–3, 2)

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23. x 2 + 4 y 2 − 2 x + 16 y + 1 = 0

26. x 2 − 4 y 2 − 14 x − 32 y − 11 = 0

( x 2 − 2 x + 1) + 4( y 2 + 4 y + 4) = −1 + 1 + 16

( x 2 − 14 x + 49) − 4( y 2 + 8 y + 16) = 11 + 49 − 64

( x − 1)2 + 4( y + 2) 2 = 16

4( y + 4)2 − ( x − 7) 2 = 4

( x − 1) 2 ( y + 2)2
+
=1
16
4

( y + 4) 2 −

24. 25 x 2 + 9 y 2 + 150 x − 18 y + 9 = 0

( x − 7) 2
=1
4

27. 4 x 2 + 16 x − 16 y + 32 = 0

25( x 2 + 6 x + 9) + 9( y 2 − 2 y + 1) = −9 + 225 + 9

4( x 2 + 4 x + 4) = 16 y − 32 + 16

25( x + 3) 2 + 9( y − 1) 2 = 225

4( x + 2)2 = 16( y − 1)

( x + 3) 2 ( y − 1)2
+
=1
9
25

( x + 2)2 = 4( y − 1)

2

28. x 2 − 4 x + 8 y = 0

2

25. 9 x − 16 y + 54 x + 64 y − 127 = 0
2

2

9( x + 6 x + 9) − 16( y − 4 y + 4) = 127 + 81 − 64
2

2

9( x + 3) − 16( y − 2) = 144

x 2 − 4 x + 4 = −8 y + 4
1⎞
⎛
( x − 2) 2 = −8 ⎜ y − ⎟
2⎠
⎝

( x + 3) 2 ( y − 2) 2
−
=1
16
9

608

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29. 2 y 2 − 4 y − 10 x = 0
2( y 2 − 2 y + 1) = 10 x + 2

1⎞
⎛
( y − 1) 2 = 5 ⎜ x + ⎟
5⎠
⎝
5
1⎞
⎛ ⎞⎛
( y − 1) 2 = 4 ⎜ ⎟⎜ x + ⎟
5⎠
⎝ 4 ⎠⎝

Horizontal parabola, p =

5
4

⎛ 1 ⎞
⎛ 21 ⎞
Vertex ⎜ − , 1⎟ ; Focus is at ⎜ , 1⎟ and
⎝ 5 ⎠
⎝ 20 ⎠

directrix is at x = − 29 .

20

30. −9 x 2 + 18 x + 4 y 2 + 24 y = 9
−9( x 2 − 2 x + 1) + 4( y 2 + 6 y + 9) = 9 − 9 + 36

4( y + 3)2 − 9( x − 1)2 = 36
2

2

( y + 3)
( x − 1)
−
=1
9
4
a 2 = 9, a = 3
The distance between the vertices is 2a = 6.

31. 16( x − 1)2 + 25( y + 2)2 = 400
( x − 1) 2 ( y + 2) 2
+
=1
25
16
Horizontal ellipse, center (1, –2), a = 5, b = 4,
c = 25 − 16 = 3
Foci are at (–2, –2) and (4, –2).

32. x 2 − 6 x + 4 y + 3 = 0
x 2 − 6 x + 9 = −4 y − 3 + 9
3⎞
⎛
( x − 3) 2 = −4 ⎜ y − ⎟
2⎠

⎝
Vertical parabola, opens downward, vertex
⎛ 3⎞
⎜ 3, ⎟ , p = 1
⎝ 2⎠
5
⎛ 1⎞
Focus is at ⎜ 3, ⎟ and directrix is y = .
2
⎝ 2⎠

33. a = 5, b = 4
( x − 5) 2 ( y − 1)2
+
=1
25
16
34. Horizontal hyperbola, a = 2, c = 3,
b = 9−4 = 5
( x − 2)2 ( y + 1)2
−
=1
4
5

35. Vertical parabola, opens upward, p = 5 – 3 = 2
2

( x − 2) = 4(2)( y − 3)
( x − 2) 2 = 8( y − 3)

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36. An equation for the ellipse can be written in the
( x − 2) 2 ( y − 3) 2
form
+
=1.
a2
b2
Substitute the points into the equation.
16
4
= 1,
=1
2
a
b2
Therefore, a = 4 and b = 2.
( x − 2) 2 ( y − 3) 2
+
=1
16
4
37. Vertical hyperbola, center (0, 3), 2a = 6, a = 3,
c = 5, b = 25 − 9 = 4
( y − 3) 2 x 2
−
=1
9

16

38. Vertical ellipse; center (2, 6), a = 8, c = 6,
b = 64 − 36 = 28
( x − 2) 2 ( y − 6) 2
+
=1
28
64

39. Horizontal parabola, opens to the left
10 − 2
Vertex (6, 5), p =
=4
2
( y − 5) 2 = −4(4)( x − 6)
( y − 5)2 = −16( x − 6)

40. Vertical parabola, opens downward, p = 1
( x − 2) 2 = −4( y − 6)

41. Horizontal ellipse, center (0, 2), c = 2
Since it passes through the origin and center is at
(0, 2), b = 2.
a = 4+4 = 8
x 2 ( y − 2) 2
+
=1
8
4

42. Vertical hyperbola, center (0, 2), c = 2,
b2 = 4 − a 2
An equation for the hyperbola can be written in
( y − 2)2
x2
the form
−
=1.
a2
4 − a2
Substitute (12, 9) into the equation.
49
144
−
=1
2
a
4 − a2
49(4 − a 2 ) − 144a 2 = a 2 (4 − a 2 )
a 4 − 197a 2 + 196 = 0
(a 2 − 196)(a 2 − 1) = 0
a 2 = 196, a 2 = 1

Since a < c, a = 1, b = 4 − 1 = 3
( y − 2) 2 −

x2
=1
3

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609

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43. x 2 + xy + y 2 = 6
cot 2θ = 0
π
π
2θ = , θ =
2
4
2
2
2
x=u
–v
=
(u – v )
2
2
2
2
2
2
(u + v)

y=u
+v
=
2
2
2
1
1
1
(u – v) 2 + (u – v )(u + v) + (u + v) 2 = 6
2
2
2
3 2 1 2
u + v =6
2
2
2
2
u
v
+
=1
4 12

45. 4 x 2 + xy + 4 y 2 = 56

cot 2θ = 0, 2θ =

π

π
,θ =
2
4

2
(u – v)
2
2
y=
(u + v)
2
1
2(u – v)2 + (u – v)(u + v) + 2(u + v)2 = 56
2
9 2 7 2
u + v = 56
2
2
2
u
v2
+
=1
112 16
x=

9

44. 3x 2 + 10 xy + 3 y 2 + 10 = 0

cot 2θ = 0, 2θ =

π
π
,θ =
2
4

2
x=
(u – v)
2
2
y=
(u + v)
2
3
3
(u – v) 2 + 5(u – v )(u + v) + (u + v) 2 + 10 = 0
2
2
8u 2 – 2v 2 = –10
2

2

v
u
–

=1
5
5
4

46. 4 xy – 3 y 2 = 64
3
, r=5
4
3
cos 2θ =
5

cot 2θ =

cos θ =
sin θ =
x=
y=

1
5
1
5

1 + 53
2
1–

3

5

2

=
=

2
5
1
5

(2u – v)
(u + 2v)

4
3
(2u – v)(u + 2v) – (u + 2v)2 = 64
5
5

u 2 – 4v 2 = 64
u 2 v2
–
=1
64 16

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by any means, without permission in writing from the publisher.


47.

1
1
– x 2 + 7 xy – y 2 – 6 2 x – 6 2 y = 0
2
2
π
π
cot 2θ = 0, 2θ = , θ =
2
4
2
2
(u – v) ; y =
(u + v)
x=
2
2
1
7
1
– (u – v) 2 + (u – v)(u + v) – (u + v) 2 – 6(u – v) − 6(u + v) = 0

4
2
4
3u 2 – 4v 2 –12u = 0
3(u 2 – 4u + 4) – 4v 2 = 12
(u – 2)2 v 2
=1
–
4
3

48.

3 2
3
x + xy + y 2 + 2 x + 2 y = 13
2
2
π
π
cot 2θ = 0, 2θ = , θ =
2
4
2
2
x=
(u – v) ; y =
(u + v)
2
2

3
1
3
(u – v) 2 + (u – v)(u + v) + (u + v) 2 + (u – v) + (u + v) = 13
4
2
4
2u 2 + v 2 + 2u = 13
1⎞
1
⎛
2 ⎜ u 2 + u + ⎟ + v 2 = 13 +
4⎠
2
⎝
2

1⎞
27
⎛
2 ⎜ u + ⎟ + v2 =
2⎠
2
⎝

( u + 12 )

2

27

4

v2
27
2

=1

A = 4, B = −3, C = D = E = 0, F = −18

y

4−0
4
=−
−3
3
Since 0 ≤ 2θ ≤ π , sin 2θ is positive, so cos 2θ is negative; using a 3-4-5
4
right triangle, we conclude cos 2θ = − . Thus
5

u

49.

+

cot 2θ =

sin θ =

1 − cos 2θ
=
2

1 − (− 4 5) 3 10
=
and
2
10

cos θ =

1 + cos 2θ
=
2

1 + (− 4 5)
10
=
. Rotating through the angle
2
10

1
2

v

5

5

−5

x

−5

θ = cos −1 (−0.8) = 71.6 , we have

4

(

10
10

u−

3 10
10

) − 3(
2

v

10

10

u−

3 10
10

v

)(

3 10
10

u+

10
10

)

v = 18 or

v2 u 2
−
= 1. This is a hyperbola in standard position
4 36
in the uv-system; its axis is the v-axis, and a = 2, b = 6 .

45v 2 − 5u 2 = 180 or

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by any means, without permission in writing from the publisher.


A = 11, B = 96, C = 39, D = 240, E = 570, F = 875

1 − cos 2θ
=
2

Thus sin θ =

1 − (− 7 25) 4
= and
2
5

u

11 − 39
7
=−
96

24
Since 0 ≤ 2θ ≤ π , cos 2θ is negative; using a 7-24-25 right triangle, we
7
conclude cos 2θ = − .
25
cot 2θ =

y

v

50.

x

1 + cos 2θ
=
2

1 + (− 7 25) 3
= .
2
5
1
Rotating through the angle θ = cos −1 (−0.28) = 53.13 , we have
2
cos θ =

( u − v ) + 96 ( u − v )( u + v ) +
39 ( u + v ) + 240 ( u − v ) + 570 ( u + v ) = −875

11

3
5

4
5

4
5

3
5

2

3
5

2

4
5

3
5

4
5

4
5

3
5

4
5

3
5

or
3u 2 − v 2 + 24u + 6v = −35
3(u 2 +8u+16) − (v 2 - 6v + 9) = −35 + 48 − 9
3(u + 4) 2 − (v − 3) 2 = 4
(u + 4)2
4

−

3

(v − 3)2
=1
4

This is a hyperbola in standard position in the uv-system; its axis is the u2 3
, b=2.

axis, its center is (u, v) = (−4,3) and a =
3
51.

34 x 2 + 24 xy + 41y 2 + 250 y = –325
7
, r = 25
24
7
cos 2θ = –
25
cot 2θ = –

7
1 + 25
3
4
=
; sin θ =
2
5
2
5
1
1
x = (3u – 4v) ; y = (4u + 3v)
5
5
34
24

41
(3u – 4v) 2 + (3u – 4v)(4u + 3v) + (4u + 3v) 2 + 50(4u + 3v) = –325
25
25
25

cos θ =

7
1 – 25

=

50u 2 + 25v 2 + 200u + 150v = –325
2u 2 + v 2 + 8u + 6v = –13
2(u 2 + 4u + 4) + (v 2 + 6v + 9) = –13 + 8 + 9
2(u + 2)2 + (v + 3)2 = 4
(u + 2)2 (v + 3) 2
+
=1
2
4
This is an ellipse in standard position in the uv-system, with major axis
parallel to the v-axis. Its center is ( u , v ) = ( −2, −3) and a = 2 , b = 2 .

612

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52.

16 x 2 + 24 xy + 9 y 2 – 20 x –15 y –150 = 0
7
, r = 25
24
7
cos 2θ =
25
cot 2θ =

7
1 + 25

cos θ =

2

=

4
;
5

sin θ =

7
1 – 25

2

=

3
5

1
1
x = (4u – 3v) ; y = (3u + 4v)
5
5
16
24
9
(4u − 3v)2 + (4u − 3v )(3u + 4v) + (3u + 4v) 2 25u 2 − 25u = 150
25
25
25
−4(4u − 3v) − 3(3u + 4v ) = 150
u2 − u = 6
1
1
u2 – u + = 6 +
4
4

2

1⎞
25
⎛
⎜u – ⎟ =
2⎠
4
⎝
The graph consists of the two parallel lines u = −2 and u = 3 .

53. a.

If C is a vertical parabola, the equation for C

c.

2

written in the form ( x − h)2 + ( y − k )2 = r 2 .
Substitute the three points into the equation.

can be written in the form y = ax + bx + c .
Substitute the three points into the equation.
2=a–b+c
0=c
6 = 9a + 3b + c
Solve the system to get a = 1, b = –1, c = 0.

(−1 − h) 2 + (2 − k )2 = r 2

h2 + k 2 = r 2

(3 − h)2 + (6 − k ) 2 = r 2

2

y= x −x

Solve the system to get h =

b. If C is a horizontal parabola, an equation for
C can be written in the form
x = ay 2 + by + c . Substitute the three points
into the equation.
–1 = 4a + 2b + c
0=c
3 = 36a + 6b + c
1
Solve the system to get a = , b = –1, c = 0.
4
1 2
x= y −y
4

If C is a circle, an equation for C can be

r2 =

5
5

, k = , and
2
2

25
.
2
2

2

5⎞ ⎛
5⎞
25
⎛
⎜x− ⎟ +⎜ y− ⎟ =
2⎠ ⎝
2⎠
2
⎝

54.

Let (p, q) be the coordinates of P. By
properties of similar triangles and since
KP
x− p
y−q
α=
,α =

and α =
. Solve
AP
a− p
b−q
for p and q to get
x −αa
y −αb
. Since P(p, q) is
and q =
p=
1−α
1−α
a point on a circle of radius r
centered at (0, 0), p 2 + q 2 = r 2
2

2

⎛ x −α a ⎞ ⎛ y −αb ⎞
2
Therefore, ⎜
⎟ +⎜
⎟ = r or
α
α
−
−
1
1

⎝
⎠ ⎝
⎠
( x − α a) 2 + ( y − α b) 2 = (1 − α )2 r 2 is the
equation for C.

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55.

56. Parabola: horizontal parabola, opens to the right,

y 2 = Lx + Kx 2

⎛
L
L2
K ⎜ x2 + x +
⎜
K
4K 2
⎝

⎞
L2
⎟ – y2 =
⎟
4K
⎠

2

L ⎞
L2
⎛
2
K⎜x+
⎟ –y =
2K ⎠
4K
⎝

( x + 2LK ) 2
L2
4K 2

–

y

2

L2
4K

=1

If K < –1, the conic is a vertical ellipse. If
K = –1, the conic is a circle. If –1 < K < 0,
the conic is a horizontal ellipse. If K = 0, the
original equation is y 2 = Lx, so the conic is
a horizontal parabola. If K > 0, the conic is a
horizontal hyperbola.
If −1 < K < 0 (a horizontal ellipse) the length
of the latus rectum is (see problem 45,
Section 10.2)
L2 1
2b 2
=2
= L
a
4K L
2K
From general considerations, the result for a
vertical ellipse is the same as the one just
obtained.
For K = −1 (a circle) we have
2
L
L⎞
L2
⎛

2
x
−
+
y
=
⇒2 = L
⎜
⎟
2⎠
4
2
⎝
If K = 0 (a horizontal parabola) we have
L
L
y 2 = Lx; y 2 = 4 x; p = , and the latus
4
4
rectum is
L
2 Lp = 2 L = L .
4
If K > 0 (a horizontal hyperbola) we can use
the result of Problem 46, Section 10.2. The
2b 2
length of the latus rectum is
, which is
a
equal to L .

p = c – a, y 2 = 4(c − a )( x − a )
Hyperbola: horizontal hyperbola, b2 = c 2 − a 2
x2
a2

y2
b2

−

y2

=1

b2

x2

=

a2
b2

y2 =

a2

−1
( x 2 − a 2 ) Now show that y 2

(hyperbola) is greater than y 2 (parabola).
b2
a
=

2

( x2 − a2 ) =

c2 − a2
a

(c + a )(c − a)

a2
(c + a )( x + a)
a

2

(2a )(2a )
a2

2

( x2 − a2 )

( x + a )( x − a )

(c − a )( x − a ) >

(2a )(2a )
a2

(c − a)( x − a)

(c − a )( x − a ) = 4(c − a)( x − a)

c + a > 2a and x + a > 2a since c > a and x > a
except at the vertex.

57. x = u cos α – v sin α
y = u sin α + v cos α
(u cos α – v sin α) cos α + (u sin α + v cos α) sin α = d
u (cos 2 α + sin 2 α ) = d
u=d
Thus, the perpendicular distance from the origin is d.

614

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by any means, without permission in writing from the publisher.