CAT 2e ISM Chapter 11 Part 3 - Oak Park and River Forest ...faculty.oprfhs.org/cavalos/Book Chapters...

Section 11.9

1641

60. The calculator is sampling using a more refined partition for 3πθ = than it is for

2πθ = . Using more points yields a better approximation to the graph of the curve.

61. The calculator is sampling using a more refined partition for 0.4θ π= than it is for 2πθ = . Using more points yields a better approximation to the graph of the curve.

62. The calculator is sampling using a more refined partition for 3πθ = than it is for

0.8θ π= . Using more points yields a better approximation to the graph of the curve. Section 11.9 Solutions ------------------------------------------------------------------------------- 1. In order to graph the curve defined parametrically by

1, , 0x t y t t= + = ≥ , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that since 1t x= − , substituting this into the expression for y yields

1, 1y x x= − ≥ .

The graph is as follows:

2. In order to graph the curve defined parametrically by

[ ]23 , 1, in 0, 4x t y t t= = − , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that since 3

xt = , substituting this

into the expression for y yields

[ ]2

211 1, in 0,123 9xy x x⎛ ⎞= − = −⎜ ⎟

⎝ ⎠.


Chapter 11

1642


[ ]23 , 1, in 0, 4x t y t t= − = + , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that since 3

xt = − , substituting

this into the expression for y yields

[ ]2

211 1, in 12,03 9xy x x⎛ ⎞= − + = + −⎜ ⎟

⎝ ⎠.



[ ]2 21, 1, in 3,3x t y t t= − = + − , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that since 2 1t x= + , substituting this into the expression for y yields

( ) [ ]1 1 2, in 1,8y x x x= + + = + − .



[ ]2 3, , in 2, 2x t y t t= = − , it is easiest to be guided by a sequence of tabulated points, as follows:

t 2x t= 3y t= -2 4 -8 -1 1 -1 0 0 0 1 1 1 2 4 8


Section 11.9

1643


[ ]3 31, 1, in 2, 2x t y t t= + = − − , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that since 3 1t x= − , substituting this into the expression for y yields

( ) [ ]1 1 2, in 7,9y x x x= − − = − − .



[ ], , in 0,10x t y t t= = , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, substituting y t= into the expression for x yields

[ ], in 0,10x y y= .



[ ]2, 1, in 0,10x t y t t= = + , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, substituting x t= into the expression for y yields

[ ]2 1, in 0,10y x x= + .


Chapter 11

1644


( ) ( ) [ ]2 31 , 2 , in 0,1x t y t t= + = + , it is easiest to be guided by a sequence of tabulated points, as follows:

t ( )21x t= + ( )32y t= + 0 1 8

0.25 1.5625 11.391 0.50 2.25 15.63 0.75 3.0625 20.80 1.00 4 27



( ) ( ) [ ]3 21 , 2 , in 0,4x t y t t= − = − , it is easiest to be guided by a sequence of tabulated points, as follows:

t ( )31x t= − ( )22y t= − 0 -1 4 1 0 1 2 1 0 3 8 1 4 27 4



, , ln 3 ln 3t tx e y e t−= = − ≤ ≤ it is easiest to be guided by a sequence of tabulated points, as follows:

t tx e= ty e−= -ln3 1/3 3 -ln2 1/2 2

0 1 1 ln 2 2 1/2 ln 3 3 1/3


Section 11.9

1645


2 2, 4, ln 2 ln 3t tx e y e t−= = + − ≤ ≤ it is easiest to be guided by a sequence of tabulated points, as follows:

t 2tx e−= 2 4ty e= +-ln2 4 4.25

0 1 5 ln 2 1/4 8 ln 3 1/9 9



4 82 1, 1, 0 4x t y t t= − = + ≤ ≤ it is easiest to be guided by a sequence of tabulated points, as follows:

t 42 1x t= − 8 1y t= + 0 1 5 1 0.1353 11.39 2 0.0183 58.60 3 0.0025 407.43 4 0.00003 2984.96



6 33 1, 2 , 1 1x t y t t= − = − ≤ ≤ it is easiest to be guided by a sequence of tabulated points, as follows:

t 63 1x t= − 32y t= -1 2 -2 0 -1 0 1 2 2


Chapter 11

1646


3 3( 2) , ( 2) , 0 4x t t y t t t= − = − ≤ ≤ it is easiest to be guided by a sequence of tabulated points, as follows:

t 3( 2)x t t= − 3( 2)y t t= − 0 0 0 1 -1 -1 2 0 0 3 3 3 4 32 32



83 , 5 2, 3 3x t t y t t= − = − − − ≤ ≤ it is easiest to be guided by a sequence of tabulated points, as follows:

t 3x t t= − 85 2y t= − −-3 -4.3267 -32,807 -1 -1 -7 0 0 -2 1 -1 -7 3 -4.3267 -32.807


Section 11.9

1647


[ ]3sin , 2cos , in 0, 2x t y t t π= = , consider the following sequence of tabulated points:

t 3sinx t= 2cosy t= 0 0 2

2π 3 0 π 0 -2

32

π -3 0 2π 0 2

Eliminating the parameter reveals that the graph is an ellipse. Indeed, observe that

2 2 2 29sin , 4cosx t y t= =

so that 2 2

2 2sin , cos9 4x yt t= = .

Summing then yields 2 2

19 4x y

+ = .



[ ]cos 2 , sin , in 0, 2x t y t t π= = , consider the following sequence of tabulated points:

t cos 2x t= siny t= 0 1 0

2π -1 1 π 1 0

32

π -1 -1 2π 1 0

Notice that the curve retraces itself within this interval. Eliminating the parameter reveals that the path that the graph takes is a portion of a parabola opening to the right. Indeed, observe that

2 2cos 2 2sin 1 2 1x t t y= = − = − .


Chapter 11

1648


[ ]sin 1, cos 2, in 0, 2x t y t t π= + = − , consider the following sequence of tabulated points:

t sin 1x t= + cos 2y t= − 0 1 -1

2π 2 -2 π 1 -3

32

π 0 -2 2π 1 -1

Eliminating the parameter reveals that the graph is a circle with radius 1 centered at (1,-2). Indeed, observe that

( ) ( )2 22 2

1 sin , 2 cos1 sin , 2 cos

x t y tx t y t

− = + =

− = + =

so that ( ) ( )2 21 2 1x y− + + = .



tan , 1, in ,4 4x t y t π π⎡ ⎤= = −⎣ ⎦ ,

observe that since the y-coordinate is always 1, the path is a horizontal line segment starting at the point (-1, 1) and ending at the point (1, 1).


Section 11.9

1649


[ ]1, sin , in 2 , 2x y t t π π= = − , observe that since the x-coordinate is always 1, the path is a vertical line segment starting and ending (after retracing its path) at the point (1, 0). The retracing occurs simply because of the periodicity of the sine curve.



[ ]sin , 2, in 0, 2x t y t π= = , observe that since the y-coordinate is always 2, the path is a horizontal line segment starting at the point (0, 2), goes to (1,2) and comes back through (0, 2) and visits (-1, 2), and then finally ends at (0,2).



[ ]2 2sin , cos , in 0, 2x t y t t π= = , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, simply adding the two equations yields

1x y+ = , 0 1, 0 1x y≤ ≤ ≤ ≤ , which is equivalent to

1y x= − , 0 1, 0 1x y≤ ≤ ≤ ≤ .


Chapter 11

1650


[ ]2 22sin , 2cos , in 0, 2x t y t t π= = , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that

[ ]2 2sin , cos , in 0,22 2x yt t t π= =

so that adding these two equations yields

12 2x y+ = , 0 2, 0 2x y≤ ≤ ≤ ≤ ,

which is equivalent to 2y x= − , 0 2, 0 2x y≤ ≤ ≤ ≤ .



[ ]2sin 3 , 3cos 2 , in 0, 2x t y t t π= = , consider the following sequence of tabulated points:

t 2sin 3x t= 3cos 2y t= 0 0 3

6π 2 3

2 4

π 2 0

3π 0 3

2− 2

π -2 -3 2

3π 0 3

2− 3

4π 2 0

56

π 2 32

π 0 3 The pattern replicates for [ ] in , 2t π π .


Section 11.9

1651


[ ]4cos 2 , , in 0, 2x t y t t π= = , consider the following sequence of tabulated points:

t 4 cos 2x t= y t= 0 4 0

4π 0 4

π 2

π -4 2π

34

π 0 34

π π 4 π

54

π 0 54

π 3

2π -4 3

2π

74

π 0 74

π 2π 4 2π



( ) ( )2 2cos 1, sin 1, 2 2t tx y tπ π= − = + − ≤ ≤ consider the following sequence of tabulated points:

t ( )2cos 1tx = − ( )2sin 1ty = +

2π− -2 1 π− -1 0 0 0 1 π -1 2 2π -2 1


Chapter 11

1652


( ) ( )3 3sin 3, cos 1, 0 6t tx y t π= + = − ≤ ≤ consider the following sequence of tabulated points:

t ( )2cos 1tx = − ( )2sin 1ty = +6π− -2 1 3π− -1 2 0 0 1

3π -1 0 6π -2 1



( ) ( ) 74 4 4 42sin , 2cos ,x t y t tπ π π π= + = − + − ≤ ≤

consider the following sequence of tabulated points:

t ( )42sinx t π= + ( )42cosy t π= − +

4π− 0 -2

4π 2 0

34π 0 2

54π -2 0

74π 0 -2



23 33cos (3 ), 2cos(3 ),x t y t tπ π= − = − ≤ ≤

consider the following sequence of tabulated points:

t 23cos (3 )x t= − 2cos(3 )y t=

3π− -3 -2

0 -3 2 3π -3 -2

It oscillates.


Section 11.9

1653

31. In order to express the parametrically-defined curve given by 21,x y tt

= = as a

rectangular equation, we eliminate the parameter t in the following sequence of steps: 1tx

= so that 2

2

1 1yx x

⎛ ⎞= =⎜ ⎟⎝ ⎠

.

So, the equation is 2

1yx

= .

32. In order to express the parametrically-defined curve given by 2 21, 1x t y t= − = + as a rectangular equation, we eliminate the parameter t in the following sequence of steps:

2 1t x= + so that ( 1) 1 2y x x= + + = + . So, the equation is 2y x= + .

33. In order to express the parametrically-defined curve given by 3 31, 1x t y t= + = − as a rectangular equation, we eliminate the parameter t in the following sequence of steps:

3 1t x= − so that ( 1) 1 2y x x= − − = − . So, the equation is 2y x= − .

34. In order to express the parametrically-defined curve given by 23 , 1x t y t= = − as a rectangular equation, we eliminate the parameter t in the following sequence of steps:

3xt = so that

2 2

1 13 9x xy ⎛ ⎞= − = −⎜ ⎟

⎝ ⎠.

So, the equation is 21 19

y x= − .

35. In order to express the parametrically-defined curve given by 2, 1x t y t= = + as a rectangular equation, we eliminate the parameter t in the following sequence of steps:

t x= so that 2 1y x= + .

So, the equation is 2 1y x= + .

36. In order to express the parametrically-defined curve given by 2 2sin , cosx t y t= = as a rectangular equation, we eliminate the parameter t in the following sequence of steps:

2 2sin cos 1x y t t+ = + = So, the equation is 1x y+ = .

Chapter 11

1654

37. In order to express the parametrically-defined curve given by 2 22sin , 2cosx t y t= = as a rectangular equation, we eliminate the parameter t in the following sequence of steps:

2 2sin cos 12 2x y t t+ = + = so that 2x y+ =

So, the equation is 2x y+ = .

38. In order to express the parametrically-defined curve given by 2 2sec , tanx t y t= = as a rectangular equation, we eliminate the parameter t in the following sequence of steps:

2 22 2

2 2 2

sin 1 cos 1tan 1 sec 1 1cos cos cos

t ty t t xt t t

−= = = = − = − = −

So, the equation is 1y x= − .

39. In order to express the parametrically-defined curve given by ( )2 24 1 , 1x t y t= + = − as a rectangular equation, we eliminate the parameter t in the following sequence of steps:

Note that 2 14x t= + so that 2 1

4xt = − . Hence, 1 1 2

4 4x xy ⎛ ⎞= − − = −⎜ ⎟

⎝ ⎠.

Simplifying, we see that the equation is 4 8x y+ = .

40. In order to express the parametrically-defined curve given by 1,x t y t= − = as a rectangular equation, we eliminate the parameter t in the following sequence of steps:

Note that 2 1x t= − so that 2 1t x= + . Hence, 2 1y x= + .

Squaring both sides and simplifying, we see that the equation is 2 2 1y x− = . 41. Assuming that the initial height h = 0, we need to find t such that

( )216 400sin 45 0 0y t t= − + + = . We solve this equation as follows:

( )

( )

2

2

16 400sin 45 0 0

16 200 2 0

16 200 2 0

200 20, 17.7 sec.16

t t

t t

t t

t

− + + =

− + =

− + =

= ≈

So, it hits the ground approximately 17.7 seconds later.

Section 11.9

1655

42. We know from Exercise 31 that the time traveled is 17.7 seconds. So, the horizontal distance traveled is ( ) ( )( )17.7 400cos 45 17.7 5,000 ft.x = ≈

Also, the maximum altitude must occur at 17.7 8.85 sec.2

t ≈ = The height at this time is

given by ( )2(8.85) 16(8.85) 400sin 45 (8.85) 1250 ft.y = − + ≈ 43. First, we convert all measurements to feet per second:

105 mi.1 hr.

1 hr.⎛ ⎞⎜ ⎟⎝ ⎠

5280 ft.3600 sec. 1 mi.

⎛ ⎞⎜ ⎟⎝ ⎠

ft.154 sec.⎛ ⎞

=⎜ ⎟⎝ ⎠

Now, from the given information, we know that the parametric equations are: ( )

( )2

154cos 20

16 154sin 20 3

x t

y t t

=

= − + +

We need to determine the time 0t corresponding to x = 420 ft. Then, we will compare

( )0y t to 15; if ( )0 15y t < , then it doesn’t clear the fence. Observe that

( ) 0 0420420 154cos 20 so that 2.9023 sec.

154cos 20t t= = ≈

Then, we have ( )2.9023 21.0938y ≈ , so it does clear the fence. 44. We use the same information as in Exercise 33. This time, we need to determine 0t such that ( )0 0y t = . Then, we will compute

( )0

0

total horizontal distance traveled

maximum height2

x tty

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

Observe that solving ( )216 154sin 20 3 0t t− + + = using the quadratic formula yields

( )2154sin 20 154sin 20 4( 16)(3)

2( 16)52.6711 54.4632 0.056

32

t− ± − −

=−

− ±≈ ≈ −

− or 3.3479

Now, we have ( )( )

3.3479 484.5 ft. total horizontal distance traveled1.67395 46.3 ft. maximum height

xy

≈ =≈ =

Chapter 11

1656

45. From the given information, we know that the parametric equations are: ( )

( )2

700cos 60

16 700sin 60 0

x t

y t t

=

= − + +

We need to determine 0t such that ( )0 0y t = . Then, we will compute

( )0

0


maximum height2

x tty

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

Observe that solving ( )216 700sin 60 0t t− + = yields

( )( )

216 700sin 60 0 0

16 700sin 60 0

700sin 600, 37.89 sec.16

t t

t t

t

− + + =

− + =

= ≈

Now, we have ( )( )37.89 13,261 ft. total horizontal distance traveled18.94 5742 ft. maximum height

xy

≈ =≈ =


( )2

2000cos 60

16 2000sin 60 0

x t

y t t

=

= − + +


( )0

0


maximum height2

x tty

=

⎛ ⎞ =⎜ ⎟⎝ ⎠


( )( )

216 2000sin 60 0 0

16 2000sin 60 0

2000sin 600, 108.25 sec.16

t t

t t

t

− + + =

− + =

= ≈

Now, we have ( )( )108.25 108,253 ft. total horizontal distance traveled54.127 46,875 ft. maximum height

xy

≈ == =

Section 11.9

1657


( )2

4000cos30

16 4000sin 30 20

x t

y t t

=

= − + +

We need to determine 0t such that ( )0 0y t = .

Observe that solving ( )216 4000sin 30 20 0t t− + + = using the quadratic formula yields

( )2

2

4000sin 30 4000sin 30 4( 16)(20)

2( 16)

2000 2000 1280 0.00132

t− ± − −

=−

− ± +≈ ≈ −

− or 125

So, it stays in flight for 125 seconds. 48. From the given information, we know that the parametric equations are:

( )( )2

5000cos 40

16 5000sin 40 20

x t

y t t

=

= − + +

We need to determine 0t such that ( )0 0y t = . Then, we calculate ( )0x t . If this is larger than 2(5,280) = 10,560 ft., then it can hit the target. Observe that solving ( )216 5000sin 40 20 0t t− + + = using the quadratic formula yields

( )25000sin 40 5000sin 40 4( 16)(20)

2( 16)0.006

t− ± − −

=−

≈ − or 200.877 sec.

So, ( )(200.877) 5000cos 40 (200.877) 769,405 ft.x = ≈ Since this is significantly larger than 10,560 ft., it can definitely hit a target 2 miles away.

Chapter 11

1658

49. The parametric equations are ( )

( )2

100cos35

16 100sin 35

x t

y t t

=

= − +


50. The parametric equations are ( )

( )2

150cos55

16 150sin 55

x t

y t t

=

= − +


51. The points are as follows:

t x(t) y(t) 0 A+B 0

2π 0 A+B

π -A-B 0 3

2π 0 -A-B

2π A+B 0

52. In general, multiply the arguments of the trigonometric functions by coefficients larger than 1 in order to increase speed. Note that A+B represents the length of the arm when fully extended.

53. The original domain must be 0t ≥ . Therefore, only the portion of the parabola where 0y ≥ is part of the plane curve.

54. The original domain must be 0t ≥ . Therefore, only the portion of the parabola where 0x ≥ is part of the plane curve.

55. False. They simply constitute a set of points in the plane.

56. True. By definition of parametric equation (viewed as how points are generated by action of a third parameter).

57. Given the parametric equations , 1 , 0x t y t t= = − ≥ ,

observe that since 2t x= , we have 21y x= − . Now, since x t= is always

0≥ and 21y x= − is also always 0≥ , this generates a quarter circle in QI.

58. Given the parametric equations ln , , 0x t y t t= = > ,

we see that lnx y= , so that xy e= .

Section 11.9

1659

59. Observe that

( ) ( )22 2 2tan sec 1yxa b t t− = − = .

60. Observe that

( ) ( ) ( ) ( )22 2 2

2 2csc cot 1yx t ta b− = − =

61. Observe that 2 22 sin , 2 cosx y a t y x b t+ = − = .

Hence, 2 2sin , cos

2 2x y y xt t

a b+ −

= = ,

and so,

12 2

x y y xa b+ −

+ = .

Solving for y then yields 2a b aby x

a b a b−

= ++ +

,

which is the equation of a line. 62. Observe that

2 cos , 2 sinx y a t x y a t+ = − = . Hence,

2 22 2sin , cos

2 2x y x yt t

a a+ −⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠,

and so, 2 2

12 2

x y x ya a+ −⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠.

Foiling these squared terms and then simplifying yields 2 2 22x y a+ = ,

which is the equation of a circle centered at the origin of radius 2a .

Chapter 11

1660

63. Observe that ( )ln yt

by be t= ⇒ = . Hence,

( ) ( ) ( )ln ln ayb

ay yb baatx e e e= = = = ,

and so, solving for y, we obtain 1ay bx= .

If a is even, the graph looks essentially like that of y x= , while if a is odd, the graph looks essentially like that of 3y x= . 64. Observe that

lnxax a t e t= ⇒ = .

Hence,

( ) ( )ln ln ln lnx xa a xy be b e b

a= = + = + .

This is the equation of a line with slope 1a and y-intercept lnb.

65. Consider the curve defined parametrically by: { sin sin , cos cosx a t at y a t at= − = +

We have the following graphs:

Graph for a = 2 Graph for a = 3 Graph for a = 4 Notice that as the value of a increases, the distance between the vertices and the origin gets larger, and the number of distinct vertices (and corresponding congruent arcs) increases (and = a, when a is a positive integer).

Section 11.9

1661

66. Consider the curve defined by:{ cos cos , sin sinx a t b at y a t at= − = + We have the following graphs:

Graph for a = 3, b = 1 Graph for a = 4, b = 2 Graph for a = 6, b = 2 In each case, one cycle of the curve is completed in 2π units of time. 67. Consider the curve defined by:{ cos , sinx at y bt= = We have the following graphs:

Graph for a = 2, b = 4 Graph for a = 4, b = 2

Graph for a = 1, b = 3 Graph for a = 3, b = 1

In each case, one cycle of the curve is traced out in 2π units of time.

Chapter 11

1662

68. First, we consider the curve defined by the following parametric equations: { sin sin , cos cosx a at t y a at t= − = −

Graph for a = 2 Graph for a = 3 Now, in comparison, we consider the curve defined by the following parametric equations:

{ sin sin , cos cosx a at t y a at t= − = +

Graph for a = 2 Graph for a = 3 All curves are generated in 2π units of time.

Section 11.9

1663

69. Consider the curve defined by the following parametric equations: cos sin , sin cosx a at t y a at t= − = −

Graph for a = 2 Graph for a = 3 All curves are generated in 2π units of time. 70. First, we consider the curve defined by the following parametric equations:

sin cos , cos sinx a at t y a at t= − = −

Graph for a = 2 Graph for a = 3 All curves are generated in 2π units of time. (In comparison to #69, the graphs start at a different point corresponding to t = 0, but generate the same graph for an interval that constitutes one full period.)

Chapter 11

1664

Chapter 11 Review Solutions ---------------------------------------------------------------------- 1. False. The focus is always in the region “inside” the parabola.

2. False. It opens to the right (in the direction of positive x).

3. True. 4. False. The center is (-1,3). 5. Vertex is (0,0) and focus is (3,0) . So, the parabola opens to the right. As such, the general form is 2 4y px= . In this case,

3p = , so the equation is 2 12y x= .

6. Vertex is (0,0) and focus is (0,2) . So, the parabola opens up. As such, the general form is 2 4x py= . In this case,

2p = , so the equation is 2 8x y= .

7. Vertex is (0,0) and the directrix is 5x = . So, the parabola opens to the left.

Since the distance between the vertex and the directrix is 5, and the focus must be equidistant from the vertex, we know

5p = − . Hence, the equation is 2 2( 0) 4( 5)( 0) 20 20y x x y x− = − − = − ⇒ = −

8. Vertex is (0,0) and the directrix is 4y = . So, the parabola opens down.

Since the distance between the vertex and the directrix is 4 and the focus must be equidistant from the vertex, we know

4p = − . Hence, the equation is 2 2( 0) 4( 4)( 0) 16 16x y y x y− = − − = − ⇒ = − .

9. Vertex is (2,3) and focus is (2,5) . So, the parabola opens up. As such, the general form is 2( 2) 4 ( 3)x p y− = − . In this case, 2p = , so the equation is

2( 2) 8( 3)x y− = − .

10. Vertex is ( 1, 2)− − and focus is (1, 2)− . So, the parabola opens to the right. As such, the general form is

2( 2) 4 ( 1)y p x+ = + . In this case, 2p = ,

so the equation is 2( 2) 8( 1)y x+ = + .

11. Focus is (1,5) and the directrix is 7y = . So, the parabola opens down.

Since the distance between the focus and directrix is 2 and the vertex must occur halfway between them, we know 1p = − and the vertex is (1,6) . Hence, the equation is

2( 1) 4( 1)( 6) 4( 6)x y y− = − − = − − .

12. Focus is (2, 2) and the directrix is 0x = . So, the parabola opens to the right.

Since the distance between the focus and directrix is 2 and the vertex must occur halfway between them, we know 1p = and the vertex is (1,2) . Hence, the equation is

2( 2) 4(1)( 1) 4( 1)y x x− = − = − .

Chapter 11 Review

1665

13. Equation: 2 12 4( 3)x y y= − = − (1) So, 3p = − and opens down. Vertex: (0,0) Focus: (0, 3)− Directrix: 3y = Latus Rectum: Connects x-values corresponding to 3y = − . Substituting this into (1) yields:

2 36 so that 6x x= = ± So, the length of the latus rectum is 12.

14. Equation: 2 8 4(2)x y y= = (1) So, 2p = and opens up. Vertex: (0,0) Focus: (0,2) Directrix: 2y = − Latus Rectum: Connects x-values corresponding to 2y = . Substituting this into (1) yields:

2 16 so that 4x x= = ± So, the length of the latus rectum is 8.

15. Equation: 2 1

44( )y x x= = (1) So, 1

4p = and opens to the right. Vertex: (0,0) Focus: 1

4( ,0) Directrix: 1

4x = − Latus Rectum: Connects y-values corresponding to 1

4x = . Substituting this into (1) yields:

2 1 14 2 so that y y= = ±

So, the length of the latus rectum is 1.

Chapter 11

1666

16. Equation: 2 326 4( )y x x= − = − (1)

So, 32p = − and opens to the left.

Vertex: (0,0) Focus: 3

2( ,0)− Directrix: 3

2x = Latus Rectum: Connects y-values corresponding to 3

2x = − . Substituting this into (1) yields:

2 9 so that 3y y= = ± So, the length of the latus rectum is 6.

17. Equation: 2( 2) 4(1)( 2)y x+ = − (1) So, 1p = and opens to the right. Vertex: (2, 2)− Focus: (3, 2)− Directrix: 1x = Latus Rectum: Connects y-values corresponding to 3x = . Substituting this into (1) yields:

2( 2) 4 so that 2 2 4, 0y y+ = = − ± = − So, the length of the latus rectum is 4.

18. Equation: 2( 2) 4( 1)( 1)y x− = − + (1) So, 1p = − and opens to the left. Vertex: ( 1, 2)− Focus: ( 2,2)− Directrix: 0x = Latus Rectum: Connects y-values corresponding to 2x = − . Substituting this into (1) yields:

2( 2) 4 so that 2 2 4, 0y y− = = ± = So, the length of the latus rectum is 4.

Chapter 11 Review

1667

19. Equation: 2( 3) 4( 2)( 1)x y+ = − − (1) So, 2p = − and opens down. Vertex: ( 3,1)− Focus: ( 3, 1)− − Directrix: 3y = Latus Rectum: Connects x-values corresponding to 1y = − . Substituting this into (1) yields:

2( 3) 16 so that 3 4 1, 7x x+ = = − ± = − So, the length of the latus rectum is 8.

20. Equation: 2( 3) 4( 2)( 2)x y− = − + (1) So, 2p = − and opens down. Vertex: (3, 2)− Focus: (3, 4)− Directrix: 0y = Latus Rectum: Connects x-values corresponding to 4y = − . Substituting this into (1) yields:

2( 3) 16 so that 3 4 1, 7x x− = = ± = − So, the length of the latus rectum is 8.

21. Equation: Completing the square yields:

2

2 25 254 4

25 752 4

25 752 8

25 7512 2 8

5 2 25 0( 5 ) 2 25

( ) 2

( ) 2( )

( ) 4( )( )

x x yx x y

x y

x y

x y

+ + + =

+ + = − − +

+ = − −

+ = − +

+ = − + (1)

So, 12p = − and opens down.

Vertex: 5 752 8( , )− − Focus: 5 79

2 8( , )− − Directrix: 71

8y = − Latus Rectum: Connects x-values corresponding to 79

8y = − . Substituting this into (1) yields:

25 5 3 72 2 2 2( ) 1 so that 1 ,x x+ = = − ± = − −

So, the length of the latus rectum is 2.

Chapter 11

1668

22. Equation: Completing the square yields:

2

2

2

2

2 16 1 0( 2 1) 1 16

( 1) 16( 1) 4(4)

y y xy y x

y xy x

+ − + =

+ + − =

+ =

+ = (1)

So, 4p = and opens to the right. Vertex: (0, 1)− Focus: (4, 1)− Directrix: 4x = − Latus Rectum: Connects y-values corresponding to 4x = . Substituting this into (1) yields:

2( 1) 64 so that 1 8 7, 9y y+ = = − ± = − So, the length of the latus rectum is 16.

23. Assume the vertex is at (0, 2)− . Then, the general equation of this parabola is

2 4 ( 2)x p y= + (1). Since ( 5,0)− is on the graph, we can substitute it into (1) to see:

25825 4 (2) 8 so that p p p= = =

So, the focus is at 98(0, ) , and the receiver

should be placed 258 3.125≈ feet from the

center. 24. Assume that the parabola has vertex at (0,40) and opens down. Then, the form of the equation is 2( 0) 4 ( 40)x p y− = − (1). Since the point (15,0) is on the graph, we can substitute it into (1) to find p: 225 160 so that 1.406p p= − = − . So, (1) becomes 2 5.625( 40)x y= − − . We need to determine the y-value when 4x = . If it is larger than 14, then the RV will pass under the bridge without a problem. Observe that

16 5.625( 40) 5.625 22537.16

y yy

= − − = − +=

.

Hence, the RV will pass under the bridge without a problem.

Chapter 11 Review

1669

25.

26.

27. Write the equation in standard form as

2225 1yx + = .

28. Write the equation in standard form as 22

16 8 1yx + = .

29. Since the foci are ( 3,0) and (3,0)− , we know that 3c = and the center of the ellipse is (0,0) . Further, since the vertices are ( 5,0) and (5,0)− and they lie on the x-axis, the ellipse is horizontal and 5a = . Hence,

2 2 2

2

2

9 2516

c a bb

b

= −

= −

=

Therefore, the equation of the ellipse is 22

25 16 1yx + = .

30. Since the foci are (0, 2) and (0,2)− , we know that 2c = and the center of the ellipse is (0,0) . Further, since the vertices are (0, 3) and (0,3)− and they lie on the y-axis, the ellipse is vertical and 3a = . Hence,

2 2 2

2

2

4 95

c a bb

b

= −

= −

=

Therefore, the equation of the ellipse is 22

5 9 1yx + = .

Chapter 11

1670

31. Since the ellipse is centered at (0,0) , the length of the horizontal major axis being 16 implies that 8a = , and the length of the horizontal minor axis being 6 implies that 3b = . Since the major axis is vertical, the equation of the ellipse is

22

9 64 1yx + = .

32. Since the ellipse is centered at (0,0) , the length of the horizontal major axis being 30 implies that 15a = , and the length of the vertical minor axis being 20 implies that 10b = . Since the major axis is horizontal, the equation of the ellipse is

22

225 100 1yx + = .

33.

34. Write the equation in standard form as 2 2( 3) ( 4)

6 120 1x y+ −+ = .

35. First, write the equation in standard form by completing the square:

2 2

1 14 12

2 2

2 2

2 2

( 2) ( 3)

4( 4 ) 12( 6 ) 1234( 4 4) 12( 6 9) 123 16 108

4( 2) 12( 3) 1

1x y

x x y yx x y y

x y− +

− + + = −

− + + + + = − + +

− + + =

+ =

36. First, write the equation in standard form by completing the square:

2 2

1 14 9

2 2

2 2

2 2

( 1) ( 4)

4( 2 ) 9( 8 ) 1474( 2 1) 9( 8 16) 147 4 144

4( 1) 9( 4) 1

1x y

x x y yx x y y

x y− −

− + − = −

− + + − + = − + +

− + − =

+ =

Chapter 11 Review

1671

37. Since the foci are ( 1,3), (7,3)− and the vertices are ( 2,3), (8,3)− , we know: i) the ellipse is horizontal with center at 2 8

2( ,3) (3,3)− + = , ii) 3 1 so that 4c c− = − = , iii) the length of the major axis is 8 ( 2) 10− − = . Hence, 5a = . Now, since 2 2 2c a b= − , 2 216 25 so that 9b b= − = . Hence, the equation of the ellipse is

2 2( 3) ( 3)25 9 1x y− −+ = .

38. Since the foci are (1, 3), (1, 1)− − and the vertices are (1, 4), (1,0)− , we know: i) the ellipse is vertical with center at 4 0

2(1, ) (1, 2)− + = − , ii) 2 3 so that 1c c− − = − = , iii) the length of the major axis is 0 ( 4) 4− − = . Hence, 2a = . Now, since 2 2 2c a b= − , 2 21 4 so that 3b b= − = . Hence, the equation of the ellipse is

2 2( 1) ( 2)3 4 1x y− ++ = .

39. Assume that the sun is at the origin. Then, the vertices of Jupiter’s horizontal elliptical trajectory are: 8 8( 7.409 10 ,0), (8.157 10 ,0)− × × From this, we know that: i) The length of the major axis is 8 8 98.157 10 ( 7.409 10 ) 1.5566 10× − − × = × , and so the value of a is half of this, namely 87.783 10× ; ii) The center of the ellipse is ( )8 87.409 10 (8.157 10 ) 7

2 ,0 (3.74 10 ,0)− × + × = × ;

iii) The value of c is 73.74 10× . Now, since 2 2 2c a b= − , we have ( ) ( )2 27 8 23.74 10 7.783 10 b× = × − , so that

2 176.044 10b = × . Hence, the equation of the ellipse must be ( ) ( )

27 2

17 17

(3.74 10 ) 0

6.058 10 6.044 101

x y− × −

× ×+ = .

(Note: There is more than one correct way to set up this problem, and it begins with choosing the center of the ellipse. If, alternatively, you choose the center to be at (0,0), then the actual coordinates of the vertices and foci will change, and will result in the following slightly different equation for the trajectory:

22

2 2778,300,000 777,400,0001yx + =

Both are equally correct!.)

Chapter 11

1672

40. Assume that the sun is at the origin. Then, the vertices of Mars’ horizontal elliptical trajectory are: 8 8( 2.07 10 ,0), (2.49 10 ,0)− × × From this, we know that: i) The length of the major axis is 8 8 82.49 10 ( 2.077 10 ) 4.56 10× − − × = × , and so the value of a is half of this, namely 82.28 10× ; ii) The center of the ellipse is ( )8 82.49 10 ( 2.07 10 ) 7

2 ,0 (2.1 10 ,0)× + − × = × ;


2 165.154 10b = × . Hence, the equation of the ellipse must be ( ) ( )

27 2

16 16

(2.1 10 ) 0

5.1984 10 5.154 101

x y− × −

× ×+ = .

41.

Notes on the Graph: The equations of the asymptotes are

83y x= ± .

42.


79y x= ± .

Chapter 11 Review

1673

43.

Notes on the Graph: First, write the equation in standard form as 2 2

25 1x y− = . The equations of the asymptotes are 1

5y x= ± .

44.

Notes on the Graph: First, write the equation in standard form as

2 2

8 16 1y x− = . The equations of the

asymptotes are 22y x= ± .

45. Since the foci are ( 5,0), (5,0)− and the vertices are ( 3,0), (3,0)− , we know that: i) the hyperbola opens right/left with center at (0,0) , ii) 3, 5a c= = , Now, since 2 2 2c a b= + , 2 225 9 so that 16b b= + = . Hence, the equation of the

hyperbola is 22

9 16 1yx − = .

46. Since the foci are (0, 3), (0,3)− and the vertices are (0, 1), (0,1)− , we know that: i) the hyperbola opens up/down with center at (0,0) , ii) 1, 3a c= = , Now, since 2 2 2c a b= + , 2 29 1 so that 8b b= + = . Hence, the equation of the hyperbola is 22

8 1xy − = .

47. Since the center is (0,0) and the transverse axis is the y-axis, the general form of

the equation of this hyperbola is 2 2

2 2 1y xa b− = with asymptotes a

by x= ± . In this case,

3ab = so that 3a b= . Thus, the equation simplifies to

2 2

2 2(3 )1y x

b b− = , or equivalently

2 2 29y x b− = . If we took b = 1, then this simplifies to

2 29 1y x− = .

Chapter 11

1674

48. Since the center is (0,0) and the transverse axis is the y-axis, the general form of

the equation of this hyperbola is 2 2

2 2 1y xa b− = with asymptotes a

by x= ± . In this case, 12

ab = so that 2a b= . Thus, the equation simplifies to

2 2

2 2(2 )1y x

a a− = , or equivalently

22 24xy a− = .

49.


2( 2) 1y x= ± − + .

50.


12 ( 3) 4y x= ± + + .

51.

Notes on the Graph: First, write the equation in standard form by completing the square:

( ) ( )( ) ( )

2 2

2 2

2 2

2 2

( 2) ( 3)10 8

8 4 10 6 138

8 4 4 10 6 9 138 32 90

8( 2) 10( 3) 80

1x y

x x y y

x x y y

x y− +

− − + =

− + − + + = + −

− − + =

− =

The equations of the asymptotes are

25

( 2) 3y x= ± − − .

Chapter 11 Review

1675

52.


( ) ( )( ) ( )

2 2

12

2 2

2 2

2 2

( 3) ( 1)2

2 6 8 2 6

2 6 9 8 2 1 6 18 8

2( 3) 8( 1) 4

1x y

x x y y

x x y y

x y+ −

+ − − = −

+ + − − + = − + −

+ − − =

− =

The equations of the asymptotes are 12 ( 3) 1y x= ± + + .

53. Since the vertices are (0,3), (8,3) , we know that the center is (4,3) and the transverse axis is parallel to the x-axis. Hence, 4 0 so that 4a a− = = . Also, since the foci are ( 1,3), (9,3)− , we know that 4 1 so that 5c c+ = − = − . Now, to find b, we substitute the values of c and a obtained above into 2 2 2c a b= + to see that 225 16 b= +

so that 2 9b = . Hence, the equation of the hyperbola is 2 2( 4) ( 3)

16 9 1x y− −− = .

54. Since the vertices are (4, 2), (4,0)− , we know that the center is (4, 1)− and the transverse axis is parallel to the y-axis. Hence, 1 2 so that 1a a− − = − = . Also, since the foci are (4, 3), (4,1)− , we know that 1 3 so that 2c c− − = − = . Now, to find b, we substitute the values of c and a obtained above into 2 2 2c a b= + to see that 24 1 b= + so

that 2 3b = . Hence, the equation of the hyperbola is 2 2( 1) ( 4)

1 3 1y x− −− = .

55. Assume that the stations coincide with the foci and are located at ( 110,0), (110,0)− . The difference in distance between the ship and each of the two stations must remain constantly 2a, where ( ,0)a is the vertex. Assume that the radio signal speed is 186,000

secmi and the time difference is 0.00048 sec . Then, using distance = rate × time, we obtain:

2 (186,000)(0.00048) 89.28 so that 44.64a a= = = So, the ship will come ashore between the two stations 65.36 miles from one and 154.64 miles from the other.

Chapter 11

1676

56. Assume that the stations coincide with the foci and are located at ( 200,0), (200,0)− . The difference in distance between the ship and each of the two stations must remain constantly 2a, where ( ,0)a is the vertex. Assume that the radio signal speed is 186,000 sec

mi and the time difference is 0.0008 sec . Then, using distance = rate × time, we obtain:

2 (186,000)(0.0008) 148.8 so that 74.4a a= = = So, the ship will come ashore between the two stations 125.6 miles from one and 274.4 miles from the other.

57. Solve the system 2 3

5x yx y

⎧ + = −⎨

− =⎩

(1)(2)

Add (1) and (2) to get an equation in terms of x: 2 2 0

( 2)( 1) 02, 1

x xx x

x

+ − =+ − =

= −

Now, substitute each value of x back into (2) to find the corresponding values of y: 2 : 2 5 so that 7

1: 1 5 so that 4x y yx y y= − − − = = −= − = = −

So, the solutions are ( 2, 7) and (1, 4)− − − .


24

x yx y

+ =⎧⎨

+ =⎩

(1)(2)

Solve (1) for y: 2y x= − (3) Substitute (3) into (2) to get an equation in terms of x:

2 2

2

4 4 42 4 0

2 ( 2) 00, 2

x x xx x

x xx

+ − + =

− =− =

=

Now, substitute these values of x back into (3) to find the corresponding values of y: 0 : 2 (0) 22 : 2 (2) 0

x yx y= = − == = − =

So, the solutions are (0, 2) and (2,0) .

Chapter 11 Review

1677


2

52 0x yx y

⎧ + =⎪⎨

− =⎪⎩

(1)(2)

Solve (2) for y: 22y x= (3) Substitute (3) into (1) to get an equation in x:

( )( )

2 2 2

2 4

4 2

2 2

(2 ) 54 5

4 5 04 5 1 0

1

x xx x

x xx x

x

+ =+ =

+ − =

+ − =

= ±

Now, substitute each value of x back into (2) to find the corresponding values of y: 2

2

1: 2(1) 21: 2( 1) 2

x yx y= = == − = − =

So, the solutions are (1,2) and ( 1,2)− .


2 2

166 16

x yx y

⎧ + =⎪⎨

+ =⎪⎩

(1)(2)

Multiply (1) by 1− , and then add to (2): 25 0

0xx== (3)

Substitute (3) into (1) to find the corresponding values of y: 2 20 16

4yy

+ == ±

So, the solutions are (0,4) and (0, 4)− .


34

x yx y

+ =⎧⎨

+ =⎩

(1)(2)

Solve (1) for y: 3y x= − (3) Substitute (3) into (2) to get an equation in terms of y:

2 2

2 2

2

6 36 402(2)

(3 ) 49 6 4

2 6 5 0

which are not real numbers

x xx x x

x x

x ± −

+ − =

+ − + =

− + =

=

Hence, the system has no solution.

Chapter 11

1678


416

xyx y

=⎧⎨

+ =⎩

(1)(2)

Solve (1) for y: 4xy = (3)

Substitute (3) into (2) to get an equation in terms of x:

( )2

22 4

2 16

4

2

4 2

2 2

4 2

2

4 2

16

16

16 16

16 16 0

16 16 0

16 16 0

x

x

x

x

xx

x xx x

x xx

x x

+ =

+ =

+=

+− =

− +=

− + = (4)

Let 2u x= (5) . Then, (4) can be written in the equivalent form 2 16 16 0u u− + = . Solve this equation using the quadratic formula:

16 256 4(16) 16 192 16 8 32 2 2 8 4 3u ± − ± ±= = = = ±

Substituting both of these values into (5) now leads to the following two equations in x:

2 8 4 33.864

xx= +≅ ±

and 2 8 4 3

1.035xx= −≅ ±

Now, substitute each of these values of x back into (3) to find the corresponding value of y:

43.864

43.864

41.035

41.035

3.864: 1.0353.864: 1.035

1.035: 3.8651.035: 3.865

x yx yx yx y

−

−

= = ≅

= − = ≅ −

= = ≅

= − = ≅ −

So, the approximate solutions are ( ) ( )( ) ( )3.864,1.035 , 3.864, 1.035 ,

1.035,3.865 , and 1.035, 3.865

− −

− −.

63. Solve the system 2 2 12

2x xy y

x y⎧ + + = −⎨

− =⎩

(1)(2)

Solve (2) for x: 2x y= + (3) Substitute (3) into (1) to get an equation in terms of y:

2 2

2 2 2

6 36 4(3)(16)22

( 2) ( 2) 124 4 2 12

3 6 16 0 so that which are not real numbers

y y y yy y y y y

y y y − ± −

+ + + + = −

+ + + + + = −

+ + = =

Hence, the system has no solution.

Chapter 11 Review

1679


3 3x yx y

⎧ − = −⎨

+ =⎩

(1)(2)

Solve (1) for x: 2 9x y= − (3) Substitute (3) into (1) to get an equation in terms of y:

2

2

103

3( 9) 33 30 0

(3 10)( 3) 0 so that , 3

y yy y

y y y

− + =

+ − =+ − = = −

Now, substitute these values of y back into (3) to find the corresponding values of x: ( )

( )

210 10 193 3 9

2

: 9

3: 3 9 0

y x

y x

= − = − − =

= = − =

So, the solutions are 10 193 9(0,3) and ( , )− .

65. Solve the system 3 3 19

1x y

x y⎧ − = −⎨

− = −⎩

(1)(2)

Solve (2) for x: 1x y= − (3) Substitute (3) into (1) to get an equation in terms of x:

2

3 3

( 1)( 2 1)

3 2 2 3

2

2

( 1) 19

( 2 2 1) 19 03 3 18 0

6 0( 2)( 3) 0 so that 2, 3

y y y

y y

y y y y y yy y

y yy y y

− − +

− − = −

− + − + − − + =

− + + =

− − =+ − = = −

Now, substitute these values of y back into (3) to find the corresponding values of x: 2 : 2 1 3, 3 : 3 1 2y x y x= − = − − = − = = − =

So, the solutions are (2,3) and ( 3, 2)− − .

Chapter 11

1680

66. Solve the system 2

2

2 4 92 0

x xyx xy

⎧ + =⎪⎨

− =⎪⎩

(1)(2)

Multiply (2) by 2 and then add to (1) to get an equation in terms of x: 2

2 94

32

4 9xxx

=

=

= ±

Now, substitute these values of x back into (2) to find the corresponding values of y: 23 3 3

2 2 29 34 4

23 3 32 2 2

9 34 4

: ( ) 2( )( ) 03 so that

: ( ) 2( )( ) 03 so that

x yy y

x yy y

= − =

= =

= − − − − =

= − = −

So, the solutions are 3 3 3 32 4 2 4( , ) and ( , )− − .


2 2

2 1 15

1 1 3

x y

x y

⎧ + =⎪⎪⎨⎪ − = −⎪⎩

(1)

(2)

Add (1) and (2) to get an equation in terms of x:

2

2

2 1 14 2

3 12

12 3so that .

xxx x

=

=

= = ±

Now, substitute these values of x back into (2) to find the corresponding value of y: 12 2 21

2

2

2

2

2 17

17

1 1: 3( )

14 3

17

7 1

xy

y

yyy

y

= − = −

− = −

=

=

=

= ±

12 2 21

2

2

2

2

2 17

17

1 1: 3( )

14 3

17

7 1

xy

y

yyy

y

= − − = −−

− = −

=

=

=

= ±

So, the solutions are 1 1 1 1 1 1 1 12 2 2 27 7 7 7

( , ), ( , ), ( , ), and ( , )− − − − .

Chapter 11 Review

1681


2 2

24

x yx y

⎧ + =⎨

+ =⎩

(1)(2)

Add (1) and (2) to obtain the false statement 0 = 6. Hence, there are no real values of x and y that can satisfy both equations simultaneously. Hence, there is no solution of this system. 69.

70.

71.

72.

73.

74.

Chapter 11

1682

75.

76.

77.

78.

79.

80.

Chapter 11 Review

1683

81. The coordinates of the point obtained by rotating ( )3, 2− by an angle of 60 counterclockwise are:

cos sin3cos 60 2sin 603 322 23 32

X x yθ θ= += − +

⎛ ⎞= − + ⎜ ⎟⎜ ⎟

⎝ ⎠

= − +

sin cos3sin 60 2cos 60

3 13 22 2

3 3 12

Y x yθ θ= − += +

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

= +

So, the corresponding coordinates of this point in the XY-system are 3 3 33, 12 2

⎛ ⎞− + +⎜ ⎟⎜ ⎟⎝ ⎠

.

82. The coordinates of the point obtained by rotating ( )4, 3− by an angle of 45 counterclockwise are:

cos sin4cos 45 3sin 45

2 24 32 2

22

X x yθ θ= += −

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

=

sin cos4sin 45 3cos 45

2 24 32 2

7 22

Y x yθ θ= − += − −

⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= −

So, the corresponding coordinates of this point in the XY-system are 2 7 2,2 2

⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠.

Chapter 11

1684

83. Consider the equation 2 22 4 3 2 16 0x xy y+ − − = . In order to obtain the equation of this curve in the XY-system, we make the following substitution into the original equation with 30θ = :

( )

( )

1cos sin 321sin cos 32

x X Y X Y

y X Y X Y

θ θ

θ θ

⎧ ⎛ ⎞= − = − ⎜ ⎟⎪⎪ ⎝ ⎠⎨

⎛ ⎞⎪ = + = + ⎜ ⎟⎪ ⎝ ⎠⎩

Doing so yields the following, after collecting like terms and simplifying:

( ) ( )( ) ( )( ) ( ) ( )

2 2 22

2 2 2 2 2 2

2 2

2 2

1 1 12 3 4 3 3 3 2 3 16 02 2 2

2 3 2 3 4 3 3 3 3 2 2 3 3 64 0

16 16 64 0

14 4

X Y X Y X Y X Y

X XY Y X XY XY Y X XY Y

X YX Y

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + + − − + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

− + + − + − − + + − =

− − =

− =

Now, the graph of 2 22 4 3 2 16 0x xy y+ − − = , along with the new axes

( ) 1tan 30 ,3

y x x⎛ ⎞= = ⎜ ⎟⎝ ⎠

3y x= − is as follows:

Chapter 11 Review

1685

84. Consider the equation 2 225 14 25 288 0x xy y+ − − = . In order to obtain the equation of this curve in the XY-system, we make the following substitution into the original equation with 45θ = :

( )

( )

2cos sin2

2sin cos2

x X Y X Y

y X Y X Y

θ θ

θ θ

⎧ ⎛ ⎞= − = −⎪ ⎜ ⎟⎜ ⎟⎪ ⎝ ⎠

⎨⎛ ⎞⎪ = + = + ⎜ ⎟⎪ ⎜ ⎟⎝ ⎠⎩


( ) ( )( ) ( )2 2 2

2 2

2 2

2 2

2 2 225 14 25 288 02 2 2

14 14 576 07 7 1288 288

X Y X Y X Y X Y

X YX Y

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + − + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠− − =

− =

(c) The graph of 2 225 14 25 288 0x xy y+ − − = , along with the new axes

( )tan 45 ,y x x y x= = = − is as follows:

85. To find the amount of rotation needed to transform the given equation into one in

the XY-system with no XY term, we seek the solution θ of the equation cot 2 A CB

θ −=

in [ ]0,π . In this case, 4, 2 3, 6A B C= = = , so that this equation becomes 1cot 23

θ = − . Hence, we need to find [ ]in 0,θ π such that

31cos 2 and sin 22 2θ θ= − = .

But, this occurs precisely when 2 120 , so that 60θ θ= = .

Chapter 11

1686

86. To find the amount of rotation needed to transform the given equation into one in the XY-system with no XY term, we seek the solution θ of the equation cot 2 A C

Bθ −= in

[ ]0,π . In this case, 4, 5, 4A B C= = = , so that solving this equation yields:

cot 2 0 cos 2 0 2 90 45θ θ θ θ= ⇒ = ⇒ = ⇒ = .

87. Consider the equation 2 22 2 2 8 0x xy y x y+ + + − + = . Here, 1, 2, 1A B C= = = , so that 2 4 0B AC− = . Thus, the curve is a parabola. In order to obtain its graph, follow these steps: Step 1: Find the angle of counterclockwise rotation necessary to transform the equation into the XY-system with no XY term. We seek the solution θ of the equation cot 2 0A C

Bθ −= = in [ ]0,π . Hence, we need to

find [ ] in 0,θ π such that cos 2 0 and sin 2 1θ θ= = .

But, this occurs precisely when 2 90 , so that 45θ θ= = . Step 2: Now, determine the equation of this curve in the XY-system making use of the following substitution into the original equation with 45θ = :

( ) ( )2 2cos sin sin cos2 2

x X Y X Y y X Y X Yθ θ θ θ⎛ ⎞ ⎛ ⎞

= − = − = + = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


( ) ( )( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( )

2 2 22 2

2 2 2 2 2 2

2

2 2 222 2 2

2 22 2 8 02 2

2 2 2 2 2 16 0

4

X Y X Y X Y X Y

X Y X Y

X XY Y X Y X XY Y X Y X Y

X Y

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞

+ − − + + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

− + + − + + + + − − + + =

+ =

Step 3: The graph of 2 25 2 2 2 8 0x xy y x y+ + + − + = , along with the new axes

( )tan 45 ,y x x= = y x= − is as follows:

Chapter 11 Review

1687

88. Consider the equation 2 276 48 3 28 100 0x xy y+ + − = . Here, 76, 48 3, 28A B C= = = , so that 2 4 1600 0B AC− = − < . Thus, the curve is an ellipse In order to obtain its graph, follow these steps: Step 1: Find the angle of counterclockwise rotation necessary to transform the equation into the XY-system with no XY term.

We seek the solution θ of the equation 1cot 23

A CB

θ −= = in [ ]0,π . Hence, we need

to find [ ] in 0,θ π such that 31cos 2 and sin 22 2θ θ= = .


( ) ( )1 1cos sin 3 sin cos 32 2

x X Y X Y y X Y X Yθ θ θ θ⎛ ⎞ ⎛ ⎞= − = − = + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


( ) ( )( ) ( )( ) ( ) ( )

2 2 22 2

2 2 2 2 2 2

2 2

2 2

1 1 176 3 48 3 3 3 28 3 100 02 2 2

76 3 2 3 48 3 3 3 3 28 2 3 3 400 0

400 16 400 0

11 25

X Y X Y X Y X Y


X YX Y

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + + + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

− + + + − − + + + − =

+ − =

+ =

Step 3: The graph of 2 276 48 3 28 100 0x xy y+ + − = , along with the new axes

( ) 1tan 30 ,3

y x x⎛ ⎞= = ⎜ ⎟⎝ ⎠

3y x= − is as follows:

89. The directrix is horizontal and 7 units below the pole, so the equation is

( )( )( )37

3 37 7

7 31 sin 1 sin

rθ θ

= =− −

, which simplifies to 217 3sin

rθ

=−

.

Chapter 11

1688

90. Since the curve is a parabola, 1e = . Further, since the directrix is vertical and 2

units to the right of the pole, the equation is ( )( )( )1 2 2

1 1 cos 1 cosr

θ θ= =

+ +.

91. Observe that ( )

32

5 54 4

6 64 5cos 4 1 cos 1 sin

rθ θ θ

= = =− − −

, so that 54 1e = > . Hence,

this conic is a hyperbola.

92. Observe that ( )

25

3 35 5

2 25 3sin 5 1 sin 1 sin

rθ θ θ

= = =+ + +

, so that 35 1e = < . Hence, this

conic is an ellipse.

93. Consider the equation 42 cos

rθ

=+

.

(a) Since ( )

( )12

1 12 2

44 42 cos 2 1 cos 1 cos

rθ θ θ

= = =+ + +

, 12 1e = < . So, this conic is an ellipse.

(b) From (a), we see that 12 and 4e p= = . From the general form of the equation, we

know the directrix is vertical and is 4 units to the right of the origin. Since the directrix is perpendicular to the major axis, we know that the major axis lies along the x-axis. Hence, the vertices occur when 0,θ π= . Observe that ( ) 4

30r = and ( ) 4r π = . So, in

rectangular coordinates, the vertices are ( ) ( )43 ,0 and 4,0− .

(c) In order to obtain the graph, we need to find the center and y-intercepts. Center = midpoint of vertices =

43 4 4,0 ,0

2 3−⎛ ⎞ ⎛ ⎞= −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

The length of the major axis = ( ) 1643 32 4a = − − = , so that 8

3a = . Now, using cea

= , we

see that ( )( )8 1 43 2 3c ae= = = . Since 2 2 2b c a= − , we further find that 2 16

3b = , so 4 33b = .

Hence, the length of the minor axis (which is perpendicular to the major axis through the center) = 2b. So, the y-intercepts are ( )4 34

3 3 ,0− ± . The graph is as follows:

Chapter 11 Review

1689

94. Consider the equation 61 sin

rθ

=−

.

(a) Observe that 1e = . So, this conic is a parabola. (b) Inspecting the equation in (a), we see that 1 and 6e p= = . From the general form of the equation, we know the directrix is horizontal and is below the origin. As such, since the parabola opens up, the vertex must lie on the y-axis. Further, since the right-side of the equation is undefined at 2

πθ = , the vertex must occur at 3

2πθ = . Evaluating the equation at 3

2πθ = yields 3r = . So, the vertex is rectangular

coordinates is ( )0, 3− .

(c) Consider the following table of points: θ 6

1 sinr

θ=

− ( ),r θ

0 6 ( )6,0

2π undefined --- π 6 ( )6,π 32π 3 ( )3

23, π 2π 6 ( )6, 2π



[ ]sin , 4cos , in ,x t y t t π π= = − , consider the following sequence of tabulated points:

t sinx t= 4cosy t= π− 0 -4

2π− -1 0 0 0 4

2π 1 0 π 0 -4

Eliminating the parameter reveals that the graph is an ellipse. Indeed, observe that

22 2 2sin , cos

4yx t t⎛ ⎞= =⎜ ⎟

⎝ ⎠

so that summing then yields 2

2 116yx + = .


Chapter 11

1690


[ ]2 25sin , 2cos , in ,x t y t t π π= = − , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed,

2 2sin , cos5 2x yt t= =

so that adding the two equations yields

15 2x y+ = , 0 5, 0 2x y≤ ≤ ≤ ≤ ,

which is a line segment.



[ ]2 24 , , in 3,3x t y t t= − = − , it is easiest to eliminate the parameter t and write the equation in rectangular form since the result is a known graph. Indeed, observe that since 2t y= , substituting this into the expression for x yields

[ ]4 in 5, 4x y x= − − .



[ ]3, 4, in 4, 4x t y t= + = − , observe that since the y-coordinate is always 4, the path is a horizontal line segment starting at the point (-1, 4) and ends at (7, 4).


Chapter 11 Review

1691

99. In order to express the parametrically-defined curve given by 24 ,x t y t= − = as a rectangular equation, we eliminate the parameter t by simply substituting y = t into the equation for x. So, the equation is 24x y= − . 100. In order to express the parametrically-defined curve given by

2 25sin , 2cosx t y t= = as a rectangular equation, we eliminate the parameter t. Indeed,

observe that 2 2sin , cos5 2x yt t= = , so that adding the two equations yields 1

5 2x y+ = ,

which is equivalent to 2 5 10x y+ = . 101. In order to express the parametrically-defined curve given by

2 22 tan , 4secx t y t= = as a rectangular equation, we eliminate the parameter t in the following sequence of steps:

( ) ( )

2 22

2 2 2

2 2

sin 1 cos 12 tan 2 2 2 2cos cos cos

2 12 sec 2 4sec 2 24 2

t tx tt t t

t t y

⎛ ⎞ ⎛ ⎞− ⎛ ⎞= = = = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

= − = − = −

So, the equation is 2 4y x= + . 102. In order to express the parametrically-defined curve given by

2 23 4, 3 5x t y t= + = − as a rectangular equation, we eliminate the parameter t in the following sequence of steps:

23 4t x= − so that ( 4) 5 9y x x= − − = − . So, the equation is 9y x= − . 103. The vertex is at (0.6, -1.2), and it should open down. The graph below confirms this:

104. The vertex is at (2.8, 0.2) and it should open to the right. The graph below confirms this:

Chapter 11

1692

105. a. Completing the square yields

2

2 2 2

2

2.8 3 6.852.8 1.4 3 6.85 1.4

( 1.4) 3 8.81

1.4 3 8.81

1.4 3 8.81

y y xy y x

y x

y x

y x

+ = − +

+ + = − + +

+ = − +

+ = ± − +

= − ± − +


b. From a, we obtain

( )

2

2 2 2

2

2

2 34

2.8 3 6.852.8 1.4 3 6.85 1.4

( 1.4) 3 8.81( 1.4) 3( 2.937)

( ( 1.4)) 4 ( 2.937)

y y xy y x

y xy x

y x

+ = − +

+ + = − + +

+ = − +

+ = − −

− − = − −

The vertex is (2.937, -1.4). The parabola opens to the left. c. Yes, parts a and b agree.

106. a. The equation is equivalent to

2 10.2 24.8y x x= − + . The graph is as follows:

b. Completing the square yields

( )( ) ( )

2

2

2 2 2

2

2 14

10.2 24.810.2 24.8

10.2 5.1 24.8 5.1

5.1 1.21

5.1 4 ( ( 1.21))

x x yx x y

x x y

x y

x y

− + =

− = −

− + = − +

− = +

− = − −

The vertex is (5.1, -1.21). The parabola opens up. c. Yes, parts a and b agree.

Chapter 11 Review

1693

107. From the graphs, we see that as c increases the minor axis (along the x-axis) of the ellipse whose equation is given by

2 24( ) 1cx y+ = decreases.

108. From the graphs, we see that as c increases the minor axis (along the y-axis) of the ellipse whose equation is given by

2 24( ) 1x cy+ = decreases.

109. From the graphs, we see that as c increases, the vertices of the hyperbola whose equation is given by 2 24( ) 1cx y− = are at ( )1

2 ,0c± are moving towards the origin:

110. From the graphs, we see that as c increases, the vertices of the hyperbola whose equation is given by 2 24( ) 1x cy− = remain at ( )1,0± , but the graphs open more narrowly:

Chapter 11

1694

111. The graph of this nonlinear system is as follows:

The points of intersection are approximately (0.635, 2.480), (–0.635, 2.480), (–1.245, 0.645), and (1.245, 0.645).

112. The graph of this nonlinear system is as follows:

The points of intersection are approximately (0.876, 1.458) and (1.350, –1.781).

113. The darker region below is the one desired:

114. The darker region below is the one desired:

115. a.

The angle of rotation is computed by

2 53

34

38

cot 2 12

1.2 rad

A CB

π

π

θθθ

− −−= = = −

=

= ≈

b.


( )

2 ( 5) 73 3

1 73

cot 2

2 cot0.2 rad

A CBθ

θ

θ

− −−

−

= = =

=

≈

Chapter 11 Review

1695

116. a.


1 ( 1)2

34

38

cot 2 12

1.2 rad

A CBπ

π

θθθ

− −−−= = = −

=

= ≈

b.


1 12

2

4

cot 2 02

0.8 rad

A CB

π

π

θθθ

− −= = =

=

= ≈

117. The calculator is sampling using to wide a step size, and is jumping over parts of the graph. Using more points yields a better approximation to the graph of the curve. 118. The calculator is sampling using to wide a step size, and is jumping over parts of the graph. Using more points yields a better approximation to the graph of the curve. 119. First, we consider the curve defined by the following parametric equations:

sin cos , cos sinx a at b bt y a at b bt= + = +

Graph for a = 2, b=3 Graph for a = 3, b=2 All curves are generated in 2π units of time.

Chapter 11

1696

120. First, we consider the curve defined by the following parametric equations: sin cos , cos sinx a at b bt y a at b bt= − = −

Graph for a = 1, b=2 Graph for a = 2, b=1 All curves are generated in 2π units of time. Chapter 11 Practice Test Solutions--------------------------------------------------------------- 1. c Parabola opens to the right 2. b Parabola opens up 3. d Ellipse is horizontal 4. e Hyperbola whose transverse axis is the

x-axis 5. f Ellipse is vertical 6. a Hyperbola whose transverse axis is y-

axis. 7. Vertex is (0,0) and focus is ( 4,0)− . So, the parabola opens to the left. As such, the general form is 2 4y px= . In this case, 4p = − , so the equation is

2 16y x= − .

8. Vertex is (0,0) and the directrix is 2y = . So, the parabola opens down. Since the distance between the vertex and directrix is 2, we know 2p = − . Hence, the equation is

2 4( 2) 8x y y= − = − .

9. Vertex is ( 1,5)− and focus is ( 1, 2)− . So, the parabola opens down. As such, the general form is 2( 1) 4 ( 5)x p y+ = − . In this case, 3p = − , so the equation is

2( 1) 12( 5)x y+ = − − .

10. Vertex is (2, 3)− and the directrix is 0x = . So, the parabola opens to the right.

Since the distance between the vertex and directrix is 2, we know 2p = . Hence, the

equation is 2( 3) 4(2)( 2) 8( 2)y x x+ = − = − .

Chapter 11 Practice Test

1697

11. Since the foci are (0, 3), (0,3)− and the vertices are (0, 4), (0,4)− , we know: i) the ellipse is vertical with center at (0,0) , ii) 3c = , iii) the length of the major axis is 4 ( 4) 8− − = . Hence, 4a = . Now, since 2 2 2c a b= − , 2 29 16 so that 7b b= − = . Hence, the equation of the ellipse is

22

7 16 1yx + = .

12. Since the foci are ( 1,0), (1,0)− and the vertices are ( 3,0), (3,0)− , we know: i) the ellipse is horizontal with center at (0,0) , ii) 1c = , iii) the length of the major axis is 3 ( 3) 6− − = . Hence, 3a = . Now, since 2 2 2c a b= − , 2 21 9 so that 8b b= − = . Hence, the equation of the ellipse is

22

9 8 1yx + = .

13. Since the foci are (2, 4), (2, 4)− and the vertices are (2, 6), (2,6)− , we know: i) the ellipse is vertical with center at 6 6

2(2, ) (2,0)− + = , ii) 4c = , iii) the length of the major axis is 6 ( 6) 12− − = . Hence, 6a = . Now, since 2 2 2c a b= − , 2 216 36 so that 20b b= − = . Hence, the equation of the ellipse is

2 2( 2)20 36 1x y− + = .

14. Since the foci are ( 6, 3), ( 5, 3)− − − − and the vertices are ( 7, 3), ( 4, 3)− − − − , we know: i) the ellipse is horizontal with center at 7 4 11

2 2( , 3) ( , 3)− +− − = − − , ii) 1

2c = , iii) the length of the major axis is 4 ( 7) 3− − − = . Hence, 3

2a = . Now, since 2 2 2c a b= − , 2 2 2 231

2 2( ) ( ) so that 2b b= − = . Hence, the equation of the ellipse

is 211 2

2( ) ( 3)2.25 2 1x y+ ++ = .

Chapter 11

1698

15. Since the vertices are ( 1,0), (1,0)− , and the center is the midpoint of the segment connecting them, the center must be (0,0). Moreover, since the form of the vertices is ( , )h a k± ( which in this case is (0 ,0)a± ) we see that 1a = . It remains to find b. At this point, we need to use the fact that we are given that the equations of the asymptotes are

2y x= ± . Since the transverse axis is parallel to the x-axis, we know that the slopes of the asymptotes for such a hyperbola are b

a± . Thus, 1 2b± = ± , which implies that 2b = ± .

Hence, the equation of the hyperbola must be 22

4 1yx − = .

16. Since the vertices are (0, 1), (0,1)− , and the center is the midpoint of the segment connecting them, the center must be (0,0). Moreover, since the form of the vertices is ( , )h a k± ( which in this case is (0 ,0)a± ) we see that 1a = . It remains to find b. At this point, we need to use the fact that we are given that the equations of the asymptotes are

13y x= ± . Since the transverse axis is parallel to the y-axis, we know that the slopes of

the asymptotes for such a hyperbola are ab± . Thus, 1

3ab± = ± . Since 1a = , this implies

that 3b = ± . Hence, the equation of the hyperbola must be 229 1xy − = .

17. Since the vertices are (2, 4), (2, 4)− , we know that the center is (2,0) and the transverse axis is parallel to the y-axis. Hence, 0 4 so that 4a a− = − = . Also, since the foci are (2, 6), (2,6)− , we know that 0 6 so that 6c c− = − = . Now, to find b, we substitute the values of c and a obtained above into 2 2 2c a b= + to see that 236 16 b= +

so that 2 20b = . Hence, the equation of the hyperbola is 2 2( 2)

16 20 1y x−− = .

18. Since the vertices are ( 6, 3), ( 5, 3)− − − − , we know that the center is ( 5.5, 3)− − and the transverse axis is parallel to the x-axis. Hence, 5.5 6 so that 0.5a a− − = − = . Also, since the foci are ( 7, 3), ( 4, 3)− − − − , we know that 5.5 7 so that 1.5c c− − = − = . Now, to find b, we substitute the values of c and a obtained above into 2 2 2c a b= + to see that

2 2 2(1.5) (0.5) b= + so that 2 2b = . Hence, the equation of the hyperbola is 2 2( 5.5) ( 3)

0.25 2 1x y+ +− = .


1699

19.


( ) ( )( ) ( )

2 2

2 2

2 2

2 2

( 1) ( 2)4 9

9 2 4 4 43

9 2 1 4 4 4 43 9 16

9( 1) 4( 2) 36

1x y

x x y y

x x y y

x y+ −

+ − − =

+ + − − + = + −

+ − − =

− =

The equations of the asymptotes are 32 ( 1) 2y x= ± + + .

20.

First, write the equation in standard form by completing the square:

2

14

2 2

2 2

2 2

( 1) 2

4( 2 ) ( 10 ) 284( 2 1) ( 10 25) 28 4 25

4( 1) ( 5) 1

( 5) 1x

x x y yx x y y

x y

y−

− + + = −

− + + + + = − + +

− + + =

+ + =

21.

Equation: Completing the square yields: 2

2

2

2

2

4 16 20( 4 4) 16 20 4

( 2) 16 16( 2) 16( 1)( 2) 4(4)( 1)

y y xy y x

y xy xy x

+ = −

+ + = − +

+ = −

+ = −

+ = − (1)

So, 4p = and opens to the right and its vertex is (1, 2)− .

Chapter 11

1700

22.

23. Assume that the vertex is at the origin and that the parabola opens up. We are given that 1.5p = . Hence, the equation is

2 4(1.5) 6 , 2 2x y y x= = − ≤ ≤ .

24. Assume that the sun is at the origin. Then, the vertices of Uranus’s horizontal elliptical trajectory are: 9 9( 2.739 10 ,0), (3.003 10 ,0)− × × From this, we know that: i) The length of the major axis is 9 9 93.003 10 ( 2.739 10 ) 5.742 10× − − × = × , and so the value of a is half of this, namely 92.871 10× ; ii) The center of the ellipse is ( )9 93.003 10 ( 2.739 10 ) 8

2 ,0 (1.32 10 ,0)× + − × = × ;


2 188.225 10b = × . Hence, the equation of the ellipse must be ( )28 2

18 18

(1.32 10 )

8.24 10 8.225 101

x y− ×

× ×+ = .

25.

26.


1701

27.

28.

29. Observe that ( )2 2

3 3

12 12 43 2sin 3 1 sin 1 sin

rθ θ θ

= = =+ + +

, so that 23 1e = < . Hence, this

conic is an ellipse. 30. Consider the equation 2 26 3 6 4 3 21 3 0x xy y+ + − = . Here, 6 3, 6, 4 3A B C= = = . In order to obtain its graph, follow these steps: Step 1: Find the angle of counterclockwise rotation necessary to transform the equation into the XY-system with no XY term.

We seek the solution θ of the equation 1cot 23

A CB

θ −= = in [ ]0,π . Hence, we need

to find [ ] in 0,θ π such that 31cos 2 and sin 22 2θ θ= = .


( )

( )

1cos sin 321sin cos 32

x X Y X Y

y X Y X Y

θ θ

θ θ

⎧ ⎛ ⎞= − = − ⎜ ⎟⎪⎪ ⎝ ⎠⎨

⎛ ⎞⎪ = + = + ⎜ ⎟⎪ ⎝ ⎠⎩


( ) ( )( ) ( )( ) ( ) ( )

2 2 22 2

2 2 2 2 2 2

2 2

2 2

1 1 16 3 3 6 3 3 4 3 3 21 3 02 2 2

6 3 3 2 3 6 3 3 3 4 3 2 3 3 84 3 0

28 3 12 3 84 3 0

13 7

X Y X Y X Y X Y


X YX Y

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + + + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

− + + + − − + + + − =

+ + =

+ =

Chapter 11

1702


( )2

120cos 45

16 120sin 45 0

x t

y t t

=

= − + +


( )0 total horizontal distance traveledx t =


( )( )

216 120sin 45 0

16 120sin 45 0

120sin 450, 5.3 sec.16

t t

t t

t

− + =

− + =

= ≈

Now, we conclude that ( )5.3 450 ft. total horizontal distance traveledx = =

32. Given the parametric equations [ ]1 , , in 0,1x t y t t= − = ,

observe that since 2t y= , we have 21x y= − . Now, since y t= is always 0≥ and 21x y= − is also always 0≥ , this

generates a quarter circle in QI. 33.

34.

So, the approximate solutions of this system are (2.457, 3.001) and (-2.457, -3.001).


1703

35. a. The equation is equivalent to 2 4.2 5.61y x x= + + .


b. Completing the square yields

( )( )

2

2 2 2

2

2 14

4.2 5.614.2 2.1 5.61 2.1

( 2.1) 1.2( ( 1.2)) 4 1.2

x x yx x yx yx y

+ = −

+ + = − +

+ = −

− − = −

The vertex is (-1.2, 1.2). The parabola opens up. c. Yes, parts a and b agree.

36. a.


( )2 1 1

3 3

1 13

cot 2

2 cot

.052 rad

A CBθ

θ

θ

− −−

−

= = = −

= −

≈

b.


( )2 ( 1) 3

3 3

1 33

cot 2

2 cot

.26 rad

A CBθ

θ

θ

− −−

−

= = =

=

≈

Chapter 11

1704

Chapter 11 Cumulative Review ------------------------------------------------------------------- 1.

( )( )( )( )

2( 2) ( 2) 20 0( 2) 5 ( 2) 4 0

3 6 0

6,3

x xx x

x x

x

+ − + − =

+ − + + =

− + =

= −

2. The equation is of the general form 2 2 2( ) ( )x h y k r− + − = .

Since ( , ) (5,1)h k = , this becomes 2 2 2( 5) ( 1)x y r− + − = .

In order to find r, substitute the point (6,-2) into the above equation to obtain

2 2 2(6 5) ( 2 1) 1 9 10 r− + − − = + = = . Hence, the desired equation is

2 2( 5) ( 1) 10x y− + − = . 3. ( ) ( )8 7( ) 8 7 8 7 7 8 7

7 7

x h x x h xh h

hh

− + − − − − − +=

−= = −

4. The general equation is I kPt= . We need to find k. To this end, we substitute in the given information to obtain

90 (1500)(2)k= , so that 3

100k = . Hence, the equation is 3

100I Pt= . 5. The general form is 2( )y a x h k= − + . We know that the vertex ( , ) (7,7)h k = , so that this equation becomes 2( 7) 7y a x= − + . In order to find a, substitute the point (10,10) into this equation to obtain

2 1310 (10 7) 7 3 9a a a= − + ⇒ = ⇒ = .

Hence, the desired equation is 213 ( 7) 7y x= − + .

6. Use ( )1 ntrnA P= + . Here, 0.047, 52, 17, 65,000r n t A= = = = . We need to

determine P. Substitute all of this information into the equation to obtain ( )

( )

52(17)0.04752

52(17)0.04752

65,000 165,000 29,246.16

1

P

P

= +

= ≈+

So, you should invest $29,246.16. 7.

2 log 412log log16 0 2log log16 log log16 log 4 10 4x x x x− = ⇒ = ⇒ = = ⇒ = =

9. ( ) ( )1cot 27 1.9626

tan 27− = ≈

−

8. Let x = the shortest leg in a 30 60 90− − triangle. If x = 8 in., then the other leg has length 8 3 in. and the hypotenuse has length 16 in.

10. Amplitude is 0.007, and frequency is 2 1

850 425ππ = .

Chapter 11 Cumulative Review

1705

11.

( ) sin 1 1tan csc cos cos sincos sin cos

θθ θ θ θ θθ θ θ⎛ ⎞+ = + = +⎜ ⎟⎝ ⎠

12.

( ) ( )11 311 1166 6 211

12

1 cos 1 2 3cos cos cos2 2 2 2 2

ππ ππ

+ + +⎛ ⎞ ⎛ ⎞− = − = = − = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

13. Observe that

( )2

2 2

2

2 14

12

4cos 4cos 2 1 0

4cos 4 2cos 1 1 0

12cos 3 0coscos

x x

x x

xxx

+ + =

+ − + =

− =

=

= ±

This holds when 2 4 53 3 3 3, , ,x π π π π=

14. First, note that at the end of two hours, one plane flew due north for 900 miles, and the other plane flew for 750 miles at an angle 135 clockwise from due north. Consider the following diagram:

Using the Law of Cosines yields

( )2 2 2900 750 2(900)(750)cos 135 2327094.155

1525.48

d

d

= + − ≈

≈\

So, the two planes are approximately 1525 miles apart after two hours. 15. cos , sin 15cos110 , 15sin110 5.13,14.10u u uθ θ= = ≈ −

135

Chapter 11

1706

16. Observe that ( ) ( )15 75

1 25 , 2i iz e z e= = Then, we have

( ) ( )

( )

( )

15 751 2

15 75

90

5 2

(2)(5)

10

10

i

i

i

iz z e e

e

e

i

+

⋅ = ⋅

=

=

=

17. Let x = cost of a soda and y = cost of a soft pretzel. Solve the system:

3 2 6.77 ( )5 4 12.25 ( )

x yx y+ =⎧

⎨ + =⎩

12

Multiply (1) by -5 and (2) by 3, and add:

15 10 33.8515 12 36.75

2 2.901.45

x yx y

yy

− − = −+ + =

==

Substitute this value of y into (1) to solve for x:

3 2(1.45) 6.77 1.29x x+ = ⇒ = So, a soda costs $1.29 and a soft pretzel costs $1.45.

18. The partial fraction decomposition is of the form

( ) 22

3 53 5( 3) 5

x A Bx Cx xx x

+ += +

− +− +

Multiply by the LCD to obtain ( )2 23 5 5 ( )( 3) ( ) ( 3 ) (5 3 )x A x Bx C x A B x C B x A C+ = + + + − = + + − + − .

We must solve the following system: 0 ( )

3 35 3 5 ( )

A BC BA C

+ =⎧⎪ − =⎨⎪ − =⎩

1(2)3

Solve (1) for B: ( )B A= − 4 Substitute (4) into (2): 3 3 ( )C A+ = 5 Solve the 2 2× system:

5 3 5 ( )3 3 ( )A CA C− =⎧

⎨ + =⎩

35

CONTINUED ONTO NEXT PAGE.

Chapter 11 Cumulative Review

1707

Multiply (5) by 3 and add to (3) to see that 1A = . Substitute this back into (5) to obtain 0C = . Finally, substitute the value of A into (1) to see that 1B = − . Hence, the partial fraction decomposition is

( ) 22

3 5 13 5( 3) 5

x xx xx x

+ −= +

− +− +

19.

20. 2 1 2

12 25

31 2 1 7 1 2 1 73 1 2 11 0 5 5 10

1 2 1 70 1 1 2

R R R

R R

+ →

− →

⎡ − ⎤ ⎡ − ⎤⎯⎯⎯⎯⎯→⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦

⎡ − ⎤⎯⎯⎯⎯⎯→⎢ ⎥− −⎣ ⎦

Let y a= . Then, 2x a= − and by substituting these two values into the equation corresponding to the first row of the last matrix, we see that ( 2) 2 7a a z− − + = , so that

9z a= + .

21. 16 4 12 9 12 21

2 318 0 2 0 3 15

7 16 3318 3 17

B A− −⎡ ⎤ ⎡ ⎤

− = −⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦−⎡ ⎤

= ⎢ ⎥− −⎣ ⎦

22. Write the system in matrix form as 25 40 1275 105 69

xy

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Observe that 25 40

5,62575 105

12 401,500

69 105

25 122,625

75 69

x

y

D

D

D

= = −−

−= = −

−

−= =

So, 1500 2625 745625 15 5625 15,x y= = = − = − .

Chapter 11

1708

23. Since the vertices are (6,3) and (6,-7), the center is ( )3 7

26, (6, 2)− = − . Using the foci, we see that

2 2 and 2 6c c− + = − − = − , so that 4c = . Also, we have

2 3 and 2 7a a− + = − − = − , so that 5a = . As such, 2 2 2 3c a b b= − ⇒ = . Thus, the equation of the ellipse is

2 2( 6) ( 2)9 25 1x y− ++ = .

24. Since the vertices are (5,-2) and (5,0), the center of the hyperbola is (5,-1). Also, we have a = 1. Thus, the equation so far is

22

2

( 5)( 1) 1xyb−

+ − = .

To find b, we use the fact that c = 2, so that 2 2 2 3c a b b= + ⇒ =

Thus, the desired equation is 2

2 ( 5)( 1) 13

xy −+ − =

25. Solve the system:

2 2

6 ( )20

x yx y

+ =⎧⎨

+ =⎩

1(2)

Solve (1) for y: 6 ( )y x= − 3 Substitute (3) into (2) and solve for x:

2 2

2

2

(6 ) 202 12 36 20

6 8 0( 4)( 2) 0

2,4

x xx x

x xx x

x

+ − =

− + =

− + =− − =

=

Substitute each value into (1) to find the corresponding y-value. This yields two solutions, namely (2,4) and (4,2). 26. The graph from the calculator is as follows:

27. The dark region is the one desired:

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