Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 1

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 2

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 3

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 4

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 5

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 6

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 7

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 8

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 9

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 10

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 11

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 12

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 13

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

eSolutions Manual - Powered by Cognero Page 14

7-3 Hyperbolas

Graph the hyperbola given by each equation.

2.  − = 1

ANSWER:  

4.  − = 1

ANSWER:  

6.  −  = 1

ANSWER:  

8.  − = 1

ANSWER:  

10. 3y2 − 5x

2 = 15

ANSWER:  

Graph the hyperbola given by each equation.

12.  − = 1

ANSWER:  

14.  − = 1

ANSWER:  

16.  − = 1

ANSWER:  

18. x2 − 4y2 − 6x − 8y = 27

ANSWER:  

20. 13x2 − 2y

2 + 208x + 16y = −748

ANSWER:  

22. EARTHQUAKES  Shortly after a seismograph detects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies on a branch of the hyperbola

represented by  –  = 1, where the seismographs are located at the foci. Graph the

hyperbola.

ANSWER:  

Write an equation for the hyperbola with the given characteristics.24. vertices (7, 5), (−5, 5); foci (11, 5), (−9, 5)

ANSWER:  

−  = 1

26. vertices (−1, 9), (−1, 3); asymptotes y = ± x +

ANSWER:  

−  = 1

28. foci (9, 7), (−17, 7); asymptotes y = ± x +

ANSWER:  

− =1

30. center (0, −5); asymptotes y = ± − 5, conjugate axis length of 12 units

ANSWER:  

−  = 1

32. vertices (2, 10), (2, −2); conjugate axis length of 16 units

ANSWER:  

−  = 1

Determine the eccentricity of the hyperbola given by each equation.

34.  − = 1

ANSWER:  1.52

36.  − = 1

ANSWER:  1.06

38.  − = 1

ANSWER:  2.03

Determine the eccentricity of the hyperbola given by each equation.

40. 11x2 − 2y

2 − 110x + 24y = −181

ANSWER:  2.55

42. 3x2 − 2y

2 + 12x − 12y = 42

ANSWER:  1.58

Use the discriminant to identify each conic section.

44. 14y + y2 = 4x − 97

ANSWER:  parabola

46. 14 +4y + 2x2 = −12x − y

2

ANSWER:  ellipse

48. 2x + 8y + x2 + y

2 = 8

ANSWER:  circle

50. x2 + y2 + 8x − 6y + 9 = 0

ANSWER:  circle

52. −8x + 16 = 8y + 24 − x2

ANSWER:  parabola

54. PHYSICS    An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the other edge are dipped in a thick liquid.  The liquid will rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

ANSWER:  

−  = 1 or  −  = 1

56. ASTRONOMY   While each of the planets in the solar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.  

  The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a.  3x2 − 18x − 580850 = 4.84y

2 − 38.72y

b.  −360x − 8y = −y2 − 1096

c.  −24.88y + x2 = 6x − 3.11y

2 + 412341

ANSWER:  a. hyperbola b. parabola

c. ellipse  

Derive the general form of the equation for a hyperbola with each of the following characteristics.58. horizontal transverse axis centered at the origin

ANSWER:  

 

Solve each system of equations. Round to the nearest tenth if necessary.

60. y = x + 3 and  −  = 1

ANSWER:  

(−6.2, 4.6), (16.5, −1.1)

62. 3x – y = 9 and  +   = 1

ANSWER:  

(1.9, −3.4), (4.3, 4.0)

64.  –  = 1 and  +   = 1

ANSWER:  

(−2.0, −0.2), (−6.7, −3.2), (5.6, −2.6), (2.0, −0.2)

66. ARCHITECTURE  The Kobe Port Tower is a hyperboloid structure in Kobe, Japan. This means that the shapeis generated by rotating a hyperbola around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the tower has an eccentricity of 19. Refer to the photo on Page 450. a. If the tower is 8 meters wide at its narrowest point, determine an equation of the hyperbola used to generate the hyperboloid. b. If the top of the tower is 32 meters above the center of the hyperbola and the base is 76 meters below the center, what is the radius of the top and the radius of the base of the tower?   

ANSWER:  

a.  –  = 1

b. top: about 4.3 m; base: about 5.7 m

Write an equation for each hyperbola.

68. 

ANSWER:  

Write an equation for the hyperbola with the given characteristics.70. The center is at (5, 1), a vertex is at (5, 9), and an equation of an asymptote is 3y = 4x – 17.

ANSWER:  

–  = 1

72. The foci are at (0, 2 ) and (0, −2 ). The eccentricity is .

ANSWER:  

–  = 1

74. The hyperbola has foci at (−1, 9) and (−1, −7) and the slopes of the asymptotes are ± .

ANSWER:  

–  = 1

76. MULTIPLE REPRESENTATIONS   In this problem, you will explore a special type of hyperbola called a conjugate hyperbola. This occurs when the conjugate axis of one hyperbola is the transverse axis of another.

a. GRAPHICAL Sketch the graphs of  −  = 1 and  − = 1 on the same coordinate plane.

b. ANALYTICAL Compare the foci, vertices, and asymptotes of the graphs.

c. ANALYTICAL Write an equation for the conjugate hyperbola for  − = 1.

d. GRAPHICAL Sketch the graphs of the new conjugate hyperbolas. e. VERBAL Make a conjecture about the similarities of conjugate hyperbolas.

ANSWER:  a.

b. The foci of the first graph are (−10, 0) and (10, 0). The foci of the second graph are (0, −10) and (0, 10). The vertices of the first graph are (−6, 0) and (6, 0). The vertices of the second graph are (0, −8) and (0, 8). The graphshave the same asymptotes.

c.  −  = 1

d.

e . Sample answer: Conjugate hyperbolas have the same asymptotes, and the distance from the center to each focus is the same.

78. REASONING  Consider rx2 = −sy

2 – t. Describe the type of conic section that is formed for each of the following. Explain your reasoning. a.  rs = 0 b.  rs > 0 c.  r = s d.  rs < 0

ANSWER:  

a. Parabola; if rs = 0, then either r = 0 or s = 0. So, either the x2 term is 0 or the y

2 term is 0. Since the equation

will have only one squared term, it will be the equation of a parabola.

b. Ellipse; if rs > 0, then either r and s are both greater than 0, or r and s are both less than 0. In both cases, the equation will have squared terms that are added. So, it will be the equation of an ellipse. c. Circle; if r = s then the equation will have squared terms that are added and it can be written so that the coefficient of both terms is 1. So, it will be the equation of a circle.

d. Hyperbola; if rs < 0, then either r < 0 or s < 0. In both cases, the equation will have squared terms that are subtracted. So, it will be the equation of a hyperbola.

80. REASONING  Suppose you are given two of the following characteristics: vertices, foci, transverse axis, conjugate axis, or asymptotes. Is it sometimes, always, or never possible to write the equation for the hyperbola?

ANSWER:  Sometimes; for example, when you are given the vertices and foci, it is possible to write the equation for the hyperbola. When you are given the vertices and transverse axis, it is not possible to write the equation for the hyperbola.

82. PROOF  An equilateral hyperbola is formed when a = b in the standard form of the equation for a hyperbola.

Prove that the eccentricity of every equilateral hyperbola is .

ANSWER:  

In an equilateral hyperbola, a = b and c2 = a

2 + b

2.

c2 = a2 + a

2                 a = b

c2 = 2a2

c = a

Since e = , we have

 

Thus, the eccentricity of any equilateral hyperbola is .

Graph the ellipse given by each equation.

84. (x – 8)2 +  = 1

ANSWER:  

86.  + = 1

ANSWER:  

Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on the augmented matrix to solve the system.

88. 3x1 + 11x2 – 9x3 = 25

−8x1 + 5x2 + x3 = −31

x1 – 9x2 + 4x3 = 13

ANSWER:  

·   =  ; (−2, −7, −12)

90. 2x1 − 5x2 + x3 = 28

3x1 + 4x2 + 5x3 = 17

7x1 − 2x2 + 3x3 = 33

ANSWER:  

·   =  ; (1, −4, 6)

Solve each equation for all values of .

92. sin  + cos   = 0

ANSWER:  

+ nπ,

Find the exact values of the six trigonometric functions of .

94. 

ANSWER:  

sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ =

Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function.

96. f (x) = 2x5 – 11x

4 + 69x

3 + 135x

2 – 675x; 3 – 6i

ANSWER:  

, 0, −3, 3 + 6i, 3 – 6i; x(2x – 5)(x + 3)(x – 3 – 6i)(x – 3 + 6i)

98. REVIEW What is the equation of the graph?

A  y = x

2 + 1

B  y – x = 1

C  y2 – x

2 = 1

D  x2 + y2 = 1

E  xy = 1

ANSWER:  C

100. The foci of the graph are at ( , 0) and (− , 0). Which equation does the graph represent?

A    −  = 1

B    −  = 1

C    −  = 1

D    −  = 1

ANSWER:  A

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7-3 Hyperbolas