Graph the hyperbola given by each equation.
1. − = 1
SOLUTION:
The equation is in standard form, where h = 0, k = 0, a = or4,b = or3,andc = or5.
In the standard form of the equation, the y-termisbeingsubtracted.Therefore,theorientationofthehyperbolais
horizontal.
center: (h, k) = (0, 0)
vertices: (h ± a, k) = (4, 0) and (−4, 0)
foci: (h ± c, k) = (5, 0) and (−5, 0)
Graph the center, vertices, foci, and asymptotes. Then make a table of values to sketch the hyperbola.
3. − = 1
SOLUTION:
The equation is in standard form, where h = 0, k = 0, a = or7,b = orabout5.48,andc = or
about 8.89. In the standard form of the equation, the y-termisbeingsubtracted.Therefore,theorientationofthe
hyperbola is horizontal.
center: (h, k) = (0, 0)
vertices: (h ± a, k) = (7, 0) and (−7, 0)
foci: (h ± c, k) = (8.89, 0) and (0, −8.89)
Graph the center, vertices, foci, and asymptotes. Then use a table of values to sketch the hyperbola.
5. − = 1
SOLUTION:
The equation is in standard form, where h = 0, k = 0, a = or3,b = orabout4.58,andc = or
about 5.48. In the standard form of the equation, the y-term is being subtracted. Therefore, the orientation of the
hyperbola is horizontal.
center: (h, k) = (0, 0)
vertices: (h ± a, k) = (3, 0) and (−3, 0 )
foci: (h ± c, k) = (5.48, 0) and (−5.48, 0)
Graph the center, vertices, foci, and asymptotes. Then use a table of values to sketch the hyperbola.
7. − = 1
SOLUTION:
The equation is in standard form, where h = 0, k = 0, a = or9,b = orabout2.83,andc = or
about 9.43. In the standard form of the equation, the x−term is being subtracted. Therefore, the orientation of the
hyperbola is vertical.
center: (h, k) = (0, 0)
vertices: (h, k ± a) = (0, 9) and (0, −9)
foci: (h, k ± c) = (0, 9.43) and (0, −9.43)
Graph the center, vertices, foci, and asymptotes. Then use a table of values to sketch the hyperbola.
9.3x2 − 2y2 = 12
SOLUTION:
First, divide each side by 12 to write the equation in standard form.
The equation is now in standard form, where h = 0, k = 0, a = or2,b = orabout2.45,andc = or
about 3.16. In the standard form of the equation, the y−term is being subtracted. Therefore, the orientation of the
hyperbola is horizontal.
center: (h, k) = (0, 0)
vertices: (h ± a, k) = (2, 0) and (−2, 0 )
foci: (h ± c, k) = (3.16, 0) and (−3.16, 0)
Graph the center, vertices, foci, and asymptotes. Then use a table of values to sketch the hyperbola.
11.LIGHTINGThelightprojectedonawallbyatablelampcanberepresentedbyahyperbola.Thelightfroma
certain table lamp can be modeled by – =1. Graph the hyperbola. Refer to the photo on Page 449.
SOLUTION:
The equation is in standard form, where h = 0, k = 0, a = or15,b = or9,andc = or
about 17.49. In the standard form of the equation, the x−term is being subtracted. Therefore, the orientation of the
hyperbola is vertical.
center: (h, k) = (0, 0)
vertices: (h, k ± a) = (0, 15) and (0, −15)
foci: (h, k ± c) = (0, 17.49) and (0, −17.49)
Graph the center, vertices, foci, and asymptotes. Then use a table of values to sketch the hyperbola.
Graph the hyperbola given by each equation.
13. − = 1
SOLUTION:
The equation is in standard form, where h = 0, k = 7, a = or2,b = orabout5.74,andc = or
about 6.08. In the standard form of the equation, the x−term is being subtracted. Therefore, the orientation of the
hyperbola is vertical.
center: (h, k) = (0, 7)
vertices: (h, k ± a) = (0, 9) and (0, 5)
foci: (h, k ± c) = (0, 13.08) and (0, 0.92)
asymptote:
Graph the center, vertices, foci, and asymptotes. Then use a table of values to sketch the hyperbola.
15. − = 1
SOLUTION:
The equation is in standard form, where h = 5, k = 1, a = or7,b = orabout4.12,andc = or
about 8.12. In the standard form of the equation, the y−term is being subtracted. Therefore, the orientation of the
hyperbola is horizontal.
center: (h, k) = (5, 1)
vertices: (h ± a, k) = (12, 1) and (−2, 12)
foci: (h ± c, k) = (13.12, 1) and (−3.12, 1)
asymptotes:
Graph the center, vertices, foci, and asymptotes. Then use a table of values to sketch the hyperbola.
17. − = 1
SOLUTION:
The equation is in standard form, where h = −6, k = −5, a = or8,b = orabout7.62,andc =
orabout 11.05. In the standard form of the equation, the y−term is being subtracted. Therefore, the
orientation of the hyperbola is horizontal.
center: (h, k) = (−6, −5)
vertices: (h ± a, k) = (2, −5) and (−14, −5)
foci: (h ± c, k) = (5.05, −5) and (−17.05, −5)
asymptotes:
Graph the center, vertices, foci, and asymptotes. Then use a table of values to sketch the hyperbola.
19.−x2 + 3y2 − 4x + 6y = 28
SOLUTION:
First, write the equation in standard form.
The equation is now in standard form, where h = −2, k = −1, a = or3,b = orabout5.2,andc =
or6. In the standard form of the equation, the x−term is being subtracted. Therefore, the orientation of
the hyperbola is vertical.
center: (h, k) = (−2, −1)
vertices: (h, k ± a) = (−2, 2) and (−2, −4)
foci: (h, k ± c) = (−2, 5) and (−2, −7)
asymptotes:
Graph the center, vertices, foci, and asymptotes. Then use a table of values to sketch the hyperbola.
21.−5x2 + 2y2 − 70x − 8y = 287
SOLUTION:
First, write the equation in standard form.
The equation is now in standard form, where h = −7, k = 2, a = or5,b = orabout3.2,andc =
orabout 5.92. In the standard form of the equation, the x−term is being subtracted. Therefore, the
orientation of the hyperbola is vertical.
center: (h, k) = (−7, 2)
vertices: (h, k ± a) = (−7, 7) and (−7, −3)
foci: (h, k ± c) = (−7, 7.92) and (−7, −3.92)
asymptotes:
Graph the center, vertices, foci, and asymptotes. Then use a table of values to sketch the hyperbola.
Write an equation for the hyperbola with the given characteristics.
23.foci (−1, 9), (−1, −7); conjugate axis length of 14 units
SOLUTION:
Because the x−coordinates of the foci are the same, the transverse axis is vertical, and the standard form of the
equation is – =1.
The center is the midpoint of the segment between the foci, or (−1, 1). So, h = −1 and k = 1. The length of the
conjugate axis of a hyperbola is 2b. So, 2b = 14, b = 7, and b2 = 49. You can find c by determining the distance
from a focus to the center. One focus is located at (−1, 9), which is 8 units from (−1, 1). So, c = 8.
Now you can use the values of b and c to find a.
Using the values of h, k, a, and b, the equation for the hyperbola is – =1.
a2 = c2 – b2
a2 = 82 – 72
a2 = 64 – 49
a2 = 15
a =
25.foci (9, −1), (−3, −1); conjugate axis length of 6 units
SOLUTION:
Because the y−coordinates of the foci are the same, the transverse axis is horizontal, and the standard form of the
equation is – =1.
The center is the midpoint of the segment between the foci, or (3, −1). So, h = 3 and k = −1. The length of the
conjugate axis of a hyperbola is 2b. So, 2b = 6, b = 3, and b2 = 9. You can find c by determining the distance from
a focus to the center. One focus is located at (9, −1), which is 6 units from (3, −1). So, c = 6.
Now you can use the values of b and c to find a.
Using the values of h, k, a, and b, the equation for the hyperbola is – =1.
a2 = c2 – b2
a2 = 62 – 32
a2 = 36 – 9
a2 = 27
a =
27.vertices (−3, −12), (−3, −4); foci (−3, −15), (−3, −1)
SOLUTION:
Because the x−coordinates of the vertices are the same, the transverse axis is vertical, and the standard form of
the equation is – =1.
The center is the midpoint of the segment between the foci, or (−3, −8). So, h = −3 and k = −8. You can find c by
determining the distance from a focus to the center. One focus is located at (−3, −15) which is 7 units from (−3,
−8). So, c = 7. You can find a by determining the distance from a vertex to the center. One vertex is located at
(−3, −12) which is 4 units from (−3, −8). So, a = 4 and a2 = 16.
Now you can use the values of c and a find b.
Using the values of h, k, a, and b, the equation for the hyperbola is – =1.
b2 = c2 – a2
b2 = 72 – 42
b2 = 49 – 16
b2 = 33
b =
29.center (−7, 2); asymptotes y = ±x + , transverse axis length of 10 units
SOLUTION:
The center is (−7, 2). Therefore, h = −7 and k = 2.
The transverse axis has a length of 2a units. So, 10 = 2a, a = 5, and a2 = 25. The slopes of the asymptotes are ,
and the standard form of the equation is – =1. So, the positive slope isequalto ,
where b = 7 and b2 = 49.
Using the values of h, k, a, and b, the equation for the hyperbola is – =1.
31.vertices (0, −3), (−4, −3); conjugate axis length of 12 units
SOLUTION:
Because the y−coordinates of the vertices are the same, the transverse axis is horizontal, and the standard form of
the equation is – =1.
The center is located at the midpoint of the vertices, or (–2, –3). So, h = −2 and k = −3. Because the vertices are 4
units apart, 2a = 4, a = 2, and a2 = 4. The length of the conjugate axis is 12 units, so 2b = 12, b = 6, and b2 = 36.
Using the values of h, k, a, and b, the equation for the hyperbola is − =1.
33.ARCHITECTUREThefloorplanforanofficebuildingisshownbelow.
a. Write an equation that could model the curved sides of the building.
b. Each unit on the coordinate plane represents 15 feet. What is the narrowest width of the building?
SOLUTION:
a. From the graph, you can see that the transverse axis is vertical, so the standard form of the equation is
– =1.
The center is located at the midpoint of the vertices, or (5, 4). So, h = 5 and k = 4. Because the vertices are 6 units
apart, 2a = 6, a = 3, and a2 = 9. The length of the conjugate axis is 2b = 10. So, b = 5 and b2 = 25.
Using the values of h, k, a, and b, the equation for the hyperbola is – =1.
b. The length of the narrowest width of the building is 6 units. Since each units represents 15 feet, the length of the
narrowest width is (15)(6) or 90 feet.
Determine the eccentricity of the hyperbola given by each equation.
35. − = 1
SOLUTION:
First, find c.
Use the eccentricity equation and the values of c and a to determine the eccentricity of the hyperbola.
c2 = a2 + b2
c2 = 24 + 15
c =
e =
=
≈1.27
37. − = 1
SOLUTION:
First, find c.
Use the eccentricity equation and the values of c and a to determine the eccentricity of the hyperbola.
c2 = a2 + b2
c2 = 32 + 25
c =
e =
=
≈1.33
39. − = 1
SOLUTION:
First, find c.
Use the eccentricity equation and the values of c and a to determine the eccentricity of the hyperbola.
c2 = a2 + b2
c2 = 16 + 29
c =
e =
=
≈1.68
Determine the eccentricity of the hyperbola given by each equation.
41.−4x2 + 3y2 + 72x − 18y = 321
SOLUTION:
First, write the equation in standard form.
Next, find c.
Use the eccentricity equation and the values of c and a to determine the eccentricity of the hyperbola.
c2 = a2 + b2
c2 = 8 + 6
c =
e =
=
≈1.32
43.−x2 + 7y2 + 24x + 70y = −24
SOLUTION:
First, write the equation in standard form.
Next, find c.
Use the eccentricity equation and the values of c and a to determine the eccentricity of the hyperbola.
c2 = a2 + b2
c2 = 1 + 7
c =
e =
=
≈2.83
Use the discriminant to identify each conic section.
45.18x − 3x2 + 4 = −8y2 + 32y
SOLUTION:
First, write the equation in the general form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
Next, find the discriminant.
The discriminant is greater than 0, so the conic is a hyperbola.
18x − 3x2 + 4 = –8y2 + 32y
–3x2 + 8y2 + 18x – 32y +4 = 0
B2 − 4AC = 02 – 4(–3)(8)
= 96
47.12y − 76 − x2 = 16x
SOLUTION:
First, write the equation in the general form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
Next, find the discriminant.
The discriminant is 0, so the conic is a parabola.
12y − 76 − x2 = 16x
–x2 – 16x + 12y – 76 = 0
B2 − 4AC = 02 – 4(–1)(0)
= 0
49.5y2 − 6x + 3x2 − 50y = −3x2 − 113
SOLUTION:
First, write the equation in the general form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
Next, find the discriminant.
The discriminant is less than 0, so the conic must be either a circle or an ellipse. Because A ≠ C, the conic is an
ellipse.
5y2 − 6x + 3x2 − 50y = –3x2 − 113
6x2 + 5y2 – 6x – 50y + 113 = 0
B2 − 4AC = 02 – 4(6)(5)
= –120
51.−56y + 5x2 = 211 + 4y2 + 10x
SOLUTION:
First, write the equation in the general form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
Next, find the discriminant.
The discriminant is greater than 0, so the conic must be a hyperbola.
–56y + 5x2 = 211 + 4y2 + 10x
5x2 – 4y2 – 10x – 56y – 211 = 0
B2 − 4AC = 02 – 4(5)( –4)
= 80
53.x2 − 4x = −y2 + 12y − 31
SOLUTION:
First, write the equation in the general form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
Next, find the discriminant.
The discriminant is less than 0, so the conic must be either a circle or an ellipse. Because A = C, the conic is a
circle.
x2 − 4x = –y2 + 12y − 31
x2 + y2 – 4x – 12y + 31 = 0
B2 − 4AC = 02 – 4(1)(1)
= –4
55.AVIATIONTheFederalAviationAdministrationperformsflighttrialstotestnewtechnologyinaircraft.When
one of the test aircraft collected its data, it was 18 kilometers farther from Airport B than Airport A. The two
airports are 72 kilometers apart along the same highway, with Airport B due south of Airport A.
a. Write an equation for the hyperbola centered at the origin on which the aircraft was located when the data were
collected.
b. Graph the equation, indicating on which branch of the hyperbola the plane was located.
c. When the data were collected, the plane was 40 miles from the highway. Find the coordinates of the plane.
SOLUTION:
a. The common difference is 18 kilometers and the absolute value of the difference of the distances from any point
on a hyperbola to the foci is 2a, so 2a = 18, a = 9, and a2 = 81. The two airports are the foci of the hyperbola and
are 72 kilometers apart, so 2c = 72, c = 36, and c2 = 1296. c2 = a2 + b2, so b2 = 1296 − 81 or 1215. Airport B is
due south of airport A, so the transverse axis is vertical and a2−term goes with the y2−term. The equation for the
hyperbola is – =1.
b.
The plane is on the top branch because it is closer to airport A.
c. Because the highway is the transverse axis and the plane is 40 kilometers from the highway, x = 40. Substitute
40 for x in the equation.
– =1
– 1.317 = 1
y2 = 187.67
y ≈13.7
The coordinates of the plane are (40, 13.7)
Derive the general form of the equation for a hyperbola with each of the following characteristics.
57.vertical transverse axis centered at the origin
SOLUTION:
From the definition of a hyperbola, you know that the absolute value of the difference of distances from any point P
(x, y) on a hyperbola to the foci is constant, so |PF1 − PF2| = 2a. Since you want to determine the equation for a
hyperbola with a vertical transverse axis centered at the origin, use the distance formula d =
andletF1 = (0, c) and F2 = (0, –c).
Use the definition of a hyperbola.
Isolate one radical and then square both sides.
Square both sides again. Simplify, and then substitute for c2.
Solve each system of equations. Round to the nearest tenth if necessary.
59.2y = x − 10 and − = 1
SOLUTION:
Solve the first equation for y.
Substitute Equation 1 into Equation 2 and simplify.
Use the quadratic formula to solve for x.
Substitute the values for x back into Equation 1 to find that the solutions of the system.
The solutions are (−1.3, −5.7) and (7, −1.5).
61.y = 2x and – =1
SOLUTION:
Substitute Equation 1 into Equation 2 and simplify.
Use the quadratic formula to solve for x.
Substitute the values for x back into Equation 1 to find that the solutions of the system.
The solutions are (−5, −10) and (6.9, 13.8).
x =
=
=
= −5, 6.879
63. + =1and – =1
SOLUTION:
Add the two equations together.
Substitute and solve for x.
x = 0 when y = −6 as well.
The solutions of the system are (0, 6) and (0, −6).
65.FIREWORKS A fireworks grand finale is heard by Carson and Emmett, who are 3 miles apart talking on their
cell phones. Emmett hears the finale about 1 second before Carson does. Assume that sound travels at 1100 feet
per second.
a. Write an equation for the hyperbola on which the fireworks were located. Place the locations of Carson and
Emmett on the x−axis, with Carson on the left and the midpoint between them at the origin.
b. Describe the branch of the hyperbola on which the fireworks display was located.
SOLUTION:
a. Because Emmett and Carson are located on the x−axis, the standard form of the equation for the hyperbola is
– =1.The distance between Emmett and Carson is 3 miles, which is equivalent to 15,840
feet. They are located at the foci of the hyperbola, so c = (15840) or 7920. The absolute value of the difference
of the distances from any point on a hyperbola to the foci is 2a. Since Emmett hears the finale about 1 second
before Carson does, and sound travels 1100 feet per second, 2a = 1100 and a = 550.
Use the values of a and c to find b.
Using the values of h, k, a, and b, the equation for the hyperbola is – =1.
b. Since Emmett heard the finale before Carson did and Emmitt's location is to the right of the origin on the x−axis,
the fireworks display was located on the right branch of the hyperbola.
b2 = c2 – a2
b2 = 79202 – 5502
b2 = 62726400 – 302500
b2 = 62423900
b = 7900.88
Write an equation for each hyperbola.
67.
SOLUTION:
From the vertices, you can determine that the center is located at (0, 0) and the length of the transverse axis is 10
units. So, 2a = 10, a = 5, and a2 = 25.
From the definition of a hyperbola, you know that the absolute value of the difference of distances from any point
on the hyperbola to the foci is constant. So, PF1 – PF2 = 2a. Let F1 = (−c, 0), F2 = (c, 0), and P = .
Use the values of a and c to find b2.
Using the values of a and b, the equation for the hyperbola is .
69.SOUND When a tornado siren goes off, three people are located at J, K, and O, as shown on the graph below.
The person at J hears the siren 2 seconds before the person at O. The person at K hears the siren 1 second before
the person at O. Find each possible location of the tornado siren. Assume that sound travels at 1100 feet per
second. (Hint: Each location of the siren is a point of intersection between a hyperbola that has foci at O and J and
a hyperbola that has foci at O and K.)
SOLUTION:
First, find the equation that corresponds to each hyperbola. Let Srepresentthelocationofthesirenandt be the
time it takes for the sound to travel from point S to point O.
The distance between points J and O is 3500 feet. So, the center of the hyperbola with foci at J and O is located at
(0, 1750). Thus, h = 0, k = 1750, and c = 1750. Since the person at J hears the siren 2 seconds before the person at
O, the distance from the siren to point O is 1100t and the distance from the siren to point J is 1100(t – 2). For the
hyperbola, PF1 – PF2 = 2a. Substitute and find the value of a.
Use the values of a and c to find b2.
Since the foci are on a vertical axis, the equation of the hyperbola will have the form .
Using the values of h, k, a2, and b2, the equation for the hyperbola with foci at points J and O is
.
The distance between points K and O is 2600 feet. So, the center is located at (1300, 0). Thus, h = 1300, k = 0, and
c = 1300. Since the person at K hears the siren 1 second before the person at O, the distance from the siren to
b2 = c2 – a2
b2 = 17502 – 11002
b2 = 3062500 – 1210000
b2= 1852500
point O is 1100t and the distance from the siren to point K is 1100(t – 1). For the hyperbola, PF1 – PF2 = 2a.
Substitute and find the value of a.
Use the values of a and c to find b2.
Since the foci are on a horizontal axis, the equation of the hyperbola will have the form
Using the values of h, k, a2, and b2, the equation for the hyperbola with foci at points K and O is
.
Solving each of these equations for y will give you the following two equations.
and
Separate the ± and enter the equations in Y1 - Y4 on a graphing calculator and graph the two hyperbolas. Use
the intersect function to find the four points of intersections. So, the possible locations of the tornado siren, are at
(–142.5, 2856.0), (3653.6, 4901.0), (700.1, 513.0), (1852.3, –107.7).
b2 = 13002 – 5502
b2 = 1690000 – 302500
b2 = 1387500
Write an equation for the hyperbola with the given characteristics.
71.The hyperbola has its center at (−4, 3) and a vertex at (1, 3). The equation of one of its asymptotes is 7x + 5y =
−13.
SOLUTION:
The center is located at (−4, 3), so h = −4 and k = 3. One vertex is located at (1, 3), so a = 5 and a2 = 25. Because
the y−coordinates of the center and a vertex are the same, the transverse axis is horizontal, and the standard form
of the equation is – =1. The slopes of the asymptotes are . An equation of an
asymptote is 7x + 5y = −13 or . So, the slope of the other asymptote is , b = 7, and b2 = 49.
Using the values of h, k, a, and b, the equation for the hyperbola is – =1.
73.The eccentricity of the hyperbola is andthefociareat(−1, −2) and (13, −2).
SOLUTION:
Because the y−coordinates of the foci are the same, the transverse axis is horizontal, and the standard form of the
equation is – =1. The center is the midpoint of the segment between the foci, or (6, −2).
So, h = 6 and k = −2. The distance from a focus to the center is 7 units, so c = 7.
Substitute c into the eccentricity equation to find a.
So, a2 = 36.
Now you can use the values of a and c to find b.
Using the values of h, k, a, and b, the equation for the hyperbola is – =1.
e =
=
a = 6
b2 = c2 – a2
= 72 – 62
= 49 – 36 or 13
75.For an equilateral hyperbola, a = b when the equation of the hyperbola is written in standard form. The
asymptotes of an equilateral hyperbola are perpendicular. Write an equation for the equilateral hyperbola shown
below.
SOLUTION:
From the graph, you can see that the center of the hyperbola is (0, 0), so h = 0 and k = 0. The distance from a
focus to the center is 11 units, so c = 1.
Substitute c = 11 and a = b into c2 = a2 + b2 to find a.
Using the values of h, k, a, and b, the equation for the hyperbola is – =1.
c2 = a2 + b2
112 = a2 + a2
121 = 2a2
= a2
77.OPEN ENDED Write an equation for a hyperbola where the distance between the foci is twice the length of the
transverse axis.
SOLUTION:
Sample answer: Let the location of the center of the hyperbola be the origin, so that h = 0 and k = 0.
The distance between the foci of a hyperbola is equal to 2c, and the length of the transverse axis is equal to 2a. So,
if the distance between the foci is twice the length of the transverse axis, 2c = 2(2a) or c = 2a.
Substitute c = 2a into c2 = a2 + b2.
Let a = 4.
So, one equation for a hyperbola in which the distance between the foci is twice the length of the transverse axis is
− =1.
c2 = a2 + b2
(2a)2 = a2 + b2
4a2 = a2 + b2
3a2 = b2
3a2 = b2
3(42) = b2
48 = b2
79.Writing in Math Explain why the equation for the asymptotes of a hyperbola changes from ± to ±depending
on the location of the transverse axis.
SOLUTION:
Sample answer: If the transverse axis is horizontal, the distance between the two vertices a is a horizontal distance
and b is a vertical distance. So, the slopes of the asymptotes are equal to the rise over the run or . If the
transverse axis is vertical, then a is a vertical distance and b is a horizontal distance. So, the slopes of the
asymptotes are .
81.CHALLENGE A hyperbola has foci at F1(0, 9) and F2(0, −9) and contains point P. The distance between P and
F1 is 6 units greater than the distance between P and F2. Write the equation of the hyperbola in standard form.
SOLUTION:
Let P(x, y) be the point on the hyperbola. From the definition of a hyperbola, you know that the absolute value of
the difference of distances from any point on the hyperbola to the foci is constant, so |PF1 − PF2| = 2a. In the
problem, you are told that the distance between P and F1 is 6 units greater than the distance between P and F2, so
PF1 – PF2 = 6. If you substitute PF1 – PF2 = 6 into |PF1 − PF2| = 2a, you get 2a = 6, a = 3, and a2 = 9.
The center is the midpoint of the segment between the foci, or (0, 0). So, h = 0 and k = 0. You can find c by
determining the distance from a focus to the center. One focus is located at (0, 9) which is 9 units from (0, 0). So, c
= 9.
You can use the values of a and c to solve for b.
Using the values of h, k, a, and b, the equation for the hyperbola is – =1.
b2 = c2 – a2
b2 = 92– 32
b2 = 81 – 9
b2 = 72
b =
83.Writing in Math Describe the steps for finding the equation of a hyperbola if the foci and length of the transverse
axis are given.
SOLUTION:
Sample answer: First, determine whether the orientation of the hyperbola is vertical or horizontal. Then use the foci
to locate the center of the hyperbola and determine the values of h and k. Use the transverse axis length to find a2.
Find c, the distance from the center to a focus. Use the equation b2 = c2 – a2 to find b2. Finally, use the correct
standard form to write the equation, depending on whether the transverse axis is parallel to the x−axis or to the
y−axis.
Graph the ellipse given by each equation.
85. + =1
SOLUTION:
The ellipse is in standard form, where h = 0, k = −5, a = or8,b = or7,andc = . The
orientation is horizontal because the x−term contains a2.
center: (h, k) = (0, −5)
foci: (h ± c, k) = (3.87, −5) and (−3.87, −5)
vertices: (h ± a, k) = (8, −5) and (−8, −5)
covertices: (h, k ± b) = (0, 2) and (0, −12)
Graph the center, vertices, foci, and axes. Then use a table of values to sketch the ellipse.
87.PROJECTILE MOTIONTheheightofabaseballhitbyabatterwithaninitialspeedof80feetpersecond
can be modeled by h=−16t2 + 80t + 5, where t is the time in seconds.
a. How high above the ground is the vertex located?
b. If an outfielder's catching height is the same as the initial height of the ball, about how long after the ball is hit
will the player catch the ball?
SOLUTION:
a. Find the x-coordinate of the axis of symmetry.
Substitute 2.5 for t in the equation.
The vertex is located 105 feet above the ground.
b. First, find the initial height of the ball.
Substitute h = 5 into the original equation.
Solve for t.
Therefore, the player will catch the ball 5 seconds after it was hit.
h =−16t2 + 80t + 5
= −16(0) + 80(0) + 5
= 5
h =−16t2 + 80t + 5
5 = −16t2 + 80t + 5
0 = −16t2 + 80t
0 = t(−16t + 80)
−16t + 80 = 0
−16t = −80
t = 5
Write each system of equations as a matrix equation, AX = B. Then use Gauss−Jordan elimination on
the augmented matrix to solve the system.
89.x1 – 7x2 + 8x3 = −3
6x1 + 5x2 – 2x3 = 2
3x1 – 4x2 + 9x3 = 26
SOLUTION:
Write the system matrix in form AX = B. Make sure you align the variables. For the first matrix, the first column
should include x1, the second column x2, and the third column x3. The column matrix and the matrix of constant
terms should be listed in order.
Write the augmented matrix .Attachthematrixofconstanttermstotheendofthe3×3matrix.
Use Gauss-Jordan elimination to solve the system. First, use elementary row operations to transform A into I.
The solution of the system is given by X
X = =
Therefore, the solution of the system of equations is (−5, 10, 9).
Solve each equation for all values of .
91.tan =sec – 1
SOLUTION:
On the interval [0, 2π), secθ = 1 when θ = 0 or 2π. Therefore, the solutions on the interval (−∞, ∞)havethe
general form 2nπ, where n is an integer.
tan θ = secθ – 1
tan2 θ = (secθ – 1)(secθ – 1)
sec2θ– 1 = sec2θ – 2secθ + 1
−1 = −2secθ + 1
−2 = −2secθ
1 = secθ
93.csc – cot =0
SOLUTION:
cos θ = 1 when θ = 0. However, when θ = 0, csc θ doesnotexist.
Thisequationhasnosolution.
Find the exact values of the six trigonometric functions of .
95.
SOLUTION:
The length of the side opposite θ is 24 and the length of the hypotenuse is 25. Use the Pythagorean Theorem to
find the length of the side adjacent to θ .
Since the length of each side has been found, you can find the exact values of the trigonometric functions.
sin θ = = csc θ = =
cos θ = = sec θ = =
tan θ = = tan θ = =
a2 + b2 = c2
a2 + 242 = 252
a2 = 625 – 576
a = or7
Use the given zero to find all complex zeros of each function. Then write the linear factorization of the
function.
97.
f(x) = 2x5 – 9x4 + 146x3 + 618x2 + 752x + 291; 4 + 9i
SOLUTION:
Use synthetic substitution to verify that 4 + 9i is a zero of f(x).
Because x = 4 + 9i is a zero of f, x = 4 – 9i is also a zero of f. Divide the depressed polynomial by 4 – 9i.
Using these two zeros and the depressed polynomial from this last division, you can write f(x) = [x – (4 + 9i)][x –
(4 – 9i)](2x3 + 7x2 + 8x + 3).
Factor the remaining depressed polynomial.
(2x3 + 7x2 + 8x + 3)
= (2x + 3)(x2 + 2x + 1)
= (2x + 3)(x + 1)(x + 1)
So, remaining zeros are x = andx = −1withmultiplicity2.
The linear factorization of f is f(x) = (2x + 3)(x + 1)2(x – 4 + 9i)(x – 4 – 9i).
99.REVIEW The graph of =1isahyperbola.Whichsetofequationsrepresentstheasymptotesofthe
hyperbola’s graph?
F y = x, y =
G y = x, y =
H y = x, y =
J y = x, y =
SOLUTION:
The standard form of the equation is – =1, where h = 0, k = 0, a = 4, and b = 5. Since the y2−term is
being subtracted, the orientation of the hyperbola is horizontal, and the asymptotes are
or .
Therefore, the answer is H.
101.SAT/ACT If z = ,thenwhatistheeffectonthevalueofz when y is multiplied by 4 and x is doubled?
F z is unchanged.
G z is halved.
H z is doubled.
J z is multiplied by 4.
SOLUTION:
Multiply y by 4 and x by 2.
Because isequalto · or ·z, the answer is G.
z =
=
=
=
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7-3 Hyperbolas
