Graph the ellipse given by each equation.
1. + =1
SOLUTION:
The ellipse is in standard form, where h = –2, k = 0, a = or7,b = or3,andc = orabout6.3.The
orientation is vertical because the y–term contains a2.
center: (h, k) = (–2, 0)
foci: (h, k ± c) = (–2, 6.3) and (–2, –6.3)
vertices: (h, k ± a) = (–2, 7) and (–2, –7)
covertices: (h ± b, k) = (1, 0) and (–5, 0)
Graph the center, vertices, foci, and axes. Then use a table of values to sketch the ellipse.
2. + =1
SOLUTION:
The ellipse is in standard form, where h = –4, k = –3, a = or3,b = or2,andc = . The
orientation is horizontal because the x–term contains a2.
center: (h, k) = (–4, –3)
foci: (h ± c, k) = (–1.8, –3) and (–6.2, –3)
vertices: (h ± a, k) = (–1, –3) and (–7, –3)
covertices: (h, k ± b) = (–4, –1) and (–4, –5)
Graph the center, vertices, foci, and axes. Then use a table of values to sketch the ellipse.
3.x2 + 9y2 – 14x + 36y + 49 = 0
SOLUTION:
Complete the square for each variable to write the equation in standard form.
x2 + 9y2 – 14x + 36y + 49 = 0
x2 − 14x + 9y2 + 36y + 49 = 0
x2 − 14x + 49 + 9(y2 + 4y + 4) = 0 + 36
(x − 7)2 + 9(y + 2)2=36
+ = 1
The ellipse is in standard form, where h = 7, k = –2, a = or6,b = or2,andc = . The
orientation is horizontal because the x–term contains a2.
center: (h, k) = (7, –2)
foci: (h ± c, k) = (12.7, –2) and (1.3, –2)
vertices: (h ± a, k) = (13, –2) and (1, –2)
covertices: (h, k ± b) = (7, 0) and (7, –4)
Graph the center, vertices, foci, and axes. Then use a table of values to sketch the ellipse.
4.4x2 + y2 – 64x – 12y + 276 = 0
SOLUTION:
Complete the square for each variable to write the equation in standard form.
4x2 + y2 – 64x – 12y + 276 = 0
4x2 − 64x + y2 − 12y + 276 = 0
4x2 − 64x + 256 + y2 − 12y + 36 + 276 = 0 + 256 + 36
4(x2 − 16x + 64) + y2 − 12y + 36 = 0 + 256 + 36 – 276
4(x − 8)2 + (y − 6)2 = 16
+ = 1
The ellipse is in standard form, where h = 8, k = 6, a = or4,b = or2,andc = . The
orientation is vertical because the y–term contains a2.
center: (h, k) = (8, 6)
foci: (h, k ± c) = (8, 9.5) and (8, 2.5)
vertices: (h, k ± a) = (8, 10) and (8, 2)
covertices: (h ± b, k) = (10, 6) and (6, 6)
Graph the center, vertices, foci, and axes. Then use a table of values to sketch the ellipse.
5.9x2 + y2 + 126x + 2y + 433 = 0
SOLUTION:
Complete the square for each variable to write the equation in standard form.
9x2 + y2 + 126x + 2y + 433 = 0
9x2 + 126x + y2 + 2y + 433 = 0
9(x2 + 14x + 49) + y2 + 2y + 1 + 433 − 9(49) − 1 = 0
9(x + 7)2 + (y + 1)2 − 9 = 0
(x + 7)2 + = 1
The ellipse is in standard form, where h = –7, k = –1, a = or3,b = 1, and c = . The orientation is
vertical because the y–term contains a2.
center: (h, k) = (–7, –1)
foci: (h, k ± c) = (–7, 1.8) and (–7, –3.8)
vertices: (h, k ± a) = (–7, 2) and (–7, –4)
covertices: (h ± b, k) = (–6, –1) and (–8, –1)
Graph the center, vertices, foci, and axes. Then use a table of values to sketch the ellipse.
6.x2 + 25y2 – 12x – 100y + 111 = 0
SOLUTION:
Complete the square for each variable to write the equation in standard form.
x2 + 25y2 – 12x – 100y + 111 = 0
x2 − 12x + 25y2 − 100y + 111 = 0
x2 − 12x + 36 + 25y2 − 100y + 100 + 111 = 0 + 36 + 100
x2 − 12x + 36 + 25(y2 − 4y + 4) + 111 = 136
(x − 6)2 + 25(y − 2)2 = 25
+(y − 2)2 = 1
The ellipse is in standard form, where h = 6, k = 2, a = or5,b = 1, and c = . The orientation is
horizontal because the x–term contains a2.
center: (h, k) = (6, 2)
foci: (h ± c, k) = (10.9, 2) and (1.1, 2)
vertices: (h ± a, k) = (11, 2) and (1, 2)
covertices: (h, k ± b) = (6, 3) and (6, 1)
Graph the center, vertices, foci, and axes. Then use a table of values to sketch the ellipse.
Write an equation for the ellipse with each set of characteristics.
7.vertices (–7, –3), (13, –3); foci (–5, –3), (11, –3)
SOLUTION:
Because the y–coordinates of the vertices are the same, the major axis is horizontal, and the standard form of the
equation is + =1.
The center is the midpoint of the segment between the vertices, or (3, –3). So, h = 3 and k = –3.
The distance between the vertices is equal to 2a units. So, 2a = 20, a = 10, and a2 = 100. The distance between the
foci is equal to 2c units. So, 2c = 16 and c = 8.
Use the values of a and c to find b.
So, b = 6 and b2 = 36.
Using the values of h, k, a, and b, the equation for the ellipse is + =1.
c2 = a2 – b2
82 = 102 – b2
b = 6
8.vertices (4, 3), (4, –9); length of minor axis is 8
SOLUTION:
Because the x–coordinates of the vertices are the same, the major axis is vertical, and the standard form of the
equation is + =1.
The center is the midpoint of the segment between the vertices, or (4, –3). So, h = 4 and k = –3.
The distance between the vertices is equal to 2a units. So, 2a = 12, a = 6, and a2 = 36. The length of the minor axis
is equal to 2b. So, 2b = 8, b = 4, and b2 = 16.
Using the values of h, k, a, and b, the equation for the ellipse is + =1
9.vertices (7, 2), (–3, 2); foci (6, 2), (–2, 2)
SOLUTION:
Because the y–coordinates of the vertices are the same, the major axis is horizontal, and the standard form of the
equation is + =1.
The center is the midpoint of the segment between the vertices, or (2, 2). So, h = 2 and k = 2.
The distance between the vertices is equal to 2a units. So, 2a = 10, a = 5, and a2 = 25. The distance between the
foci is equal to 2c units. So, 2c = 8 and c = 4.
Use the values of a and c to find b.
So, b = 3 and b2 = 9.
Using the values of h, k, a, and b, the equation for the ellipse is + =1.
c2 = a2 – b2
42 = 52 – b2
b = 3
10.major axis (–13, 2) to (1, 2); minor axis (–6, 4) to (–6, 0)
SOLUTION:
Because the y–coordinates of the major axis are the same, the major axis is horizontal, and the standard form of the
equation is + =1.
The length of the major axis is equal to 2a units. So, 2a = 14, a = 7, and a2=49.Thelengthoftheminoraxisis
equal to 2b units. So, 2b = 4, b = 2, and b2 = 4. The center of the ellipse is at the midpoint of the major axis, or (–6, 2).
So, h = –6 and k = 2.
Using the values of h, k, a, and b, the equation for the ellipse is + =1.
11.foci (–6, 9), (–6, –3); length of major axis is 20
SOLUTION:
Because the x–coordinates of the foci are the same, the major axis is vertical, and the standard form of the equation
is + =1.
The center is located at the midpoint of the foci, or (–6, 3). So, h = –6 and k = 3. The length of the major axis is 2a
units. So, 2a = 20, a = 10, and a2 = 100. The distance between the foci is equal to 2c units. So, 2c = 12 and c = 6.
Use the values of a and c to find b.
So, b = 8 and b2 = 64.
Using the values of h, k, a, and b, the equation for the ellipse is + =1.
c2 = a2 – b2
62 = 102 – b2
b = 8
12.co–vertices (–13, 7), (–3, 7); length of major axis is 16
SOLUTION:
Because the y–coordinates of the co–vertices are the same, the major axis is vertical, and the standard form of the
equation is + =1.
The center is located at the midpoint of the co–vertices, or (–8, 7). So, h = –8 and k = 7. The distance between the
co–vertices is equal to 2b units. So, 2b = 10, b = 5, and b2 = 25. The length of the major axis is equal to 2a units. So,
2a = 16, a = 8, and a2 = 64.
Using the values of h, k, a, and b, the equation for the ellipse is + =1.
13.foci (–10, 8), (14, 8); length of major axis is 30
SOLUTION:
Because the y–coordinates of the foci are the same, the major axis is horizontal, and the standard form of the
equation is + =1.
The midpoint is the distance between the foci, or (2, 8). So, h = 2 and k = 8. The distance between the foci is equal
to 2c units. So, 2c = 24 and c = 12. The length of the major axis is equal to 2a. So, 2a = 30, a = 15, and a2 = 225.
Use the values of a and c to find b.
So, b = 9 and b2 = 81.
Using the values of h, k, a, and b, the equation for the ellipse is + =1.
c2 = a2 – b2
122 = 152 – b2
b = 9
Determine the eccentricity of the ellipse given by each equation.
14. + =1
SOLUTION:
The ellipse is in standard form. So, a = orabout8.49andb = .
First, determine the value of c.
Use the values of a and c to find the eccentricity of the ellipse.
c2 = a2 − b2
c2 = 72 − 54
c = orabout4.24
e =
e = orabout0.5
15. + =1
SOLUTION:
The ellipse is in standard form. So, a = orabout6.32andb = .
First, determine the value of c.
Use the values of a and c to find the eccentricity of the ellipse.
c2 = a2 − b2
c2 = 40 − 12
c = orabout5.29
e =
e = orabout0.837
16. + =1
SOLUTION:
The ellipse is in standard form. So, a = orabout7.55andb = .
First, determine the value of c.
Use the values of a and c to find the eccentricity of the ellipse.
c2 = a2 − b2
c2 = 57 – 14
c = orabout6.56
e =
e = orabout0.869
17. + = 1
SOLUTION:
The ellipse is in standard form. So, a = orabout5.74andb = .
First, determine the value of c.
Use the values of a and c to find the eccentricity of the ellipse.
c2 = a2 − b2
c2 = 33 – 27
c = orabout2.45
e =
e = orabout0.426
18. + =1
SOLUTION:
The ellipse is in standard form. So, a = orabout3.46andb = 3.
First, determine the value of c.
Use the values of a and c to find the eccentricity of the ellipse.
c2 = a2 − b2
c2 = 12 – 9
c = orabout1.73
e =
e = orabout0.5
19. + =1
SOLUTION:
The ellipse is in standard form. So, a = orabout4.8andb = .
First, determine the value of c.
Use the values of a and c to find the eccentricity of the ellipse.
c2 = a2 − b2
c2 = 23 – 17
c = orabout2.45
e =
e = orabout0.511
20. + = 1
SOLUTION:
The ellipse is in standard form. So, a = orabout6.16andb = .
First, determine the value of c.
Use the values of a and c to find the eccentricity of the ellipse.
c2 = a2 − b2
c2 = 38 – 13
c = or5
e =
e = orabout0.811
21. + = 1
SOLUTION:
The ellipse is in standard form. So, a = orabout3.16andb = .
First, determine the value of c.
Use the values of a and c to find the eccentricity of the ellipse.
c2 = a2 − b2
c2 = 10 – 8
c = or1.41
e =
e = orabout0.447
22.RACING Thedesignofanellipticalracetrackwithaneccentricityof0.75isshown.
a. What is the maximum width w of the track?
b. Write an equation for the ellipse if the origin x is located at the center of the racetrack.
SOLUTION:
a. The length of the track is 1000 feet. So, 2a = 1000, a = 500, and a2 = 250,000. Because the eccentricty is given,
you can use the eccentricty equation to find c.
Use the values of a and c to find b.
So, the maximum width of the track is 2(330.72) or about 661.44 feet.
b. Use the values of a and b to write the equation of the ellipse.
e =
0.75 =
375 = c
b2 = a2 – c2
b2 = 5002 – 3752
b =
b ≈330.72
+
+
23.CARPENTRY A carpenter has been hired to construct a sign for a pet grooming business. The plans for the sign
call for an elliptical shape with an eccentricity of 0.60 and a length of 36 inches.
a. What is the height of the sign?
b. Write an equation for the ellipse if the origin is located at the center of the sign.
SOLUTION:
a. The length of the sign is 36 inches. So, 2a = 36, a = 18, and a2 = 324. Because the eccentricty is given, you can
use the eccentricty equation to find c.
Use the values of a and c to find b.
So, the height of the sign is 2(14.4) or about 28.8 inches.
b. Use the values of a and b to write the equation of the ellipse. b2 = 14.42.
e =
0.60 =
10.8 = c
b2 = a2 – c2
b2 = 182 – 10.82
b =
b ≈14.4
+
= 1
+ = 1
Write each equation in standard form. Identify the related conic.
24.x2 + y2 + 6x − 4y − 3 = 0
SOLUTION:
x2 + y2 + 6x − 4y − 3 = 0
x2 + 6x + y2 − 4y − 3 = 0
x2 + 6x + 9 + y 2 − 4y + 4 − 3 = 0 + 9 + 4
(x + 3)2 + (y − 2)2 = 16
Because a = b and the graph is of the form (x − h)2 + (y − k)2 = r2, the conic is a circle.
25.4x2 + 8y2 − 8x + 48y + 44 = 0
SOLUTION:
4x2 + 8y2 − 8x + 48y + 44 = 0
4x2 − 8x + 8y2 + 48y + 4 = 0
4(x2 − 2x + 1) + 8(y2 + 6y + 9) + 44 = 0 + 4 + 72
4(x − 1)2 + 8(y + 3)2 = 32
+ = 1
The related conic is an ellipse because a≠b and the graph is of the form + =1.
26.x2 − 8x − 8y − 40 = 0
SOLUTION:
x2 − 8x − 8y − 40 = 0
x2 − 8x + 16 − 8y − 40 = 0 + 16
(x − 4)2 = 8y + 56
(x − 4)2 = 8(y + 7)
Because only one term is squared, the graph is a parabola.
27.y2 − 12x + 18y + 153 = 0
SOLUTION:
y2 − 12x + 18y + 153 = 0
y2 + 18y + 153 = 12x
y2 + 18y + 81 + 153 = 12x + 81
(y + 9)2 = 12x − 72
(y + 9)2= 12(x − 6)
Because only one term is squared, the graph is a parabola.
28.x2 + y2 − 8x − 6y − 39 = 0
SOLUTION:
x2 + y2 − 8x − 6y − 39 = 0
x2 − 8x + y2 − 6y − 39 = 0
x2 − 8x + 16 + y2 − 6y + 9 − 39 = 0 + 16 + 9
(x − 4)2 + (y − 3)2 = 64
Because a = b, and the graph is of the form (x − h)2 + (y − k)2 = r2, the conic is a circle.
29.3x2 + y2 − 42x + 4y + 142 = 0
SOLUTION:
3x2 + y2 − 42x + 4y + 142 = 0
3x2 − 42x + y2 + 4y + 142 = 0
3(x2 − 14x + 49) + y2 + 4y + 4 + 142 = 0 + 147 + 4
3(x − 7)2 + (y + 2)2 = 9
+ = 1
The related conic is an ellipse because a≠b and the graph is of the form + =1.
30.5x2 + 2y2 + 30x − 16y + 27 = 0
SOLUTION:
5x2 + 2y2 + 30x − 16y + 27 = 0
5x2 + 30x + 2y2 − 16y + 27 = 0
5(x2 + 6x + 9) + 2(y2 − 8y + 16) + 27 = 0 + 45 + 32
5(x + 3)2 + 2(y − 4)2= 50
+ = 1
The related conic is an ellipse because a≠b, and the graph is of the form + =1.
31.2x2 + 7y2 + 24x + 84y + 310 = 0
SOLUTION:
The related conic is an ellipse because a≠b, and the graph is of the form + =1.
32.HISTORY The United States Capitol has a room with an elliptical ceiling. This type of room is called a whispering
gallery because sound that is projected from one focus of an ellipse reflects off the ceiling and back to the other
focus. The room in the Capitol is 96 feet in length, 45 feet wide, and has a ceiling that is 23 feet high.
a. Write an equation modeling the shape of the room. Assume that it is centered at the origin and that the major axis
is horizontal.
b. Find the location of the two foci.
c. How far from one focus would one have to stand to be able to hear the sound reflecting from the other focus?
SOLUTION:
a. The length of the room is 96 feet. So, 2a = 96, a = 48, and a2 = 2304. The width of the room is 45 feet. So, 2b =
45, b = 22.5, and b2 = 506.25.
Use the values of a and b to write the equation of the ellipse.
b. To find the location of the foci, find c.
So, the foci are about 42 ft on either side of the center along the major axis.
c. The length of the segment between the foci is 2c units. So, 2(42) is about 84 feet.
+
= 1
+ = 1
c2 = a2 – b2
c2 = 2304 – 506.25
c =
c ≈42.4
Write an equation for a circle that satisfies each set of conditions. Then graph the circle.
33.center at (3, 0), radius 2
SOLUTION:
The center is located at (3, 0), so h = 3 and k = 0. The radius is 2, so r2 = 4.
Use the values of h, k, and r to write the equation of the circle.
(x – h)2 + (y – k)2 = r2
(x – 3)2 + y2 = 4
34.center at (–1, 7), diameter 6
SOLUTION:
The center is located at (–1, 7), so h = –1 and k = 7. The diameter is 6, so r = 3, and r2 = 9.
Use the values of h, k, and r to write the equation of the circle.
(x – h)2 + (y – k)2 = r2
(x + 1)2 + (y – 7)2 = 9
35.center at (–4, –3), tangent to y = 3
SOLUTION:
The center is located at (–4, –3), so h = –4 and k = –3. The line y = 3 is a horizontal line. Because the circle is
tangent to this line and its center is at (–4, –3), one endpoint of the circle is at (–4, 3). So, the radius is 3 – (–3) = 6,
and r2 = 36.
36.center at (2, 0), endpoints of diameter at (–5, 0) and (9, 0)
SOLUTION:
The center is located at (2, 0), so h = 2 and k = 0. The diameter is | –5 – 9 | or 14, so r = 7, and r2 = 49.
Use the values of h, k, and r to write the equation of the circle.
(x – h)2 + (y – k)2 = r2
(x – 2)2 + y2 = 49
37.FORMULA Derive the general form of the equation for an ellipse with a vertical major axis centered at the origin.
SOLUTION:
38.MEDICAL TECHNOLOGY Indoor Positioning Systems (IPS) use ultrasound waves to detect tags that are
linked to digital files containing information regarding a person or item being monitored. Hospitals often use IPS to
detect the location of moveable equipment and patients.
a. If the tracking system receiver must be centrally located for optimal functioning, where should a receiver be
situated in a hospital complex that is 800 meters by 942 meters?
b. Write an equation that models the sonar range of the IPS.
SOLUTION:
a. The receiver should be situated at the central point of the hospital, which is 471 meters within the long side of the
building.
b. The building is shaped like a rectangle. In order to reach every part of the building, a circle must be drawn with a
center at the central point of the building. From the central point, the farthest point of the building is at a corner of
the rectangle. Use the Pythagorean Theorem to find this distance.
This distance is the radius of the circle that encloses the entire building. So, an equation that models the sonar range
of the IPS is x2 + y2 = 381,841.
a2 + b2 = c2
4002 + 4712 = c2
381,841 = c2
Write an equation for each ellipse.
39.
SOLUTION:
From the graph, you can see that the vertices are located at (–3, 3) and (5, 3) and the co–vertices are located at (1,
4) and (1, 2). The major axis is horizontal, so the standard form of the equation is + =1.
The center is the midpoint of the segment between the vertices, or (1, 3). So, h = 1 and k = 3.
The distance between the vertices is equal to 2a units. So, 2a = 8, a = 4, and a2 = 16. The distance between the co–
vertices is equal to 2b units. So, 2b = 2, b = 1, and b2 = 1.
Using the values of h, k, a, and b, the equation for the ellipse is + =1.
40.
SOLUTION:
From the graph, you can see that the vertices are located at (4, 6) and (4, –4) and the co–vertices are located at (1,
1) and (7, 1). The major axis is vertical, so the standard form of the equation is + =1.
The center is the midpoint of the segment between the vertices, or (4, 1). So, h = 4 and k = 1.
The distance between the vertices is equal to 2a units. So, 2a = 10, a = 5, and a2 = 25. The distance between the
co–vertices is equal to 2b units. So, 2b = 6, b = 3, and b2 = 9.
Using the values of h, k, a, and b, the equation for the ellipse is + =1.
41.
SOLUTION:
From the graph, you can see that the vertices are located at (2, 9) and (2, –5) and the co–vertices are located at (–2,
2) and (6, 2). The major axis is vertical, so the standard form of the equation is + =1.
The center is the midpoint of the segment between the vertices, or (2, 2). So, h = 2 and k = 2.
The distance between the vertices is equal to 2a units. So, 2a = 14, a = 7, and a2 = 49. The distance between the
co–vertices is equal to 2b units. So, 2b = 8, b = 4, and b2 = 16.
Using the values of h, k, a, and b, the equation for the ellipse is + =1.
42.
SOLUTION:
From the graph, you can see that the vertices are located at (–9, 5) and (7, 5) and the co–vertices are located at (–1,
11) and (–1, 1). The major axis is horizontal, so the standard form of the equation is + =1.
The center is the midpoint of the segment between the vertices, or (–1, 5). So, h = –1 and k = 5.
The distance between the vertices is equal to 2a units. So, 2a = 16, a = 8, and a2 = 64. The distance between the
co–vertices is equal to 2b units. So, 2b = 12, b = 6, and b2 = 36.
Using the values of h, k, a, and b, the equation for the ellipse is + = 1.
43.PLANETARY MOTIONEachoftheplanetsinthesolarsystemmovearoundtheSuninanellipticalorbit,where
the Sun is one focus of the ellipse. Mercury is 43.4 million miles from the Sun at its farthest point and 28.6 million
milesatitsclosest,asshownbelow.ThediameteroftheSunis870,000miles.
a. Find the length of the minor axis.
b. Find the eccentricity of the elliptical orbit.
SOLUTION:
a. Thelengthofthemajoraxisis43.4+28.6+0.87=72.87.
The value of ais72.87÷2=36.435.Thedistancefromthefocus(thesun)tothevertexis28.6+ =29.035.
Therefore, the value of c, the focus to the center, is 36.435 − 29.035 = 7.4.
In an ellipse, c2 = a2 − b2. Use the values of a and c to find b.
The value of 2b,thelengthoftheminoraxis,isabout35.676·2≈71.35millionmiles.
b. Use the values of a and c to find the eccentricity of the orbit.
So, the orbit has an eccentricity of about 0.203.
e =
=
≈0.203
Find the center, foci, and vertices of each ellipse.
44. + =1
SOLUTION:
The ellipse is in standard form, where h = –5 and k = 0. So, the center is located at (–5, 0). The ellipse has a
horizontal orientation, so a2 = 16, a = 4, and b2 = 7.
Use the values of a and b to find c.
The foci are c units from the center, so they are located at (–8, 0) and (–2, 0). The vertices are a units from the
center, so they are located at (–9, 0) and (–1, 0).
c2 = a2 − b2
c2 = 16 – 7
c2 = 9
c = 3
45. + =1
SOLUTION:
The ellipse is in standard form, where h = 0 and k = –6. So, the center is located at (0, –6). The ellipse has a
horizontal orientation, so a2 = 100, a = 10, and b2 = 25.
Use the values of a and b to find c.
The foci are c units from the center, so they are located at (±5, –6). The vertices are a units from the center, so
they are located at (±10, –6).
c2 = a2 − b2
c2 = 100 – 25
c2 = 75
c = or5
46.9y2 – 18y + 25x2 + 100x – 116 = 0
SOLUTION:
First, write the equation in standard form.
The equation is now in standard form, where k = 1 and h = –2. So, the center is located at (–2, 1). The ellipse has a
vertical orientation, so a2 = 25, a = 5, and b2 = 9.
Use the values of a and b to find c.
The foci are c units from the center, so they are located at (–2, 5), (–2, –3). The vertices are a units from the
center, so they are located at (–2, 6), (–2, –4).
c2 = a2 − b2
c2 = 25 – 9
c2 = 16
c = 4
47.65x2 + 16y2 + 130x – 975 = 0
SOLUTION:
First, write the equation in standard form.
The equation is now in standard form, where h = –1 and k = 0. So, the center is located at (–1, 0). The ellipse has a
vertical orientation, so a2 = 65, a = , and b2 = 16.
Use the values of a and b to find c.
The foci are c units from the center, so they are located at (–1, ±7). The vertices are a units from the center, so
they are located at (–1, ±).
c2 = a2 − b2
c2 = 65 – 16
c2 = 49
c = 7
48.TRUCKS Ellipticaltankertrucksliketheoneshownareoftenusedtotransportliquidsbecausetheyaremore
stable than circular tanks and the movement of the fluid is minimized.
a. Draw and label the elliptical cross–section of the tank on a coordinate plane.
b. Write an equation to represent the elliptical shape of the tank.
c. Find the eccentricity of the ellipse.
SOLUTION:
a. The major axis of the ellipse is 6.8 ft, so a = 3.4. The minor axis is 4.2 ft, so b = 2.1. The ellipse is centered about
the origin, so h = k = 0. The major axis is along the x–axis, so the ellipse is horizontal. The vertices are located at (0
± 3.4, 0) and the co–vertices are located at (0, 0 ± 2.1).
Plot the vertices and co–vertices to sketch the ellipse.
b. The orientation of the ellipse is horizontal, so the standard form of the equation is + =1.
Substitute the values of a and b into the equation.
c. Use the values of a and b to find c.
Substitute a and c into the eccentricity equation.
So, the eccentricity of the ellipse is about 0.79.
c2 = a2 − b2
c2 = 11.56 – 4.41
c2 = 7.15
c ≈2.67
e =
=
≈0.79
Write the standard form of the equation for each ellipse.
49.The vertices are at (–10, 0) and (10, 0) and the eccentricity e is .
SOLUTION:
The vertices are located at (–10, 0) and (10, 0), so 2a = 20, a = 10,and a2 = 100.
The eccentricity of the ellipse is given, so you can use the eccentricity equation to find c.
e =
=
6 = c
Use the values of a and c to find b
The y–term is the same for each vertex, so the ellipse is oriented horizontally. The center is the midpoint of the
vertices or (0, 0). So, h = 0 and k = 0. Therefore, the equation of the ellipse is + =1.
c2 = a2 − b2
62 = 102 − b2
36 = 100 – b2
64 = b2
50.The co–vertices are at (0, 1) and (6, 1) and the eccentricity e is .
SOLUTION:
The co–vertices are located at (0, 1) and (6, 1), so 2b = 6, b = 3, and b2 = 9. The center is located at the midpoint of
the co–vertices, or (3, 1). So, h = 3 and k = 1. The y–term is the same for each co–vertex, so the ellipse is oriented
vertically and the standard form of the equation is + =1.
Substitute e = intotheeccentricityequation,andsolveforc.
Substitute b and c into c2 = a2 – b2 to find a.
Therefore, the equation of the ellipse is + = 1
e =
=
a = c
c2 = a2 − b2
= a2 − 9
= a2 − 9
9 =
25 = a2
51.The center is at (2, –4), one focus is at (2, –4 + 2 ), and the eccentricity e is .
SOLUTION:
The center is located at (2, –4), so h = 2 and k = –4. The distance from a focus to the center is equal to c. Since one
focus is located at (2, –4 + 2 ), c = 2 . Since the x–coordinates of the center and a focus are the same, the
ellipse is oriented horizontally and the standard form of the equation is + =1.
Find a.
So, a = 6 and a2 = 36.
Find b.
Using the values of h, k, a, and b, the equation for the ellipse is + = 1.
e =
=
=
a = 6
c2 = a2 – b2
(2 )2 = 62 – b2
20 = 36 – b2
b2 = 16
52.ROLLER COASTERS Theshapeofarollercoasterloopinanamusementparkcanbemodeledby
+ =1.
a. What is the width of the loop along the horizontal axis?
b. Determine the height of the roller coaster from the ground when it reaches the top of the loop, if the lower rail is
20 feet from ground level.
c. Find the eccentricity of the ellipse.
SOLUTION:
a. The equation is in standard form and 2025 is associated with the x2–term and is less than 3306.25. So, b2 = 2025
and b=45.Thewidthoftheloopalongthehorizontallineis2·45or90ft.
b. Find a.
So, a = 57.5 and 2a = 115. Since the lower rail is 20 ft above ground level, the maximum height is 115 + 20 or 135 ft.
c. Find c.
Now you can find the eccentricity of the ellipse.
a2 =3306.25
a = orabout57.5
c2 = a2 – b2
c2 = 3306.25 – 2025
c = orabout35.79
e =
=
≈0.62
53.FOREST FIRESTheradiusofaforestfireisexpandingatarateof4milesperday.Thecurrentstateofthefire
is shown below, where a city is located 20 miles southeast of the fire.
a. Write the equation of the circle at the current time and the equation of the circle at the time the fire reaches the
city.
b. Graph both circles.
c. If the fire continues to spread at the same rate, how many days will it take to reach the city?
SOLUTION:
a. The radius of the small circle is 8. So, the equation is x2 + y2 = 64. When the fire reaches the city, the radius will
be 8 + 20 or 28. The equation for the circle is x2 + y2 = 784.
b. Use the center of (0, 0), r = 8, and r = 28 to graph both circles.
c. Theradiusofthefireisgrowing4milesperday.Itwillreachthecityif20÷4=5days.
54.The latus rectum of an ellipse is a line segment that passes through a focus, is perpendicular to the major axis of the
ellipse, and has endpoints on the ellipse. The length of each latus rectum is units,wherea is half the length of
the major axis and b is half the length of the minor axis.
Write the equation of a horizontal ellipse with center at (3, 2), major axis is 16 units long, and latus rectum 12 units
long.
SOLUTION:
The length of the major axis is equal to 16. So, 2a = 16 and a = 8, and a2 = 64. The length of the latus rectum is 12,
so 12 = .
Find b2.
The ellipse is oriented horizontally, so the standard form of the equation is + =1.Usingthe
values of h, k, a, and b, the equation for the ellipse is + = 1.
12 =
12 =
48 = b2
Find the coordinates of points where a line intersects a circle.
55.y = x – 8, (x – 7)2 + (y + 5)2 = 16
SOLUTION:
Solve the system of equations.
Substitute x = 3 and x = 7 back into the first equation to find the y–coordinate for each point of intersection.
So, the coordinates of the intersection are (7, –1) and (3, –5).
(x – 7)2 + (y + 5)2 = 16
(x − 7)2 + (x − 8 + 5)2 = 16
x2 − 14x + 49 + x2 − 6x + 9 = 16
2x2 − 20x + 42 = 0
x2 − 10x + 21 = 0
(x − 7)(x − 3) = 0
x = 7 or 3
y = x – 8
= 7 – 8 or –1
y = x – 8
= 3 – 8 or –5
56.y = x + 9, (x – 3)2 + (y + 5)2 = 169
SOLUTION:
Solve the system of equations.
(x – 3)2 + (y + 5)2 = 169
(x − 3)2 + (x + 9 + 5)2 = 169
x2 − 6x + 9 + x2 + 28x + 196 = 169
2x2 + 22x + 36 = 0
x2 + 11x + 18 = 0
(x + 9)(x + 2) = 0
x= –9 or –2
Substitute x = –9 and x = –2 back into the first equation to find the y–coordinate for each point of intersection.
So, the coordinates of the intersection are (–2, 7) and (–9, 0).
y = x + 9
= –9 + 9 or 0
y = x + 9
= –2 + 9 or 7
57.y = –x + 1, (x – 5)2 + (y – 2)2 = 50
SOLUTION:
Solve the system of equations.
(x – 5)2 + (y – 2)2 = 50
(x − 5)2 + (–x + 1 − 2)2 = 50
x2 − 10x + 25 + x2 + 2x + 1 = 50
2x2 − 8x − 24 = 0
x2 − 4x − 12 = 0
(x − 6)(x + 2) = 0
x= 6 or –2
Substitute x = 6 and x = –2 back into the first equation to find the y–coordinate for each point of intersection.
So, the coordinates of the intersection are (–2, 3) and (6, –5).
y = –x + 1
= –6 + 1 or –5
y = –x + 1
= – (–2) + 1 or 3
58. , (x + 3)2 + (y – 3)2 = 25
SOLUTION:
Solve the system of equations.
(x + 3)2 + (y – 3)2 = 25
(x + 3)2 + = 25
x2 + 6x + 9 + x2 + 4x + 36 = 25
x2 + 10x + 20 = 0
10x2 + 90x + 180 = 0
x2 + 9x + 18 = 0
(x + 6)(x + 3) = 0
x= –6 or –3
Substitute x = –6 and x = –3 back into the first equation to find the y–coordinate for each point of intersection.
So, the coordinates of the intersection are (–3, –2) and (–6, –1).
y =
= 2 – 3 or –1
y =
= 1 – 3 or –2
59.REFLECTION Silvering is the process of coating glass with a reflective substance. The interior of an ellipse can
be silvered to produce a mirror with rays that originate at the ellipse's focus and then reflect to the other focus, as
shown. If the segment V1F1 is 2 cm long and the eccentricity of the mirror is 0.5, find the equation of the ellipse in
standard form.
SOLUTION:
If V1F1 is 2 cm, then the distance from the vertex to the center is 2 cm greater than the distance from the focus to
the center, and a = c + 2. If the eccentricity is 0.5, then =0.5.
Solve the system of equations.
By substituting c = 2 back into one of the original equations, you can find that a = 4. Use the values of a and c to
find b.
So, the equation of the ellipse is + = 1.
= 0.5
c = 0.5c + 1
0.5c = 1
c = 2
c2 = a2 – b2
22 = 42 – b2
4 = 16 – b2
b2 = 12
60.CHEMISTRY Distillationcolumnsareusedtoseparatechemicalsubstancesbasedonthedifferencesintheir
rates of evaporation. The columns may contain plates with bubble caps or small circular openings.
a. Write an equation for the plate shown, assuming that the center is at (−4, −1).
b. What is the surface area of the plate not covered by bubble caps if each cap is 2 inches in diameter?
SOLUTION:
a. The plate is circular with center (–4, –1). The diameter is 18 inches, so the radius r is 9 and r2 is 81. The equation
of the circle is (x + 4)2 + (y + 1)2 = 81.
b. The area of the plate is π·92 or 81π. The area of a cap is π·12 or π. There are 27 caps, so the area not
covered by caps is 81π − 27πor about 169.6 in2.
61.GEOMETRY The graphs of x – 5y = –3, 2x + 3y = 7, and 4x – 7y = 27 contain the sides of a triangle. Write the
equation of a circle that circumscribes the triangle.
SOLUTION:
Graph the triangle.
The equation of the circle that circumscribes the triangle is found by locating the circumcenter of the triangle.
Locate the perpendicular bisectors of each side of the triangle.
Find the midpoints of each segment
(2, 1) and (12, 3)
(5, –1) and (12, 3)
(2, 1) and (5, –1)
Thus, the midpoints of the segments are (8.5, 1), (3.5, 0), and (7, 2).
Find the equations for the perpendicular bisectors.
Find the slope of segment through points (2, 1) and (12, 3).
m⊥ is –5.
Find equation of line perpendicular through (7, 2)
Find the slope of segment through points (5, –1)and(12,3).
m⊥ is .
Find equation of line perpendicular through (8.5, 1).
Find the slope of segment through points (2, 1) and (5, –1).
m⊥ is .
Find equation of line perpendicular through (3.5, 0).
Thus, the equations of the perpendicular bisectors are:
y = –x +
y = x −
y = –5x + 37
Find the intersection of the perpendicular bisectors to find the circumcenter.
y = –x + &y = x −
y = x − & y = –5x + 37
y = –x + and y = –5x + 37.
Thus, these three lines intersect at the circumcenter of the triangle, (6.5, 4.5).
(6.5, 4.5) is the center of the circle.
The radius of the circle is the distance between this point and any vertex
Usethedistanceformulatofindtheradius.
(6.5, 4.5) and (12, 3)
(6.5, 4.5) and (5, –1)
(6.5, 4.5) and (2, 1)
Thus, the radius is andr2 = 32.5. Then equation of the circle is (x – 6.5)2 + (y – 4.5)2 = 32.5.
Write the standard form of the equation of a circle that passes through each set of points. Then identify
the center and radius of the circle.
62.(2, 3), (8, 3), (5, 6)
SOLUTION:
The standard form of the equation for a circle is (x − h)2 + (y − k)2 = r2. Substitute each point into the equation to
obtain three equations.
(2 − h)2 + (3 − k)2 = r2
(8 – h)2 + (3 − k)2 = r2
(5 – h)2 + (6 − k)2 = r2
Since (2 − h)2 + (3 − k)2 = r2 and (8 – h)2 + (3 − k)2 = r2 both have the same y–coordinate, you can solve that
system of equations to find h.
Next, substitute h = 5 into two of the equations to get a system of two equations with k and r.
Solve these two new equations for k.
Substitute h = 5 and k = 3 into any of the original equations to find r.
Since the radius must be positive, r = 3.
The equation of the circle is (x – 5)2 + (y – 3)2 = 9, with center (5, 3) and radius r = 3.
63.(1, –11), (–3, –7), (5, –7)
SOLUTION:
The standard form of the equation for a circle is (x − h)2 + (y − k)2 = r2. Substitute each point into the equation to
obtain three equations.
(1 − h)2 + (–11 − k)2 = r2
(–3 – h)2 + (–7 − k)2 = r2
(5 – h)2 + (–7 − k)2 = r2
Since (–3 – h)2 + (–7 − k)2 = r2 and (5 – h)2 + (–7 − k)2 = r2 both have the same y–coordinate, you can solve that
system of equations to find h.
Next, Substitute h = 1 into tow equations to get a system of two equations with kandr.
Solve these two new equations for k.
Substitute h = 1 and k = –7 into any of the equations to find r.
Since the radius must be positive, r = 4.
So, the equation of the circle is (x – 1)2 + (y + 7)2 = 16, with center (1, –7) and radius r = 4.
64.(0, 9), (0, 3), (–3, 6)
SOLUTION:
The standard form of the equation for a circle is (x − h)2 + (y − k)2 = r2. Substitute each point into the equation to
obtain three equations.
(0 − h)2 + (9 − k)2 = r2 or h2 + (9 − k)2 = r2
(0 − h)2 + (3 − k)2 = r2 or h2 + (3 − k)2 = r2
(–3 – h)2 + (6 − k)2 = r2
Since h2 + (9 − k)2 = r2 and h2 + (9 − k)2 = r2 both have 0 as a y–coordinate, you can solve that system of
equations to find k.
Next, substitute k = 6 into two of the equations to get a system of two equations in h and r.
Solve these two new equations for h.
Substitute h = 0 and k = 6 into any of the original equations to find r.
Since the radius must be positive, r = 3.
So, the equation of the circle is x2 + (y – 6)2 = 9, with center (0, 6) and radius r = 3.
65.(7, 4), (–1, 12), (–9, 4)
SOLUTION:
The standard form of the equation for a circle is (x − h)2 + (y − k)2 = r2. Substitute each point into the equation to
obtain three equations.
(7 − h)2 + (4 − k)2 = r2
(–1 − h)2 + (12 − k)2 = r2
(–9 – h)2 + (4 − k)2 = r2
Since (7 − h)2 + (4 − k)2 = r2 and (–9 – h)2 + (4 − k)2 = r2 both have the same y–coordinate, you can solve that
system of equations to find h.
Next, substitute h = –1 into two equations to get a system of two equations with k and r.
Solve these two new equations for k.
Substitute h = –1 and k = 4 into any of the original equations to find r.
Since the radius must be positive, r = 8.
So, the equation of the circle is (x + 1)2 + (y – 4)2 = 64, with center (–1, 4) and radius r = 8.
66.ERRORANALYSIS Yori and Chandra are graphing an ellipse that has a center at (–1, 3), a major axis of length
8, and a minor axis of length 4. Is either of them correct? Explain your reasoning.
SOLUTION:
Both Yori and Chandra are correct. In Yori’s graph, the major axis is horizontal. In Chandra’s graph, the major axis
is vertical.
67.REASONING Determine whether an ellipse represented by + =1,wherer > 0, will have the same
foci as the ellipse represented by + =1.Explainyourreasoning.
SOLUTION:
No, they will not have the same foci.
Sample answer:
For , p > 0 and r > 0, so p + r > p. Then is of the form and is
elongated vertically with a2 = p + r and b2 = p. Use these values to find c and the foci.
So, the foci are found upanddownfromthecenter(0,0).Therefore,thefociare .
For ,
p > 0 and r > 0, so p + r > p. Then is of the form and is
elongated horizontally with a2 = p + r and b2 = p. Use these values to find c and the foci.
So, the foci are found leftandrightfromthecenter(0,0).Therefore,thefociare .
CHALLENGE The area A of an ellipse of the form + =1isA = πab. Write an equation of an ellipse with
each of the following characteristics.
68.b + a = 12, A = 35π
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7-2 Ellipses and Circles
