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656 CHAPTER 10 Analytic Geometry
In Problems 79– 82, use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion of a planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from the Sun is the length of the semimajor axis of the elliptical orbit. See the illustration.
Major axis Sun
Center
Aphelion Perihelion
Mean distance
79. Earth The mean distance of Earth from the Sun is 93 million miles. If the aphelion of Earth is 94.5 million miles, what is the perihelion? Write an equation for the orbit of Earth around the Sun. 80. Mars The mean distance of Mars from the Sun is 142 million miles. If the perihelion of Mars is 128.5 million miles, what is the aphelion? Write an equation for the orbit of Mars about the Sun. 81. Jupiter The aphelion of Jupiter is 507 million miles. If the distance from the center of its elliptical orbit to the Sun is 23.2 million miles, what is the perihelion? What is the mean distance? Write an equation for the orbit of Jupiter around the Sun. 82. Pluto The perihelion of Pluto is 4551 million miles, and the distance from the center of its elliptical orbit to the Sun is 897.5 million miles. Find the aphelion of Pluto. What is the mean distance of Pluto from the Sun? Write an equation for the orbit of Pluto about the Sun. 83. Show that an equation of the form Ax^2 + Cy^2 + F = 0, A 0, C 0, F 0
where A and C are of the same sign and F is of opposite sign, (a) is the equation of an ellipse with center at 1 0, 0 2 if A C. (b) is the equation of a circle with center 1 0, 0 2 if A = C.
84. Show that the graph of an equation of the form
Ax^2 + Cy^2 + Dx + Ey + F = 0, A 0, C 0 where A and C are of the same sign,
(a) is an ellipse if
D^2 4 A
E^2 4 C
(b) is a point if
D^2 4 A
E^2 4 C
(c) contains no points if
D^2 4 A
E^2 4 C
- F is of opposite sign to A.
Explaining Concepts: Discussion and Writing
85. The eccentricity e of an ellipse is defined as the number
c a
, where a is the distance of a vertex from the center and c is the distance of a focus from the center. Because a 7 c , it follows that e 6 1. Write a brief paragraph about the general shape of each of the following ellipses. Be sure to justify your conclusions. (a) Eccentricity close to 0 (b) Eccentricity = 0.5 (c) Eccentricity close to 1
’Are You Prepared?’ Answers
1. 213 2.
9 4
3. 1 - 2, 0 2 , 1 2, 0 2 , 1 0, - 42 , 1 0, 4 2 4. 1 2, 5 2 5. left; 1; down: 4 6. 1 x - 222 + 1 y + 322 = 1
Now Work the ‘Are You Prepared?’ problems on page 667.
OBJECTIVES 1 Analyze Hyperbolas with Center at the Origin (p. 657) 2 Find the Asymptotes of a Hyperbola (p. 662 ) 3 Analyze Hyperbolas with Center at ( h, k ) (p. 663 ) 4 Solve Applied Problems Involving Hyperbolas (p. 665 )
- Distance Formula ( Section 1.1, p. 5 )
- Completing the Square (Appendix A, Section A.3, pp. A28–A29)
- Intercepts ( Section 1.2, pp. 18–19 )
- Symmetry ( Section 1.2, pp. 19–21 )
- Asymptotes (Section 4.4, pp. 218–224)
- Graphing Techniques: Transformations ( Section 2.5, pp. 104–112)
- Square Root Method (Appendix A, Section A.6, p. A47)
PREPARING FOR THIS SECTION Before getting started, review the following:
10.4 The Hyperbola
TEKS 3.H
SECTION 10.4 The Hyperbola 657
Figure 34 illustrates a hyperbola with foci F 1 and F 2. The line containing the foci is called the transverse axis. The midpoint of the line segment joining the foci is the center of the hyperbola. The line through the center and perpendicular to the transverse axis is the conjugate axis. The hyperbola consists of two separate curves, called branches, that are symmetric with respect to the transverse axis, conjugate axis, and center. The two points of intersection of the hyperbola and the transverse axis are the vertices, V 1 and V 2 , of the hyperbola.
Analyze Hyperbolas with Center at the Origin With these ideas in mind, we are now ready to find the equation of a hyperbola in the rectangular coordinate system. First, place the center at the origin. Next, position the hyperbola so that its transverse axis coincides with a coordinate axis. Suppose that the transverse axis coincides with the x -axis, as shown in Figure 35. If c is the distance from the center to a focus, one focus is at F 1 = 1 - c , 0 2 and the other at F 2 = 1 c , 0 2. Now we let the constant difference of the distances from any point P = 1 x , y 2 on the hyperbola to the foci F 1 and F 2 be denoted by { 2 a. (If P is on the right branch, the + sign is used; if P is on the left branch, the - sign is used.) The coordinates of P must satisfy the equation
d 1 F 1 , P 2 - d 1 F 2 , P 2 = { 2 a
21 x - 1 - c 222 + y^2 - 21 x - c 22 + y^2 = { 2 a
21 x + c 22 + y^2 = { 2 a + 21 x - c 22 + y^2
1 x + c 22 + y^2 = 4 a^2 { 4 a 21 x - c 22 + y^2
Next we remove the parentheses.
x^2 + 2 cx + c^2 + y^2 = 4 a^2 { 4 a 21 x - c 22 + y^2 + x^2 - 2 cx + c^2 + y^2
4 cx - 4 a^2 = { 4 a 21 x - c 22 + y^2
cx - a^2 = { a 21 x - c 22 + y^2
1 cx - a^22
2 = a^2 3 1 x - c 22 + y^2
c^2 x^2 - 2 ca^2 x + a^4 = a^2 1 x^2 - 2 cx + c^2 + y^22
c^2 x^2 + a^4 = a^2 x^2 + a^2 c^2 + a^2 y^2
1 c^2 - a^22 x^2 - a^2 y^2 = a^2 c^2 - a^4
1 c^2 - a^22 x^2 - a^2 y^2 = a^2 1 c^2 - a^22 (1)
To obtain points on the hyperbola off the x -axis, it must be that a 6 c. To see why, look again at Figure 35.
d 1 F 1 , P 2 6 d 1 F 2 , P 2 + d 1 F 1 , F 22
d 1 F 1 , P 2 - d 1 F 2 , P 2 6 d 1 F 1 , F 22 2 a 6 2 c
a 6 c
1
Difference of the distances from P to the foci equals { 2 a. Use the Distance Formula.
Isolate one radical.
Square both sides.
Simplify; isolate the radical.
Divide each side by 4.
Square both sides.
Simplify.
Distribute and simplify.
Rearrange terms.
Factor a^2 on the right side.
Use triangle F 1 PF 2.
P is on the right branch, so d ( F 1 , P ) - d ( F 2 , P ) = 2 a ; d 1 F 1 , F 22 = 2 c.
Figure 34
F 2
Transverse axis
Conjugate axis
F 1
V 1 Center
V 2
Figure 35 d 1 F 1 , P 2 - d 1 F 2 , P 2 = { 2 a
x
y
P ( x , y ) Transverse axis d ( F^2 ,^ P^ )
d ( F 1 , P )
F 1 ( c , 0) F 2 ( c , 0)
A hyperbola is the collection, or locus, of all points in the plane, the difference of whose distances from two fixed points, called the foci, is a constant.
DEFINITION
EEEEEELPELPEELPS SSSS (^) ELPS 4.F.
After reading the definition and description of a hyperbola on page 657, review Figure 2(d) on page 635. Use the figure as visual support to tell a partner the location of the transverse axis, conjugate axis, branches, and vertices. Refer back to the descriptions on page 635 for contextual support as you identify each element.
SECTION 10.4 The Hyperbola 659
y^2 =
y = {
The points above and below the foci are a {3,
b and a {3, -
b. These points
determine the “opening” of the hyperbola. See Figure 37.
Using a Graphing Utility to Graph a Hyperbola
Using a graphing utility, graph the hyperbola:
x^2 4
y^2 5
To graph the hyperbola
x^2 4
y^2 5
= 1, we need to graph the two functions
Y 1 = 25
B
x^2 4
x^2 4
- As with graphing circles, parabolas, and ellipses on a graphing utility, we use a square screen setting so that the graph is not distorted. Figure 38 shows the graph of the hyperbola.
Now Work P R O B L E M 1 9
An equation of the form of equation (2) is the equation of a hyperbola with center at the origin, foci on the x -axis at 1 - c , 0 2 and 1 c , 0 2 , where c^2 = a^2 + b^2 , and transverse axis along the x -axis. For the next two examples of this section, the direction “Analyze the equation” will mean to find the center, transverse axis, vertices, and foci of the hyperbola and graph it.
Analyzing the Equation of a Hyperbola
Analyze the equation:
x^2 16
y^2 4
The given equation is of the form of equation (2), with a^2 = 16 and b^2 = 4. The graph of the equation is a hyperbola with center at 1 0, 0 2 and transverse axis along the x -axis. Also, we know that c^2 = a^2 + b^2 = 16 + 4 = 20. The vertices are at 1 { a , 0 2 = 1 {4, 0 2 , and the foci are at 1 { c , 0 2 = 1 { 21 5, 0 2. To locate the points on the graph above and below the foci, we let x = { 21 5. Then
x^2 16
y^2 4
1 { 22522 16
y^2 4
y^2 4
y^2 4
y^2 4
y = { 1
EXAMPLE 2
Solution
EXAMPLE 3
Solution
x = { 225
Figure 37
x
y
F 1 (3, 0) F (^) 2 (3, 0)
5 5
5
5
V 2 (2, 0)
( 3,)
V 1 (2, 0)
(^5) – 2
( 3,^5 – 2 ) ( 3,^5 – 2 )
( 3,^5 – 2 )
Figure 38
4
4
6 6
660 CHAPTER 10 Analytic Geometry
Figure 40 shows the graph of a typical hyperbola defined by equation (3).
An equation of the form of equation (2),
x^2 a^2
y^2 b^2
= 1, is the equation of a hyperbola with center at the origin, foci on the x -axis at 1 - c , 0 2 and 1 c , 0 2 , where c^2 = a^2 + b^2 , and transverse axis along the x -axis.
An equation of the form of equation (3),
y^2 a^2
x^2 b^2
= 1, is the equation of a hyperbola with center at the origin, foci on the y -axis at 1 0, - c 2 and 1 0, c 2 , where c^2 = a^2 + b^2 , and transverse axis along the y -axis. Notice the difference in the forms of equations (2) and (3). When the y^2 @term is subtracted from the x^2 @term, the transverse axis is along the x -axis. When the x^2 @term is subtracted from the y^2 @term, the transverse axis is along the y -axis.
Analyzing the Equation of a Hyperbola
Analyze the equation: y^2 - 4 x^2 = 4
To put the equation in proper form, divide each side by 4:
y^2 4
Since the x^2 @term is subtracted from the y^2 @term, the equation is that of a hyperbola with center at the origin and transverse axis along the y -axis. Comparing the above equation to equation (3), we find a^2 = 4, b^2 = 1, and c^2 = a^2 + b^2 = 5. The vertices are at 1 0, { a 2 = 1 0, { 22 , and the foci are at 1 0, { c 2 = 1 0, { 152.
EXAMPLE 4
Solution
Equation of a Hyperbola; Center at (0, 0); Transverse Axis along the y -Axis
An equation of the hyperbola with center at 1 0, 0 2 , foci at 1 0, - c 2 and 1 0, c 2 , and vertices at 1 0, - a 2 and 1 0, a 2 is
y^2 a^2
x^2 b^2
= 1 where b^2 = c^2 - a^2 (3)
The transverse axis is the y -axis.
THEOREM
Figure 40
y^2 a^2
x^2 b^2
= 1, b^2 = c^2 - a^2
x V 1 (0, a )
V 2 (0, a )
y F 2 (0, c )
F (^) 1 (0, c )
Figure 39
x
V 1 = (–4, 0)
V 2 = (4, 0)
y
5
4
(–2 5 , 1)
(–2 5 , –1)
(2 , 1)
(2 5 , –1)
(a)
5.
8 8
(b)
5
F 1 = (–2 5 , 0) F 2 = (2^5 , 0)
The points above and below the foci are 1 { 21 5, 1 2 and 1 { 21 5, - 12. See Figure 39(a) for the graph drawn by hand. Figure 39(b) shows the graph obtained
using a graphing utility, where Y 1 = B
4 a
x^2 16
4 a
x^2 16
662 CHAPTER 10 Analytic Geometry
Find the Asymptotes of a Hyperbola Recall from Section 4.4 that a horizontal or oblique asymptote of a graph is a line with the property that the distance from the line to points on the graph approaches 0 as x S - or as x S^ . Asymptotes provide information about the end behavior of the graph of a hyperbola.
2
Asymptotes of a Hyperbola
The hyperbola
x^2 a^2
y^2 b^2
= 1 has the two oblique asymptotes
y =
b a
x and y = -
b a
x (4)
THEOREM
Proof We begin by solving for y in the equation of the hyperbola.
x^2 a^2
y^2 b^2
y^2 b^2
x^2 a^2
y^2 = b^2 ¢
x^2 a^2
Since x 0, we can rearrange the right side in the form
y^2 =
b^2 x^2 a^2
a^2 x^2
y = {
bx a B
a^2 x^2
Now, as x S - or as x S^ , the term
a^2 x^2
approaches 0, so the expression under the
radical approaches 1. So, as x S - or as x S^ , the value of y approaches {
bx a
that is, the graph of the hyperbola approaches the lines
y = -
b a
x and y =
b a
x
These lines are oblique asymptotes of the hyperbola.
The asymptotes of a hyperbola are not part of the hyperbola, but they do serve as a guide for graphing a hyperbola. For example, suppose that we want to graph the equation
x^2 a^2
y^2 b^2
Begin by plotting the vertices 1 - a , 0 2 and 1 a , 0 2. Then plot the points 1 0, - b 2 and 1 0, b 2 and use these four points to construct a rectangle, as shown in Figure 43.
The diagonals of this rectangle have slopes
b a
and -
b a
, and their extensions are the
asymptotes y =
b a
x and y = -
b a
x of the hyperbola. If we graph the asymptotes, we
can use them to establish the “opening” of the hyperbola and avoid plotting other points.
Figure 43
x
y
(0, b )
(0, b ) V 1 ( a , 0)
V 2 ( a , 0)
y b – ax
y b – ax
SECTION 10.4 The Hyperbola 663
Asymptotes of a Hyperbola
The hyperbola
y^2 a^2
x^2 b^2
= 1 has the two oblique asymptotes:
y =
a b
x and y = -
a b
x (5)
THEOREM
You are asked to prove this result in Problem 84. For the remainder of this section, the direction “Analyze the equation” will mean to find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola and graph it.
Analyzing the Equation of a Hyperbola
Analyze the equation: 9 x^2 - 4 y^2 = 36
Divide each side of the equation by 36 to put the equation in proper form.
x^2 4
y^2 9
We now proceed to analyze the equation. The center of the hyperbola is the origin. Since the x^2 @term is first in the equation, we know that the transverse axis is along the x -axis and the vertices and foci lie on the x -axis. Using equation (2), we find a^2 = 4, b^2 = 9, and c^2 = a^2 + b^2 = 13. The vertices are a = 2 units left and right of the center at 1 { a , 0 2 = 1 {2, 0 2 , the foci are c = 1 13 units left and right of the center at 1 { c , 0 2 = 1 { 1 13, 0 2 , and the asymptotes have the equations
y =
b a
x =
x and y = -
b a
x = -
x
To graph the hyperbola by hand, form the rectangle containing the points 1 { a , 0 2 and 1 0, { b 2 , that is, 1 - 2, 0 2 , 1 2, 0 2 , 1 0, - 32 , and 1 0, 3 2. The extensions of the diagonals of this rectangle are the asymptotes. See Figure 44(a) for the graph drawn by hand. Figure 44(b) shows the graph obtained using a graphing utility.
EXAMPLE 6
Solution
Figure 44
x
y
5 5
5 (0, 3)
(0, 3) 5
V 1 (2, 0) (^) V (2, 0)
(a)
2
y ^3 – 2 x y 3 – 2 x
F 1 F 2
Y 3 1.5 x Y 4 1.
x^2 3 4 1 Y 1
6
x
Y 1 x
2 3 4 1
9.1 9.
(b)
6 x Y 2 3 4 1
2
Seeing the Concept
Refer to Figure 44(b). Create a TABLE using Y 1 and Y 4 with x = 10, 100, 1000, and 10,000. Compare the values of Y 1 and Y 4. Repeat for Y 2 and Y 3. Now, create a TABLE using Y 1 and Y 3 with x 10, 100, 1000, and 10,000. Repeat for Y 2 and Y 4.
Now Work P R O B L E M 3 1
Analyze Hyperbolas with Center at ( h, k ) If a hyperbola with center at the origin and transverse axis coinciding with a coordinate axis is shifted horizontally h units and then vertically k units, the result is a hyperbola with center at 1 h , k 2 and transverse axis parallel to a coordinate axis. The equations of such hyperbolas have the same forms as those given in
3
SECTION 10.4 The Hyperbola 665
Complete the squares in x and in y.
- x^2 + 4 y^2 - 2 x - 16 y + 11 = 0
- 1 x^2 + 2 x 2 + 41 y^2 - 4 y 2 = - 11
- 1 x^2 + 2 x + 12 + 41 y^2 - 4 y + 42 = - 11 - 1 + 16
- 1 x + 122 + 41 y - 222 = 4
1 y - 222 -
1 x + 122 4
This is the equation of a hyperbola with center at 1 - 1, 2 2 and transverse axis parallel to the y -axis. Also, a^2 = 1 and b^2 = 4, so c^2 = a^2 + b^2 = 5. Since the transverse axis is parallel to the y -axis, the vertices and foci are located a and c units above and below the center, respectively. The vertices are at 1 h , k { a 2 = 1 - 1, 2 { 12 , or 1 - 1, 1 2 and 1 - 1, 3 2. The foci are at 1 h , k { c 2 = 1 - 1, 2 { 252. The asymptotes
are y - 2 =
1 x + 12 and y - 2 = -
1 x + 12. Figure 47(a) shows the graph drawn by hand. Figure 47(b) shows the graph obtained using a graphing utility.
Now Work P R O B L E M 5 5
Solve Applied Problems Involving Hyperbolas Look at Figure 48. Suppose that three microphones are located at points O 1 , O 2 , and O 3 (the foci of the two hyperbolas). In addition, suppose that a gun is fired at S and the microphone at O 1 records the gunshot 1 second after the microphone at O 2. Because sound travels at about 1100 feet per second, we conclude that the microphone at O 1 is 1100 feet farther from the gunshot than O 2. We can model this situation by saying that S lies on a branch of a hyperbola with foci at O 1 and O 2. (Do you see why? The difference of the distances from S to O 1 and from S to O 2 is the constant 1100.) If the third microphone at O 3 records the gunshot 2 seconds after O 1 , then S lies on a branch of a second hyperbola with foci at O 1 and O 3. In this case, the constant difference is 2200. The intersection of the two hyperbolas identifies the location of S.
Lightning Strikes
Suppose that two people standing 1 mile apart both see a flash of lightning. After a period of time, the person standing at point A hears the thunder. One second later, the person standing at point B hears the thunder. If the person at B is due west of the person at A and the lightning strike is known to occur due north of the person standing at point A , where did the lightning strike occur?
Solution
Group terms.
Complete each square.
Divide each side by 4.
4
EXAMPLE 9
Figure 47 Transverse
V 2 (1, 3) 6
( x 1) 2 Y 1 2 4 1
2
7.1 5.
4 Y ( x^ ^ 1) 2 ^2 ^ ^1
(a) (b)
x
y
5
axis
5
V 1 (1, 1)
(3, 2) (1, 2)
2
5
F 2 (1, 2 5 )
F 1 (1, 2 5 )
Figure 48
O 3 O 1 O 2
S
666 CHAPTER 10 Analytic Geometry
See Figure 49 in which the ordered pair 1 x , y 2 represents the location of the lightning strike. We know that sound travels at 1100 feet per second, so the person at point A is 1100 feet closer to the lightning strike than the person at point B. Since the difference of the distance from 1 x , y 2 to B and the distance from 1 x , y 2 to A is the constant 1100, the point 1 x , y 2 lies on a hyperbola whose foci are at A and B.
An equation of the hyperbola is
x^2 a^2
y^2 b^2
where 2 a = 1100, so a = 550. Because the distance between the two people is 1 mile (5280 feet) and each person is at a focus of the hyperbola, we have
2 c = 5280
c =
Since b^2 = c^2 - a^2 = 26402 - 5502 = 6,667,100, the equation of the hyperbola that describes the location of the lightning strike is
x^2 5502
y^2 6,667,
Refer to Figure 49. Since the lightning strike occurred due north of the individual at the point A = 1 2640, 0 2 , we let x = 2640 and solve the resulting equation.
26402 5502
y^2 6,667,
y^2 6,667,
y^2 = 146,942, y = 12,
The lightning strike occurred 12,122 feet north of the person standing at point A.
Check: The difference between the distance from 1 2640, 12,122 2 to the person at the point B = 1 - 2640, 0 2 and the distance from 1 2640, 12,122 2 to the person at the point A = 1 2640, 0 2 should be 1100. Using the distance formula, we find the difference in the distances is
as required.
Now Work P R O B L E M 7 5
Solution
Subtract
26402 5502
from both sides.
Multiply both sides by - 6,667,100. y 7 0 since the lightning strike occurred in quadrant I.
Figure 49 North
1 mile 5280 feet
East B (2640, 0) ( a,^ o)^ ( a,^ o)^ A (2640, 0)
( x, y )
332640 - 1 - 2640 2 4^2 + 1 12,122 - 022 - 312640 - 264022 + 1 12,122 - 022 = 1100
668 CHAPTER 10 Analytic Geometry
In Problems 29–36, find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation by hand. Verify your graph using a graphing utility.
29.
x^2 25
y^2 9
= 1 30.
y^2 16
x^2 4
= 1 31. 4 x^2 - y^2 = 16 32. 4 y^2 - x^2 = 16
33. y^2 - 9 x^2 = 9 34. x^2 - y^2 = 4 35. y^2 - x^2 = 25 36. 2 x^2 - y^2 = 4
In Problems 37–40, write an equation for each hyperbola.
37.
x
y
3 3
3
3
y x (^) y x 38.
x
y
3 3
3
(^3) y x
y x 39.
x
y
5 5
10
10
y 2 x y^ ^2 x 40.
x
y
5 5
5
5
y 2 x^ y^ ^2 x
In Problems 41– 48, find an equation for the hyperbola described. Graph the equation by hand.
41. Center at 1 4, - 12 ; focus at 1 7, - 12 ; vertex at 1 6, - 12 42. Center at 1 - 3, 1 2 ; focus at 1 - 3, 6 2 ; vertex at 1 - 3, 4 2 43. Center at 1 - 3, - 42 ; focus at 1 - 3, - 82 ; vertex at 1 - 3, - 22 44. Center at 1 1, 4 2 ; focus at 1 - 2, 4 2 ; vertex at 1 0, 4 2 45. Foci at 1 3, 7 2 and 1 7, 7 2 ; vertex at 1 6, 7 2 46. Focus at 1 - 4, 0 2 vertices at 1 - 4, 4 2 and 1 - 4, 2 2 47. Vertices at 1 - 1, - 12 and 1 3, - 12 ; asymptote the line y + 1 =
3 2
1 x - 12
48. Vertices at 1 1, - 32 and 1 1, 1 2 ; asymptote the line y + 1 =
3 2
1 x - 12
In Problems 49– 62, find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation by hand. Verify your graph using a graphing utility.
49.
1 x - 222 4
1 y + 322 9
= 1 50.
1 y + 322 4
1 x - 222 9
= 1 51. 1 y - 222 - 4 1 x + 222 = 4
52. 1 x + 422 - 9 1 y - 322 = 9 53. 1 x + 122 - 1 y + 222 = 4 54. 1 y - 322 - 1 x + 222 = 4 55. x^2 - y^2 - 2 x - 2 y - 1 = 0 56. y^2 - x^2 - 4 y + 4 x - 1 = 0 57. y^2 - 4 x^2 - 4 y - 8 x - 4 = 0 58. 2 x^2 - y^2 + 4 x + 4 y - 4 = 0 59. 4 x^2 - y^2 - 24 x - 4 y + 16 = 0 60. 2 y^2 - x^2 + 2 x + 8 y + 3 = 0 61. y^2 - 4 x^2 - 16 x - 2 y - 19 = 0 62. x^2 - 3 y^2 + 8 x - 6 y + 4 = 0
In Problems 63– 66, graph each function by hand. Be sure to label any intercepts.
[ Hint: Notice that each function is half a hyperbola. ]
63. f 1 x 2 = 216 + 4 x^2 64. f 1 x 2 = - 29 + 9 x^2 65. f 1 x 2 = - 2 - 25 + x^2 66. f 1 x 2 = 2 - 1 + x^2
Mixed Practice
In Problems 67–74, analyze each conic.
67.
( x - 3) 2 4
y^2 25
= 1 68.
( y + 2) 2 16
( x - 2) 2 4
= 1 69. x^2 = 16( y - 3)
70. y^2 = -12( x^ +^ 1)^ 71.^^25 x^2 +^9 y^2 -^250 x^ +^400 =^0 72.^ x^2 +^36 y^2 -^2 x^ +^288 y^ +^541 =^0 73. x^2 - 6 x - 8 y - 31 = 0 74. 9 x^2 - y^2 - 18 x - 8 y - 88 = 0
Applications and Extensions
75. Fireworks Display Suppose that two people standing 2 miles apart both see the burst from a fireworks display. After a period of time, the first person standing at point A hears the burst. One second later, the second person standing at point B hears the burst. If the person at point B is due west of the person at point A and if the display is known to occur due north of the person at point A , where did the fireworks display occur? 76. Lightning Strikes Suppose that two people standing 1 mile apart both see a flash of lightning. After a period of time, the first person standing at point A hears the thunder. Two seconds later, the second person standing at point B hears the thunder. If the person at point B is due west of the person at point A and if the lightning strike is known to occur due north of the person standing at point A , where did the lightning strike occur?
SECTION 10.4 The Hyperbola 669
77. Nuclear Power Plant Some nuclear power plants utilize “natural draft” cooling towers in the shape of a hyperboloid, a solid obtained by rotating a hyperbola about its conjugate axis. Suppose that such a cooling tower has a base diameter of 400 feet and the diameter at its narrowest point, 360 feet above the ground, is 200 feet. If the diameter at the top of the tower is 300 feet, how tall is the tower? Source: Bay Area Air Quality Management District 78. An Explosion Two recording devices are set 2400 feet apart, with the device at point A to the west of the device at point B. At a point between the devices, 300 feet from point B , a small amount of explosive is detonated. The recording devices record the time until the sound reaches each. How far directly north of point B should a second explosion be done so that the measured time difference recorded by the devices is the same as that for the first detonation? 79. Rutherford’s Experiment In May 1911, Ernest Rutherford published a paper in Philosophical Magazine. In this article, he described the motion of alpha particles as they are shot at a piece of gold foil 0.00004 cm thick. Before conducting this experiment, Rutherford expected that the alpha particles would shoot through the foil just as a bullet would shoot through snow. Instead, a small fraction of the alpha particles bounced off the foil. This led to the conclusion that the nucleus of an atom is dense, while the remainder of the atom is sparse. Only the density of the nucleus could cause the alpha particles to deviate from their path. The figure shows a diagram from Rutherford’s paper that indicates that the deflected alpha particles follow the path of one branch of a hyperbola.
x
45
y
(a) Find an equation of the asymptotes under this scenario. (b) If the vertex of the path of the alpha particles is 10 cm from the center of the hyperbola, find a model that describes the path of the particle.
80. Hyperbolic Mirrors Hyperbolas have interesting reflective properties that make them useful for lenses and mirrors. For example, if a ray of light strikes a convex hyperbolic mirror on a line that would (theoretically) pass through its rear focus, it is reflected through the front focus. This property, and that of the parabola, were used to develop the Cassegrain telescope in 1672. The focus of the parabolic mirror and the rear focus of
the hyperbolic mirror are the same point. The rays are collected by the parabolic mirror, reflected toward the (common) focus, and thus are reflected by the hyperbolic mirror through the opening to its front focus, where the eyepiece is located. If the equation of the hyperbola
is
y^2 9
x^2 16
= 1 and the focal length (distance from the vertex to the focus) of the parabola is 6, find the equation of the parabola. Source: www.enchantedlearning.com
81. The eccentricity e of a hyperbola is defined as the number c a
, where a is the distance of a vertex from the center and c is the distance of a focus from the center. Because c 7 a , it follows that e 7 1. Describe the general shape of a hyperbola whose eccentricity is close to 1. What is the shape if e is very large?
82. A hyperbola for which a = b is called an equilateral hyperbola. Find the eccentricity e of an equilateral hyperbola. [ Note: The eccentricity of a hyperbola is defined in Problem 81.] 83. Two hyperbolas that have the same set of asymptotes are called conjugate. Show that the hyperbolas
x^2 4
x^2 4
= 1
are conjugate. Graph each hyperbola on the same set of coordinate axes.
84. Prove that the hyperbola
y^2 a^2
x^2 b^2
= 1
has the two oblique asymptotes
y =
a b
x and y = -
a b
x
85. Show that the graph of an equation of the form
Ax^2 + Cy^2 + F = 0 A 0, C 0, F 0
where A and C are of opposite sign, is a hyperbola with center at 1 0, 0 2.
86. Show that the graph of an equation of the form
Ax^2 + Cy^2 + Dx + Ey + F = 0 A 0, C 0
where A and C are of opposite sign,
(a) is a hyperbola if
D^2 4 A
E^2 4 C
(b) is two intersecting lines if
D^2 4 A
E^2 4 C
‘Are You Prepared?’ Answers
1. 5 22 2.
25 4
3. 1 0, - 32 , 1 0, 3 2 4. True 5. right; 5; down; 4 6. Vertical: x = -2, x = 2; horizontal: y = 1
