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Figure 9.14 illustrates the two branches of a hyperbola. The line through the foci intersects the hyperbola at two points, called the vertices. The line segment that joins the vertices is the transverse axis. The midpoint of the transverse axis is the center of the hyperbola. Notice that the center lies midway between the vertices, as well as midway between the foci.
Standard Form of the Equation of a Hyperbola
The rectangular coordinate system enables us to translate a hyperbola’s geometric definition into an algebraic equation. Figure 9.15 is our starting point for obtaining an equation. We place the foci, and on the at the points and Note that the center of this hyperbola is at the origin. We let represent the coordinates of any point, on the hyperbola. What does the definition of a hyperbola tell us about the point in Figure 9.15? For any point on the hyperbola, the absolute value of the difference of the distances from the two foci, must be constant. We denote this constant by just as we did for the ellipse. Thus, the point is on the hyperbola if and only if
2 a, 1 x, y 2
ƒ d 2 - d 1 ƒ ,
1 x, y 2
1 x, y 2
P,
1 c, 0 2. 1 x, y 2
F 1 F 2 , x-axis 1 - c, 0 2
Center
F 1 (− c , 0) (0, 0)
P ( x , y )
F 2 ( c , 0)
d (^2) d 1 x
y
886 Chapter 9 Conic Sections and Analytic Geometry
The Hyperbola
C
onic sections are often used to create unusual architectural designs. The top of St. Mary’s Cathedral in San Francisco is a 2135-cubic-foot dome with walls rising 200 feet above the floor and supported by four massive concrete pylons that extend 94 feet into the ground. Cross sections of the roof are parabolas and hyperbolas. In this section, we study the curve with two parts known as the hyperbola.
Definition of a Hyperbola
Figure 9.13 shows a cylindrical lampshade casting two shadows on a wall. These shadows indicate the distin- guishing feature of hyperbolas: Their graphs contain two disjoint parts, called branches. Although each branch might look like a parabola, its shape is actually quite different. The definition of a hyperbola is similar to that of an ellipse. For an ellipse, the sum of the distances to the foci is a constant. By contrast, for a hyperbola the difference of the distances to the foci is a constant.
Objectives
Locate a hyperbola’s vertices
and foci.
Write equations of hyperbolas
in standard form.
Graph hyperbolas centered at
the origin.
Graph hyperbolas not
centered at the origin.
Solve applied problems
involving hyperbolas.
S e c t i o n
Figure 9.13 Casting hyperbolic shadows
Definition of a Hyperbola
A hyperbola is the set of points in a plane the difference of whose distances from two fixed points, called foci, is constant.
Vertex Vertex x
y
Transverse axis
Focus (^) Center Focus
Figure 9.14 The two branches of a hyperbola
Figure 9.
St. Mary’s Cathedral
Section 9.2 The Hyperbola 887
Use the distance formula.
After eliminating radicals and simplifying, we obtain
For convenience, let Substituting for in the preceding equation, we obtain
Divide both sides by
Simplify.
This last equation is called the standard form of the equation of a hyperbola centered at the origin. There are two such equations. The first is for a hyperbola in which the transverse axis lies on the The second is for a hyperbola in which the transverse axis lies on the y-axis.
x-axis.
x^2 a^2
y^2 b^2
a^2 b^2.
b^2 x^2 a^2 b^2
a^2 y^2 a^2 b^2
a^2 b^2 a^2 b^2
b^2 x^2 - a^2 y^2 = a^2 b^2.
b^2 = c^2 - a^2. b^2 c^2 - a^2
1 c^2 - a^22 x^2 - a^2 y^2 = a^21 c^2 - a^22.
ƒ 41 x^ +^ c^22 +^1 y^ -^022 -^41 x^ -^ c^22 +^1 y^ -^022 ƒ =^2 a
ƒ d 2 - d 1 ƒ = 2 a.
Study Tip
The form is the one you should remember. When finding the foci, this form is easy to manipulate.
c^2 = a^2 + b^2
Standard Forms of the Equations of a Hyperbola
The standard form of the equation of a hyperbola with center at the origin is
Figure 9.16(a) illustrates that for the equation on the left, the transverse axis lies on the Figure 9.16(b) illustrates that for the equation on the right, the transverse axis lies on the The vertices are units from the center and the foci are units from the center. For both equations, Equivalently, c^2 = a^2 + b^2.
c b^2 = c^2 - a^2.
y-axis. a
x-axis.
x^2 a^2
y^2 b^2
= 1 or
y^2 a^2
x^2 b^2
Study Tip
When the is preceded by a plus sign, the transverse axis is horizontal. When the y^2 -termis preceded by a plus sign, the transverse axis is vertical.
x^2 -term
y
Transverse axis (− a , 0) ( a , 0) (− c , 0) ( c , 0)
x
x − = 1 2 a^2
y^2 b^2
Figure 9.16(a) Transverse axis lies on the x-axis.
Transverse x axis
(0, a )
(0, c )
(0, − c )
(0, − a )
y
y − = 1 2 a^2
x^2 b^2
Figure 9.16(b) Transverse axis lies on the y-axis.
Section 9.2 The Hyperbola 889
We use to find the foci, which are located at and
Because The foci are located at and (0, 5). They are shown in Figure 9..
Check Point 1 Find the vertices and locate the foci for each of the following hyperbolas with the given equation:
a. b.
In Example 1, we used equations of hyperbolas to find their foci and vertices. In the next example, we reverse this procedure.
Finding the Equation of a Hyperbola from Its Foci
and Vertices
Find the standard form of the equation of a hyperbola with foci at and (0, 3) and vertices and (0, 2), shown in Figure 9..
Solution Because the foci are located at and (0, 3), on the the
transverse axis lies on the The center of the hyperbola is midway between the foci, located at (0, 0). Thus, the form of the equation is
We need to determine the values for and The distance from the center, (0, 0), to either vertex, or (0, 2), is 2, so
We must still find The distance from the center, (0, 0), to either focus, or (0, 3), is 3. Thus, Using we have
and
Substituting 5 for in gives us the standard form of the hyperbola’s
equation. The equation is
Check Point 2 Find the standard form of the equation of a hyperbola with foci at and (0, 5) and vertices and (0, 3).
The Asymptotes of a Hyperbola
As and get larger, the two branches of the graph of a hyperbola approach a pair of intersecting straight lines, called asymptotes. The asymptotes pass through the center of the hyperbola and are helpful in graphing hyperbolas.
x y
y^2 4
x^2 5
y^2 4
x^2 b^2
b^2 = 1
b^2 = 3 2 - 2 2 = 9 - 4 = 5.
3 2 = 2 2 + b^2
c = 3. c^2 = a^2 + b^2 ,
b^2. 1 0, - 32
y^2 2 2
x^2 b^2
= 1 or
y^2 4
x^2 b^2
1 0, - 22 a = 2.
a^2 b^2.
y^2 a^2
x^2 b^2
y-axis.
1 0, - 32 y-axis,
EXAMPLE 2
y^2 25
x^2 16
x^2 25
y^2 16
c^2 = 25, c = 5. 1 0, - 52
c^2 = a^2 + b^2 = 9 + 16 = 25
c^2 = a^2 + b^21 0, - c 2 1 0, c 2.
− 1
1
2
3
4
5
− 2 − 3 − 4 − 5
− 5 − 4 − 3 − 2 − 1 1 2 3 4 5
y
x
Vertex (0, 2)
Vertex (0, −2)
Focus (0, −3)
Focus (0, 3)
Figure 9.
Write equations of hyperbolas in standard form.
890 Chapter 9 Conic Sections and Analytic Geometry
Figure 9.20 shows the asymptotes for the graphs of hyperbolas centered at the origin. The asymptotes pass through the corners of a rectangle. Note that the dimen- sions of this rectangle are by The line segment of length is the conjugate axis of the hyperbola and is perpendicular to the transverse axis through the center of the hyperbola.
2 a 2 b. 2 b
y = x
x
y b a
x − = 1 2 a^2
y^2 b^2
Asymptote:
y = − bax
Asymptote:
(− a , 0) ( a , 0)
(0, b )
(0, − b )
x
y
y − = 1 2 a^2
x^2 b^2
(− b , 0) ( b , 0)
(0, a )
(0, − a )
y = (^) ba x
Asymptote:
y = − x
Asymptote: a b Figure 9.20 Asymptotes of a hyperbola
The Asymptotes of a Hyperbola Centered at the Origin
The hyperbola has a horizontal transverse axis and two asymptotes
The hyperbola has a vertical transverse axis and two asymptotes
y =
a b
x and y = -
a b
x.
y^2 a^2
x^2 b^2
y =
b a x and y = -
b a x.
x^2 a^2
y^2 b^2
Why are the asymptotes for a hyperbola whose transverse axis is
horizontal? The proof can be found in Appendix A.
Graphing Hyperbolas Centered at the Origin
Hyperbolas are graphed using vertices and asymptotes.
y = ;
b a x
Graphing Hyperbolas
1. Locate the vertices. 2. Use dashed lines to draw the rectangle centered at the origin with sides parallel to the axes, crossing one axis at and the other at 3. Use dashed lines to draw the diagonals of this rectangle and extend them to obtain the asymptotes. 4. Draw the two branches of the hyperbola by starting at each vertex and approaching the asymptotes.
;a ;b.
Graph hyperbolas centered at the origin.
892 Chapter 9 Conic Sections and Analytic Geometry
Graphing a Hyperbola
Graph and locate the foci: What are the equations of the asymptotes?
Solution We begin by writing the equation in standard form. The right side
should be 1, so we divide both sides by 36.
Simplify. The right side is now 1.
Now we are ready to use our four-step procedure for graphing hyperbolas.
Step 1 Locate the vertices. The equation that we obtained is in the form
with and
Based on the standard form of the equation with the transverse axis on the we know that the vertices are and Because Thus, the vertices are and (0, 2), shown in Figure 9.. Step 2 Draw a rectangle. Because and and We construct a rectangle to find the asymptotes, using and 2 on the (the vertices are located here) and and 3 on the The rectangle passes through these four points, shown using dashed lines in Figure 9.. Step 3 Draw extended diagonals of the rectangle to obtain the asymptotes. We draw dashed lines through the opposite corners of the rectangle, shown in Figure 9.23 , to obtain the graph of the asymptotes. Based on the standard form of the hyperbola’s equation, the equations of these asymptotes are
Step 4 Draw the two branches of the hyperbola by starting at each vertex and approaching the asymptotes. The hyperbola is shown in Figure 9..
y = ;
a b
x or y = ;
x.
a^2 = 4 b^2 = 9, a = 2 b = 3.
1 0, - a 2 1 0, a 2. a^2 = 4, a = 2.
y-axis,
y^2 4
x^2 9
a^2 = 4 (^) b^2 = 9
a^2 = 4 b^2 = 9.
y^2 a^2
x^2 b^2
y^2 4
x^2 9
9 y^2 36
4 x^2 36
9 y^2 - 4 x^2 = 36.
EXAMPLE 4
− 1
1
3
4
5
− 3 − 4 − 5
− 5 − 4 − 2 − 1 1 2 4 5
y
x
Asymptote
Asymptote Vertex (0,^ −2)
Vertex (0, 2)
Figure 9.23 Preparing to graph y^2 4
2 9 = 1
− 1
1
3
4
5
− 3 − 4 − 5
− 5 − 4 − 2 − 1 1 2 4 5
y
x
Figure 9.24 The graph of y^2 4
2 9 = 1
We now consider the foci, located at and We find using
c^2 = 4 + 9 = 13
c^2 = a^2 + b^2.
1 0, - c 2 1 0, c 2. c
Section 9.2 The Hyperbola 893
Because The foci are located at and approximately and (0, 3.6).
Check Point 4 Graph^ and^ locate^ the^ foci:^ What^ are^ the equations of the asymptotes?
Translations of Hyperbolas
The graph of a hyperbola can be centered at rather than at the origin. Horizontal and vertical translations are accomplished by replacing with and with in the standard form of the hyperbola’s equation. Table 9.2 gives the standard forms of equations of hyperbolas centered at 1 h, k 2 and shows their graphs.
y y - k
x x - h
1 h, k 2 ,
y^2 - 4 x^2 = 4.
c^2 = 13, c = 2 13. A0, - 213 B A0, 213 B,
Table 9.2 Standard Forms of Equations of Hyperbolas Centered at ( h, k )
Equation Center Transverse Axis Vertices Graph
Foci are units right and units left of center, where c^2 = a^2 + b^2.
c
c
(x-h)^2 a^2
(y-k)^2 b^2
Vertices are a units right and a units left of center.
1 h, k 2 Parallel to the x-axis; horizontal 1 h + a, k 2
1 h - a, k 2
Focus ( h − c , k ) Focus ( h + c , k )
x
y
Vertex ( h − a , k ) Vertex ( h + a , k )
Center ( h , k )
Foci are units above and units below the center, where c^2 = a^2 + b^2.
c
c
(y-k)^2 a^2
(x-h)^2 b^2
Vertices are a units above and a units below the center.
1 h, k 2 Parallel to the y-axis; vertical 1 h, k + a 2
1 h, k - a 2
x
y
Focus ( h , k + c )
Focus ( h , k − c ) Vertex ( h , k − a )
Vertex ( h , k + a )
Center ( h , k )
Graph hyperbolas not centered at the origin.
Graphing a Hyperbola Centered at
Graph: Where are the foci located? What are the
equations of the asymptotes?
1 x - 222 16
1 y - 322 9
EXAMPLE 5 1 h, k 2
Section 9.2 The Hyperbola 895
Check Point 5 Graph:^ Where are the foci located? What are the equations of the asymptotes?
In our next example, it is necessary to convert the equation of a hyperbola to standard form by completing the square on and
Graphing a Hyperbola Centered at
Graph: Where are the foci located? What are the equations of the asymptotes?
Solution We begin by completing the square on xand y.
4 x^2 - 24 x - 25 y^2 + 250 y - 489 = 0.
EXAMPLE 6 1 h, k 2
x y.
1 x - 322 4
1 y - 122 1
This is the given equation. Group terms and add 489 to both sides. Factor out 4 and respectively, so coefficients of x^2 and y^2 are 1.
41 x^2 - 6 x + n 2 - 251 y^2 - 10 y + n 2 = 489 - 25,
14 x^2 - 24 x 2 + 1 - 25 y^2 + 250 y 2 = 489
4 x^2 - 24 x - 25 y^2 + 250 y - 489 = 0
4(x^2 -6x+9)-25(y^2 -10y+25)=489+36+(–625)
We added 4 9, or 36, to the left side.
We added − 25 25, or −625, to the left side.
Add 36 + (−625) to the right side.
Complete each square by adding the square of half the coefficient of x and y ,respectively.
Factor.
Divide both sides by
Simplify.
1 x - 322
1 y - 522 4
41 x - 322
251 y - 522
41 x - 322 - 251 y - 522 = - 100
(y-5)^2 4
(x-3)^2 25
This is ( x − h )^2 , with h = 3.
This is ( y − k ) 2 , with k = 5.
Write the equation in standard form,
1 y - k 2^2 a^2
1 x - h 2^2 b^2
= 1.
We see that and Thus, the center of the hyperbola, is (3, 5). Because the is being subtracted, the transverse axis is vertical and the hyperbola opens upward and downward. We use our four-step procedure to obtain the graph of
Step 1 Locate the vertices. Based on the standard form of the equation with a vertical transverse axis, the vertices are units above and below the center. Because This means that the vertices are 2 units above and below the center, (3, 5). This puts the vertices at (3, 7) and (3, 3), shown in Figure 9.. Step 2 Draw a rectangle. Because and and The rectangle passes through points that are 2 units above and below the center (the vertices are located here) and 5 units to the right and left of the center. The rectangle is shown using dashed lines in Figure 9..
b^2 = 25, a = 2 b = 5.
a^2 = 4
a^2 = 4, a = 2.
a
(y-5)^2 4
(x-3)^2 25
a^2 = 4 b^2 = 25
x^2 -term
Study Tip h = 3 k = 5. 1 h, k 2 ,
The hyperbola’s center is (3, 5) because the last equation shows that 3 is subtracted from and 5 is subtracted from Many students tend to read the equation from left to right and get the center backward. The hyperbola’s center is not (5, 3).
y.
x
6 7 8 910
9 8 7 6
x
y
− 1 1 2 3 4 5
1
2
3
4
5
− 4 − 3 − 2 − 1
Vertex (3, 7)
Vertex (3, 3)
Center (3, 5)
Figure 9.27 The graph of 1 y - 522 4
1 x - 322 25 = 1
896 Chapter 9 Conic Sections and Analytic Geometry
6 7 8 910
9 8 7 6
x
y
− 1 1 2 3 4 5
1
2
3
4
5
− 4 − 3 − 2 − 1
Vertex (3, 7)
Vertex (3, 3)
Center (3, 5)
Figure 9.27 (repeated) The graph of 1 y - 522 4
Step 3 Draw extended diagonals of the rectangle to obtain the asymptotes. We draw dashed lines through the opposite corners of the rectangle, shown in Figure 9.27 , to obtain the graph of the asymptotes. The equations of the
asymptotes of the unshifted hyperbola are or Thus, the asymptotes for the hyperbola that is shifted three units to the right and five units up, namely
have equations that can be expressed as
Step 4 Draw the two branches of the hyperbola by starting at each vertex and approaching the asymptotes. The hyperbola is shown in Figure 9.. We now consider the foci, located units above and below the center, (3, 5). We find using
Because The foci are located at and
Check Point 6 Graph:^ Where are the foci located? What are the equations of the asymptotes?
Applications
Hyperbolas have many applications.When a jet flies at a speed greater than the speed of sound, the shock wave that is created is heard as a sonic boom. The wave has the shape of a cone. The shape formed as the cone hits the ground is one branch of a hyperbola. Halley’s Comet, a permanent part of our solar system, travels around the sun in an elliptical orbit. Other comets pass through the solar system only once, following a hyperbolic path with the sun as a focus. Hyperbolas are of practical importance in fields ranging from architecture to navigation. Cooling towers used in the design for nuclear power plants have cross sections that are both ellipses and hyperbolas. Three-dimensional solids whose cross sections are hyperbolas are used in some rather unique architectural creations, including the TWA building at Kennedy Airport in New York City and the St. Louis Science Center Planetarium.
An Application Involving Hyperbolas
An explosion is recorded by two microphones that are 2 miles apart. Microphone received the sound 4 seconds before microphone Assuming sound travels at 1100 feet per second, determine the possible locations of the explosion relative to the location of the microphones.
Solution We begin by putting
the microphones in a coordinate system. Because feet, we place 5280 feet on a horizontal axis to the right of the origin and 5280 feet on a hori- zontal axis to the left of the origin. Figure 9.28 illustrates that the two microphones are 2 miles apart.
M 2
M 1
1 mile = 5280
M 2.
M 1
EXAMPLE 7
4 x^2 - 24 x - 9 y^2 - 90 y - 153 = 0.
c^2 = 29, c = 2 29. A3, 5 + 229 B A3, 5 - 229 B.
c^2 = 4 + 25 = 29
c c^2 = a^2 + b^2.
c
y - 5 = ;
1 x - 32.
1 y - 522 4
1 x - 322 25
y = ;
y = ; x.
a b
x,
y^2 4
x^2 25
y
x
8000
6000
4000
2000
− 2000
− 6000 − 2000 2000 6000
P ( x , y ) explosion
M 2 (−5280, 0) M 1 (5280, 0)
Figure 9.28 Locating an explosion on the branch of a hyperbola
The hyperbolic shape of a sonic boom
Solve applied problems involving hyperbolas.
Sun
Elliptical orbit
Hyperbolic orbit
Orbits of comets
898 Chapter 9 Conic Sections and Analytic Geometry
10. Endpoints of transverse axis: (4, 0); asymptote: 11. Center: Focus: vertex: 12. Center: Focus: vertex:
In Exercises 13–26, use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes.
**13. 14.
- 26.**
In Exercises 27–32, find the standard form of the equation of each hyperbola.
**27. 28.
- 32.**
− 1
1
2
3
4
− 6
− 5
− 5 − 4 − 1 1 2 3
y
x
1
2
− 2 − 3
− 8
− 4
− 2 − 1 1 2 3 4
y
x 5 6
− 1
1
3
4
2
6
7
− 3 − 4
− 2
− 6 − 7
− 4 − 2 − 1 1 2 4
y
x
− 1
1
3
4
− 3 − 4
− 4 − 2 − 1 1 2 4
y
x
− 1
1
3
4
− 3 − 4
− 4 − 2 − 1 1 2 4
y
x
− 1
1
2
3
4
− 2 − 3 − 4
− 4 − 2 − 1 1 2 4
y
x
− 6
6
y = ; 3 x^2 - 2 y = ; 3 x^2 - 3
9 y^2 - 25 x^2 = 225 16 y^2 - 9 x^2 = 144
9 x^2 - 4 y^2 = 36 4 x^2 - 25 y^2 = 100
4 y^2 - x^2 = 1 9 y^2 - x^2 = 1
y^2 25
= 1
y^2 16
= 1
x^2 144
y^2 81 = 1
x^2 100
y^2 64 = 1
x^2 16
y^2 25 = 1 x^2 9
y^2 25 = 1
1 - 2, 1 2 ; 1 - 2, 6 2 ; 1 - 2, 4 2
1 4, - 22 ; 1 7, - 22 ; 1 6, - 22
1 - 4, 0 2 , y = 2 x In Exercises 33–42, use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes.
**33. 34.
- 36.**
37.
38.
**39.
42.**
In Exercises 43–50, convert each equation to standard form by completing the square on and Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. **43.
50.**
Practice Plus In Exercises 51–56, graph each relation. Use the relation’s graph to determine its domain and range.
**51. 52.
- 56.**
In Exercises 57–60, find the solution set for each system by graphing both of the system’s equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations.
**57. 58.
- 60.**
Application Exercises
61. An explosion is recorded by two microphones that are 1 mile apart. Microphone received the sound 2 seconds before microphone Assuming sound travels at 1100 feet per second, determine the possible locations of the explosion relative to the location of the microphones.
M 2.
M 1
b
4 x^2 + y^2 = 4 y^2 - 4 x^2 = 4 b
9 x^2 + y^2 = 9 y^2 - 9 x^2 = 9
b
x^2 - y^2 = 9 x^2 + y^2 = 9 b
x^2 - y^2 = 4 x^2 + y^2 = 4
y^2 4
= 1
y^2 16
= 1
x^2 25
y^2 4
= 1 x^2 9
y^2 16
= 1
x^2 25
y^2 4
= 1 x^2 9
y^2 16
= 1
9 x^2 - 16 y^2 - 36 x - 64 y + 116 = 0
4 x^2 - 25 y^2 - 32 x + 164 = 0
4 x^2 - 9 y^2 + 8 x - 18 y - 6 = 0
4 x^2 - 9 y^2 - 16 x + 54 y - 101 = 0
9 y^2 - 4 x^2 - 18 y + 24 x - 63 = 0
16 x^2 - y^2 + 64 x - 2 y + 67 = 0
4 x^2 - y^2 + 32 x + 6 y + 39 = 0
x^2 - y^2 - 2 x - 4 y - 4 = 0
x y.
1 y - 222 - 1 x + 322 = 5
1 x - 122 - 1 y - 222 = 3
1 x + 322 - 91 y - 422 = 9
1 x - 322 - 41 y + 322 = 4
1 y - 222 36
1 x + 122 49 = 1
1 y + 222 4
1 x - 122 16
= 1
1 x + 222 9
y^2 25 = 1
1 x + 322 25
y^2 16 = 1
1 x + 222 9
1 y - 122 25 = 1
1 x + 422 9
1 y + 322 16 = 1
Section 9.2 The Hyperbola 899
62. Radio towers and 200 kilometers apart, are situated along the coast, with located due west of Simultaneous radio signals are sent from each tower to a ship, with the signal from received 500 microseconds before the signal from a. Assuming that the radio signals travel 300 meters per microsecond, determine the equation of the hyperbola on which the ship is located. b. If the ship lies due north of tower how far out at sea is it? 63. An architect designs two houses that are shaped and positioned like a part of the branches of the hyperbola whose equation is where and are in yards. How far apart are the houses at their closest point? 64. Scattering experiments, in which moving particles are deflected by various forces, led to the concept of the nucleus of an atom. In 1911, the physicist Ernest Rutherford (1871–1937) discovered that when alpha particles are directed toward the nuclei of gold atoms, they are eventually deflected along hyperbolic paths, illustrated in the figure. If a particle gets as close as 3 units to the nucleus along a hyper- bolic path with an asymptote given by what is the equation of its path?
Moiré patterns, such as those shown in Exercises 65–66, can appear when two repetitive patterns overlap to produce a third, sometimes unintended, pattern.
a. In each exercise, use the name of a conic section to describe the moiré pattern. b. Select one of the following equations that can possibly describe a conic section within the moiré pattern:
x^2 + y^2 = 1; x^2 - y^2 = 1; x^2 + 4 y^2 = 4.
Nucleus (3, 0)
Alpha particle
1 y = 2 x
x
y
y = 12 x,
y
x
625 y^2 - 400 x^2 = 250,000, x y
B,
A.
B
A B.
A B, 65. 66.
Writing in Mathematics
67. What is a hyperbola? 68. Describe how to graph 69. Describe how to locate the foci of the graph of 70. Describe one similarity and one difference between the graphs of and 71. Describe one similarity and one difference between the graphs of and 72. How can you distinguish an ellipse from a hyperbola by looking at their equations? 73. In 1992, a NASA team began a project called Spaceguard Survey, calling for an international watch for comets that might collide with Earth. Why is it more difficult to detect a possible “doomsday comet” with a hyperbolic orbit than one with an elliptical orbit?
Technology Exercises
74. Use a graphing utility to graph any five of the hyperbolas that you graphed by hand in Exercises 13–26. 75. Use a graphing utility to graph any three of the hyperbolas that you graphed by hand in Exercises 33–42. First solve the given equation for by using the square root property. Enter each of the two resulting equations to produce each branch of the hyperbola. 76. Use a graphing utility to graph any one of the hyperbolas that you graphed by hand in Exercises 43–50. Write the equation as a quadratic equation in and use the quadratic formula to solve for Enter each of the two resulting equations to produce each branch of the hyperbola. 77. Use a graphing utility to graph Is the graph a
hyperbola? In general, what is the graph of
78. Graph and in the same viewing rectangle for values of and of your choice. Describe the relationship between the two graphs.
a^2 b^2
x^2 a^2
y^2 b^2
= - 1
x^2 a^2
y^2 b^2
= 1
x^2 a^2
y^2 b^2
= 0?
x^2 4
y^2 9
= 0.
y.
y
y
1 x - 322 9
1 y + 322 1 = 1. x^2 9
y^2 1 = 1
y^2 9
= 1. x^2 9
y^2 1
= 1
x^2 9
y^2 1 = 1.
x^2 9
y^2 1
= 1.
