1 Distance formula (Section 2.1 , page 178)
2 Midpoint formula (Section 2.1 , page 180)
3 Completing the square (Section 3.1 , page 321)
4 Oblique asymptotes (Section 3.6 , page 395)
5 Transformations of graphs (Section 2.7 , page 274)
6 Symmetry (Section 2.2 , page 188)
LORAN stands for LOng-RAnge Navigation. The federal government runs the LORAN system, which uses land-based radio navigation transmitters to provide users with information on position and timing. During World War II, Alfred Lee Loomis was selected to chair the National Defense Research Committee. Much of his work focused on the problem of creating a light system for plane-carried radar. As a result of this effort, he invented LORAN, the long-range navigation system whose offshoot, LORAN-C, remains in widespread use. The navigational method provided by LORAN is based on the principle of determining the points of intersection of two hyperbolas to fix the two-dimensional position of the receiver. In Example 9, we illustrate how this is done.
The LORAN-C system is now being supplemented by the global positioning system (GPS), which works on the same principle as LORAN-C. The GPS system consists of 24 satellites that orbit 11,000 miles above Earth. The GPS was created by the U.S. Department of Defense to provide precise navigation information for targeting weapons systems anywhere in the world. The GPS is available to civilian users worldwide in a degraded mode. It is now used in aviation, navigation, and automobiles. The system can provide your exact position on Earth anywhere, anytime.
The GPS and LORAN-C work together, giving a highly accurate, duplicate navigation system for the Defense Department.
1 Define a hyperbola.
Figure 7.26 shows a hyperbola in standard position, with foci F1(−c, 0)
As in the discussion of ellipses in Section 7.3 (see page 677), we take 2a as the positive constant referred to in the definition.
A point P(x, y) lies on a hyperbola if and only if
Just as in the case of an ellipse, we eliminate radicals and simplify (see Exercise 92) to obtain the equation:
Equation (2) is called the standard form of the equation of a hyperbola with center (0, 0) and foci on the x-axis. The x-intercepts of the graph of equation (2) are −a
Similarly, the standard form of the equation of a hyperbola with center (0, 0) and foci (0, −c)
Here the vertices are (0, −a)
Does the hyperbola
have its transverse axis on the x-axis or y-axis?
The orientation (left–right branches or up–down branches) of a hyperbola is determined by noting where the minus sign occurs in the standard equation. From the table above, we see that the transverse axis is on the x-axis when the minus sign precedes the y2
Does the hyperbola y28−x25=1
Find the vertices and foci for the hyperbola
First, convert the equation to the standard form.
The last equation is the equation of a hyperbola in standard form with center (0, 0). Since the coefficient of x2
Here a2=94
Now a2=94
The vertices of the hyperbola are (0, −32)
Find the vertices and foci for the hyperbola
Find the standard form of the equation of a hyperbola with
vertices (±4, 0)
center (0, 0); vertex: (1, 0); focus: (4, 0).
foci: (0,±6)
Since the foci of the hyperbola, (−5, 0)
You need to find a2
The distance a between the center (0, 0) to either vertex, (−4, 0)
Since the center (0, 0) and vertex (1, 0) are on the x-axis, the transverse axis is on the x-axis. The distance between the center (0, 0) and the vertex (1, 0) is 1, so a=1
Since the foci (0, −6)
Find the standard form of the equation of a hyperbola with
vertices at (0, ±3)
center (0, 0); vertex: (5, 0); focus: (7, 0).
foci: (0,±6)
2 Find the asymptotes of a hyperbola.
A line y=mx+b
To find the asymptotes of the hyperbola with equation x2a2−y2b2=1,
As x→∞
There is a convenient device that can be used for obtaining equations of the asymptotes of a hyperbola. Substitute 0 for the 1 on the right side of the equation of the hyperbola in standard form and then solve for y in terms of x. For example, for the hyperbola x2a2−y2b2= 1,
The lines y=±bax
Similarly, you can show that the lines y=abx
Determine the asymptotes of each hyperbola.
x24−y29=1
y29−x216=1
The hyperbola x24−y29=1
Substituting these values into y=bax and y=−bax, we get the asymptotes
The hyperbola y29−x216=1 has the form y2a2−x2b2=1; so a=3 and b=4.
Substituting these values into y=abx and y=−abx, we get the asymptotes
The hyperbolas of a and b and their asymptotes are shown in Figures 7.29(a) and 7.29(b), respectively.
Determine the asymptotes of the hyperbola y24−x29=1.
Find an equation of the hyperbola with center (0, 0) satisfying the given conditions.
Foci: (±5, 0); asymptotes y=±2x
Vertices: (0,±1); asymptotes y=±12x
Since the foci are on the x-axis, the transverse axis is on the x-axis and the asymptotes are y=±bax. We are given that the asymptotes for this hyperbola are y=±2x.
Since the vertices are (0, ±1), a=1. Because the vertices are on the y-axis, the transverse axis is on the y-axis and the asymptotes are y=±abx. We are given that the asymptotes for this hyperbola are y=±12x.
Find an equation of the hyperbola with center (0, 0) satisfying the given conditions.
Foci: (±2√2, 0); asymptotes y=±x
Vertices: (0,±1); asymptotes y=±13x
3 Graph a hyperbola.
Consider the hyperbola with equation
The vertices of this hyperbola are (−a, 0) and (a, 0). The endpoints of the conjugate axis are (0, −b) and (0, b). The rectangle with vertices (a, b), (−a, b), (−a, −b), and (a, −b) is called the fundamental rectangle of the hyperbola. See Figure 7.30 on page 698. The diagonals of the fundamental rectangle have slopes ba and −ba. Thus, the extensions of these diagonals are the asymptotes of the hyperbola.
Sketch the graph of each equation.
25x2−4y2=100
9y2−x2=1
4 Translate hyperbolas.
We can use horizontal and vertical shifts to find the standard form of the equations of hyperbolas centered at (h, k). Note that a remains the distance between the center and the vertices, and c remains the distance between the center and the foci.
Expanding the binomials in the equation of a hyperbola in standard form and collecting like terms result in an equation of the form
with A and C of opposite sign (that is, AC<0). Conversely, except in degenerate cases, any equation of this form can be put into one of the standard forms of a hyperbola by completing the squares on the x- and y-terms.
Show that 9x2−16y2+18x+64y−199=0 is an equation of a hyperbola and then graph the hyperbola.
Since we plan on completing squares, first group the x- and y-terms onto the left side and the constants onto the right side.
The last equation is the standard form of the equation of a hyperbola with center (−1, 2).
Now we sketch the graph of this hyperbola.
Steps 1–2 Locate the vertices. For this hyperbola with center (−1, 2), a2=16 and b2=9. Therefore, a=4 and b=3. From the table on page 696, with h=−1 and k=2, the vertices are (h−a, k)=(−1−4, 2)=(−5, 2) and (h+a, k)=(−1+4, 2)=(3, 2).
Step 3 Draw the fundamental rectangle. The vertices of the fundamental rectangle are (3, −1), (3, 5), (−5, 5), and (−5, −1).
Step 4 Sketch the asymptotes. Extend the diagonals of the fundamental rectangle to sketch the asymptotes:
Step 5 Sketch the graph. Draw two branches of the hyperbola opening to the left and right, starting from the vertices (−5, 2) and (3, 2) and approaching the asymptotes. See Figure 7.32 .
Show that x2−4y2−2x+16y−20=0 is an equation of a hyperbola and graph the hyperbola.
The center (h, k)=(0, 1), so h=0 and k=1. The distance, a, between the center (0, 1) and the vertex (2, 1) is 2, so a=2. The distance, c, between the center (0, 1) and the focus (4, 1) is 4, so c=4. The transverse axis lies on the horizontal line y=1.
The center (h, k)=(1, −1), so h=1 and k=−1. The distance, a, between the center (1, −1) and the vertex (1, 2) is 3, so a=3. The slope of the asymptote with positive slope is 32. The transverse axis is on the vertical line x=1.
Find an equation for each hyperbola whose graph is shown.
5 Use hyperbolas in applications.
Hyperbolas have many applications. We list a few of them here:
Comets that do not move in elliptical orbits around the sun almost always move in hyperbolic orbits. (In theory, they can also move in parabolic orbits.)
Boyle’s Law states that if a perfect gas is kept at a constant temperature, then its pressure P and volume V are related by the equation PV=c, where c is a constant. The graph of this equation is a hyperbola. In this case, the transverse axis is not parallel to a coordinate axis.
The hyperbola has the reflecting property that a ray of light from a source at one focus of a hyperbolic mirror (a mirror with hyperbolic cross sections) is reflected along the line through the other focus.
The reflecting properties of the parabola and hyperbola are combined into one design for a reflecting telescope. See Figure 7.33. The parallel rays from a star are finally focused at the eyepiece at F2.
The definition of a hyperbola forms the basis of several important navigational systems.
A three-dimensional solid called a hyperboloid is formed by rotating a hyperbola about the part of its conjugate axis from the center to one endpoint. It is often used by engineers in the construction of nuclear cooling towers and by architects in such structures as the Kobe Port Tower in Kobe, Japan, and the Saint Louis Science Center in St. Louis, Missouri.
Suppose two stations A and B (several miles apart) transmit synchronized radio signals. The LORAN receiver in your ship measures the difference in reception times of these synchronized signals. The radio signals travel at a speed of 186,000 miles per second. Using this information, you can determine the difference 2a in the distance of your ship’s receiver from the two transmitters. By the definition of a hyperbola, this information places your ship somewhere on a hyperbola with foci A and B. With two pairs of transmitters, the position of your ship can be determined at a point of intersection of the two hyperbolas. (See Exercises 86 and 87).
LORAN navigational transmitters A and B are located at (−130, 0) and (130, 0), respectively. A receiver P on a fishing boat somewhere in the first quadrant listens to the pair (A, B) of transmissions and computes the difference of the distances from the boat to A and B as 240 miles. Find the equation of the hyperbola on which P is located.
With reference to the transmitters A and B, P is located on a hyperbola with 2a=240, or a=120, and the hyperbola has foci (−130, 0) and (130, 0).
The location of the ship therefore lies on a hyperbola (see Figure 7.34) with equation
In Example 9 , the transmitters A and B are located at (−150, 0) and (150, 0), respectively, and the difference of the distances from the boat to A and B is 260 miles. Find the equation of the hyperbola on which P is located.
A hyperbola is a set of all points in the plane, the of whose distances from two fixed points is constant.
The line segment joining the two vertices of a hyperbola is called the axis.
The standard equation of the hyperbola with center (0, 0), vertices (±a, 0), and foci (±c, 0) is
For the hyperbola y2a2−x2b2=1, the vertices are and and the foci are (0, ±c), where c2= .
True or False. The graph of x2a2−y2b2=−1 is a hyperbola.
True or False. The graph of Ax2+Cy2+Dx+Ey+F=0 is a degenerate conic or a hyperbola if AC<0.
True or False. If one asymptote of a hyperbola has equation y=−2x, the other must have equation y=12x.
True or False. If one focus point of a hyperbola is (0, 5), the other must be (0, −5).
In Exercises 9–16, the equation of a hyperbola is given. Match each equation with its graph shown in (a)–(h).
x29−y24=1
x24−y225=1
y225−x24=1
y24−x29=1
4x2−25y2=100
16x2−49y2=196
16y2−9x2=100
9y2−16x2=144
In Exercises 17–28, the equation of a hyperbola is given.
Find and plot the vertices, the foci, and the transverse axis of the hyperbola.
State how the hyperbola opens.
Find and plot the vertices of the fundamental rectangle.
Sketch the asymptotes and write their equations.
Graph the hyperbola by using the vertices and the asymptotes.
x2−y24=1
y2−x24=1
x2−y2=−1
y2−x2=−1
9y2−x2=36
9x2−y2=36
4x2−9y2−36=0
4x2−9y2+36=0
y=±√4x2+1
y=±2√x2+1
y=±√9x2−1
y=±3√x2−4
In Exercises 29–42, find an equation of the hyperbola satisfying the given conditions. Graph the hyperbola.
Vertices: (±2, 0); foci: (±3, 0)
Vertices: (±3, 0); foci: (±5, 0)
Vertices: (0, ±4); foci: (0, ±6)
Vertices: (0, ±5); foci: (0, ±8)
Center: (0, 0); vertex: (0, 2); focus: (0, 5)
Center: (0, 0); vertex: (0, −1); focus: (0, −4)
Center: (0, 0); vertex: (1, 0); focus: (−5, 0)
Center: (0, 0); vertex: (−3, 0); focus: (6, 0)
Foci: (0, ±5); the length of the transverse axis is 6.
Foci: (±2, 0); the length of the transverse axis is 2.
Foci: (± √5, 0); asymptotes: y=±2x
Foci: (0, ±10); asymptotes: y=±3x
Vertices: (0, ±4); asymptotes: y=±x
Vertices: (±3, 0); asymptotes: y=±2x
In Exercises 43–62, the equation of a hyperbola is given.
Find and plot the center, vertices, transverse axis, and asymptotes of the hyperbola.
Use the vertices and the asymptotes to graph the hyperbola.
(x−1)29−(y+1)216=1
(y−1)216−(x+1)24=1
(x+2)225−y249=1
(x+1)29−(y+2)236=1
(x+4)225−(y+3)249=1
(x−3)29−(y+1)29=1
4x2−(y+1)2=25
9(x−1)2−y2=144
(y+1)2−9(x−2)2=25
6(x−4)2−3(y+3)2=4
x2−y2+6x=36
x2−4y2−4x=0
x2+4x−4y2=12
x2−2y2−8y=12
2x2−y2+12x−8y+3=0
y2−9x2−4y−30x=33
3x2−18x−2y2−8y+1=0
4y2−9x2−8y−36x=68
y2+2√2−x2+2√2x=1
x2+x=y2−14
In Exercises 63–72, identify the conic section represented by each equation and sketch the graph.
x2−6x+12y+33=0
9y2=x2−4x
y2−9x2=−1
4x2+9y2−8x+36y+4=0
x2+y2−4x+8y=16
2y2+3x−4y−7=0
2x2−4x+3y+8=0
y2−9x2+18x−4y=14
4x2+9y2+8x−54y+49=0
x2−4y2+6x+12y=0
In Exercises 73–78, find an equation (in standard form) for the hyperbola whose graph is shown.
Sound of an explosion. Points A and B are 1000 meters apart, and it is determined from the sound of an explosion heard at these points at different times that the location of the explosion is 600 meters closer to A than to B. Show that the location of the explosion is restricted to points on a hyperbola and find the equation of the hyperbola. [Hint: Let the coordinates of A be (−500, 0) and those of B be (500, 0).]
Sound of an explosion. Points A and B are 2 miles apart. The sound of an explosion at A was heard 3 seconds before it was heard at B. Assume that the sound travels at 1100 feet/second. Show that the location of the explosion is restricted to a hyperbola and find the equation of the hyperbola. [Hint: Recall that 1 mile=5280 feet.]
Thunder and lightning. Thunder is heard by Nicole and Juan, who are 8000 feet apart. Nicole hears the thunder 3 econds before Juan does. Find the equation of the hyperbola whose points are possible locations where the lightning could have struck. Assume that the sound travels at 1100 feet/second.
Locating source of thunder. In Exercise 81, Valerie is at a location midway between Nicole and Juan, and she hears the thunder 2 seconds after Juan does. Determine the location where the lightning strikes in relation to the three people involved.
LORAN. Two LORAN stations, A and B, are situated 300 kilometers apart along a straight coastline. Simultaneous radio signals are sent from each station to a ship. The ship receives the signal from A 0.0005 second before the signal from B. Assume that the radio signals travel 300,000 kilometers per second. Find the equation of the hyperbola on which the ship is located.
Target practice. A gun at G and a target at T are 1600 feet apart. The muzzle velocity of the bullet is 2000 feet per second. A person at P hears the crack of the gun and the thud of the bullet at the same time. Show that the possible locations of the person are on a hyperbola. Find the equation of the hyperbola with G at (−800, 0) and T at (800, 0). Assume that the speed of sound is 1100 feet/second. [Hint: Show that |d(P, G)−d(P, T)| is a constant.]
LORAN. Two LORAN stations, A and B, lie on an east–west line with A 250 miles west of B. A plane is flying west on a line 50 miles north of the line AB. Radio signals are sent (traveling at 980 feet per microsecond) simultaneously from A and B, and the one sent from B arrives at the plane 500 microseconds before the one from A. Where is the plane?
Using LORAN. LORAN navigational transmitters A, B, C, and D are located at (−100, 0), (100, 0), (0, −150), and (0, 150), respectively. A navigator P on a ship somewhere in the second quadrant listens to the pair (A, B) of transmitters and computes the difference of the distances from the ship to A and B as 120 miles. Similarly, by listening to the pair (C, D) of transmitters, navigator P computes the difference of the distances from the ship to C and D as 80 miles. Use a calculator to find the coordinates of the ship.
Using LORAN. LORAN navigational transmitters A, B, and C are located at (200, 0), (−200, 0), and (200, 1000), respectively. A navigator on a fishing boat listens to the pair (A, B) of transmitters and finds that the difference of the distances from the boat to A and B is 300 miles. She also finds that the difference of the distances from the boat to A and C is 400 miles. Use a calculator to find the possible locations of the boat.
Write the standard form of the equation of the hyperbola for which the difference of the distances from any point on the hyperbola to (−5, 0) and (5, 0) is equal to (a) two units; (b) four units; (c) eight units.
Sketch the graphs of all three hyperbolas in Exercise 88 on the same coordinate plane.
Write the standard form of the equation of the hyperbola for which the difference of the distances from any point on the hyperbola to the points F1 and F2 is two units, where
F1 is (0, 6) and F2 is (0, −6);
F1 is (0, 4) and F2 is (0, −4);
F1 is (0, 3) and F2 is (0, −3).
Sketch the graphs of all three hyperbolas in Exercise 90 on the same coordinate plane.
Rewrite equation
from page 691 as
Square both sides to obtain
Simplify and then isolate the radical; then square both sides again to show that the equation can be simplified to
A hyperbola for which the lengths of the transverse and conjugate axes are equal is called an equilateral hyperbola. Show that the asymptotes of an equilateral hyperbola are perpendicular to one another.
The latus rectum of a hyperbola is a line segment that passes through a focus, is perpendicular to the transverse axis, and has endpoints on the hyperbola. Show that the length of the latus rectum of the hyperbola
The eccentricity of a hyperbola, denoted by e, is defined by
Show that for every hyperbola, e>1. What happens when e=1?
In Exercises 96–100, find the eccentricity and the length of the latus rectum of each hyperbola.
x216−y29=1
36x2−25y2=900
x2−y2=49
8(x−1)2−(y+2)2=2
5x2−4y2−10x−8y−19=0
For the hyperbola x2a2−y2b2=1 with eccentricity e, show that b2=a2(e2−1).
A point P(x, y) moves in the plane so that its distance from the point (3, 0) is always twice its distance from the line x=−1. Show that the equation of the path of P is a hyperbola of eccentricity 2.
In Exercises 103–108, find an equation of a hyperbola satisfying the given conditions.
Foci are (±6, 0); length of its latus rectum is 10.
Foci are (±3√5, 0); length of its lactus rectum is 8.
Foci are (0,±12); length of its latus rectum is 36.
Foci are (0,±2√3); length of its lactus rectum is 8.
Foci are (0,±√10), and the graph passes through the point (2, 3).
Foci are (±4, 0), and the graph passes through the point (−2√2, 2√3).
In Exercises 109–112, find all points of intersection of the given curves. Make a sketch of the curves that shows the points of intersection.
y−2x−20=0 and y2−4x2=36
x−2y2=0 and x2=8y2+5
x2+y2=15 and x2−y2=1
x2+9y2=9 and 4x2−25y2=36
In Exercises 113 and 114, use the following definition. A tangent line to a hyperbola is a line that is not parallel to either asymptote of the hyperbola and intersects the hyperbola at only one point.
Find the equation of the tangent line to the hyperbola with equation x2a2−y2b2=1, with a=√53 and b=√57 at the point (2, 1).
[Hint: See the steps for Exercise 91 in Section 7.2 and (in Step 3) write the discriminant as a perfect square.]
Show that the equation of the tangent line to the hyperbola with equation x2a2−y2b2=1, at the point (x1, y1), is y=b2x1a2y1x−b2x21a2y1+y1.
[Hint: See the steps for Exercise 92 in Section 7.2 and show that (a2y1m−b2x1)2=0.]
Explain why each of the following represents a degenerate conic section. Where possible, sketch the graph.
4x2−9y2=0
x2+4y2+6=0
8x2+5y2=0
x2+y2−4x+8y=−20
y2−2x2=0
Discuss the graph of a quadratic equation in x and y of the form
Rewrite this equation in the form
Include in your discussion the types of graphs you will obtain in each of the following cases:
A>0, C>0
A>0, C<0
A<0, C>0
A<0, C<0
In each case, discuss how the classification is affected by the sign of
In Exercises 117–120, find the value of each function for the given number.
f (x)=11+x; x=3
g(x)=2x+1; x=10
h(x)=(−1)33x−1; x=5
f (x)=16xx2+12;x=6
In Exercises 121–124, find the values of the given expression for the specified integer, n.
(−1)n;n=17
(−1)n1n+1;n=8
(−1)n+12n;n=5
(−1)n−2(2n+1)+(−1)n;n=7
In Exercises 125 and 126, specify the integer that is the result of the computation, where a represents a nonzero real number.
3⋅π⋅e2⋅7⋅11⋅19⋅aa⋅3⋅π⋅e2⋅19
a⋅2⋅3⋅4⋅5⋅6⋅7⋅8⋅9⋅103⋅5⋅7⋅8⋅9⋅10⋅a
In Exercises 127 and 128, specify whether the given statement is true or false.
c(1−5a+3b)=c−5ac+3bc
3a−2b+6c−5d=(3a+6c)−(2b+5d)