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Section 7.4 The Hyperbola

Before Starting this Section, Review

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)

Objectives

1 Define a hyperbola.
2 Find the asymptotes of a hyperbola.
3 Graph a hyperbola.
4 Translate hyperbolas.
5 Use hyperbolas in applications.

What Is Loran?

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.

Definition of Hyperbola

1 Define a hyperbola.

Figure 7.26 shows a hyperbola in standard position, with foci F1(−c, 0) $F_{1} (- c, 0)$ and F2(c, 0) $F_{2} (c, 0)$ on the x-axis at equal distances from the origin. The two parts of the hyperbola are called branches.

Figure 7.26

|d(P, F1)−d(P, F2)| $| d (P, F_{1}) - d (P, F_{2}) |$ is a constant, for any point `P` on the hyperbola

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

| d (P, F 1) - d (P, F 2) | (x + c) 2 + y 2 - - - - - - - - - - - \sqrt - (x - c) 2 + y 2 - - - - - - - - - - - \sqrt = = 2 a \pm 2 a Definition of hyperbola Distance formula

$\begin{array}{rcll} | d (P, F_{1}) - d (P, F_{2}) | & = & 2 a & Definition of hyperbola \\ \sqrt{{(x + c)}^{2} + y^{2}} - \sqrt{{(x - c)}^{2} + y^{2}} & = & \pm 2 a & Distance formula \end{array}$

Just as in the case of an ellipse, we eliminate radicals and simplify (see Exercise 92) to obtain the equation:

(c 2 - a 2) x 2 - a 2 y 2 b 2 x 2 - a 2 y 2 x 2 a 2 - y 2 b 2 = = = a 2 (c 2 - a 2) a 2 b 2 1 (2) (1) Replace (c 2 - a 2) with b 2 . Divide both sides by a 2 b 2 .

$\begin{array}{rcll} (c^{2} - a^{2}) x^{2} - a^{2} y^{2} & = & a^{2} (c^{2} - a^{2}) & (1) \\ b^{2} x^{2} - a^{2} y^{2} & = & a^{2} b^{2} & Replace (c^{2} - a^{2}) with b^{2} . \\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} & = & 1 (2) & Divide both sides by a^{2} b^{2} . \end{array}$

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 $- a$ and a. The points corresponding to these x-intercepts are the vertices of the hyperbola. The distance between the vertices V1(−a, 0) $V_{1} (- a, 0)$ and V2(a, 0) $V_{2} (a, 0)$ is 2a. The line segment having the two vertices as endpoints is called the transverse axis. The midpoint of the transverse axis is the center of the hyperbola. The center is also the midpoint of the line segment joining the foci. The distance between the center and either vertex is a, and the distance between the center and either focus point is c. See Figure 7.27. The line segment having the points (0, −b) $(0, - b)$ and (0, b) as endpoints, where b2=c2−a2 $b^{2} = c^{2} - a^{2}$ and b>0, $b > 0,$ is called the conjugate axis.

Similarly, the standard form of the equation of a hyperbola with center (0, 0) and foci (0, −c) $(0, - c)$ and (0, c) on the y-axis is given by:

y 2 a 2 - x 2 b 2 = 1, where b 2 = c 2 - a 2 (3)

$\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1, where b^{2} = c^{2} - a^{2} (3)$

Here the vertices are (0, −a) $(0, - a)$ and (0, a). The transverse axis of length 2a of the graph of equation (3) lies on the y-axis, and its conjugate axis is the segment joining the points (−b, 0) $(- b, 0)$ and (0, b) of length 2b that lies on the x-axis. The distance between the center and either endpoint of the conjugate axis is b.

Summary of Main Facts

Main facts about hyperbolas centered at (0, 0)
Standard Equation	x2a2−y2b2=1; a>0, b>0 $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1; a > 0, b > 0$	y2a2−x2b2=1; a>0, b>0 $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1; a > 0, b > 0$
Transverse axis on	`x`-axis (y=0) $(y = 0)$	`y`-axis (x=0) $(x = 0)$
Length of transverse axis	2`a`	2`a`
Conjugate axis on	`y`-axis (x=0) $(x = 0)$	`x`-axis (y=0) $(y = 0)$
Length of conjugate axis	2`b`	2`b`
Vertices	(±a, 0) $(\pm a, 0)$	(0, ±a) $(0, \pm a)$
Endpoints of conjugate axis	(0, ±b) $(0, \pm b)$	(±b, 0) $(\pm b, 0)$
Foci	(±c, 0), $(\pm c, 0),$ where c2=a2+b2 $c^{2} = a^{2} + b^{2}$	(0, ±c), $(0, \pm c),$ where c2=a2+b2 $c^{2} = a^{2} + b^{2}$
Description	Hyperbola has a left branch and a right branch. (Hyperbola opens left and right.)	Hyperbola has an upper branch and a lower branch. (Hyperbola opens up and down.)
Graph

Example 1 Determining the Orientation of a Hyperbola

Does the hyperbola

x 2 4 - y 2 10 = 1

$\frac{x^{2}}{4} - \frac{y^{2}}{10} = 1$

have its transverse axis on the x-axis or y-axis?

Solution

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 $y^{2}$ -term and is on the y-axis when the minus sign precedes the x2 $x^{2}$ -term. In the equation in this example, the minus sign precedes the y2 $y^{2}$ -term; so the transverse axis is on the x-axis. See Figure 7.28.

Figure 7.28

x24−y210=1 $\frac{x^{2}}{4} - \frac{y^{2}}{10} = 1$

Practice Problem 1

Does the hyperbola y28−x25=1 $\frac{y^{2}}{8} - \frac{x^{2}}{5} = 1$ have its transverse axis on the x-axis or y-axis?

Example 2 Finding the Vertices and Foci from the Equation of a Hyperbola

Find the vertices and foci for the hyperbola

28 y 2 - 36 x 2 = 63 .

$28 y^{2} - 36 x^{2} = 63 .$

Solution

First, convert the equation to the standard form.

28 y 2 - 36 x 2 28 y 2 63 - 36 x 2 63 4 y 2 9 - 4 x 2 7 y 2 9 4 - x 2 7 4 = = = = 63111 Given equation Divide both sides by 63. Simplify . Divide numerators and denominators of each term by 4 .

$\begin{array}{rcll} 28 y^{2} - 36 x^{2} & = & 63 & Given equation \\ \frac{28 y^{2}}{63} - \frac{36 x^{2}}{63} & = & 1 & Divide both sides by 63. \\ \frac{4 y^{2}}{9} - \frac{4 x^{2}}{7} & = & 1 & Simplify . \\ \frac{y^{2}}{\frac{9}{4}} - \frac{x^{2}}{\frac{7}{4}} & = & 1 & \begin{array}{l} Divide numerators and denominators \\ of each term by 4 . \end{array} \end{array}$

The last equation is the equation of a hyperbola in standard form with center (0, 0). Since the coefficient of x2 $x^{2}$ is negative, the transverse axis lies on the y-axis.

Here a2=94 $a^{2} = \frac{9}{4}$ and b2=74. $b^{2} = \frac{7}{4} .$ The equation c2=a2+b2 $c^{2} = a^{2} + b^{2}$ gives the value of c2. $c^{2} .$

c 2 = = = a 2 + b 2 9 4 + 7 4 16 4 = 4 Replace a 2 with 9 4 and b 2 with 7 4 . Simplify .

$\begin{array}{rcll} c^{2} & = & a^{2} + b^{2} \\ = & \frac{9}{4} + \frac{7}{4} & Replace a^{2} with \frac{9}{4} and b^{2} with \frac{7}{4} . \\ = & \frac{16}{4} = 4 & Simplify . \end{array}$

Now a2=94 $a^{2} = \frac{9}{4}$ gives a=32 $a = \frac{3}{2}$ and c2=4 $c^{2} = 4$ gives c=2 $c = 2$ because we require a>0 $a > 0$ and c>0. $c > 0 .$

The vertices of the hyperbola are (0, −32) $(0, - \frac{3}{2})$ and (0, 32). $(0, \frac{3}{2}) .$ The foci of the hyperbola are (0, −2) $(0, - 2)$ and (0, 2).

Practice Problem 2

Find the vertices and foci for the hyperbola

$x 2 - 4 y 2 = 8 .$ $x^{2} - 4 y^{2} = 8 .$

Example 3 Finding the Equation of a Hyperbola

Find the standard form of the equation of a hyperbola with

vertices (±4, 0) $(\pm 4, 0)$ and foci (±5, 0). $(\pm 5, 0) .$
center (0, 0); vertex: (1, 0); focus: (4, 0).
foci: (0,±6) $(0, \pm 6)$ ; the length of the transverse axis is 8.

Solution

Since the foci of the hyperbola, (−5, 0) $(- 5, 0)$ and (5, 0), are on the x-axis, the transverse axis lies on the x-axis. The center of the hyperbola is midway between the foci, at (0, 0). The standard form of such a hyperbola is

$x 2 a 2 - y 2 b 2 = 1 .$ $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 .$

You need to find a2 $a^{2}$ and b2. $b^{2} .$

The distance a between the center (0, 0) to either vertex, (−4, 0) $(- 4, 0)$ or (4, 0), is 4; so a=4 $a = 4$ and a2=16. $a^{2} = 16 .$ The distance c between the center (0, 0) to either focus, (−5, 0) $(- 5, 0)$ or (5, 0), is 5; so c=5 $c = 5$ and c2=25. $c^{2} = 25 .$ Now use the equation b2=c2−a2 $b^{2} = c^{2} - a^{2}$ to find b2=c2−a2=25−16=9. $b^{2} = c^{2} - a^{2} = 25 - 16 = 9 .$ Substitute a2=16 $a^{2} = 16$ and b2=9 $b^{2} = 9$ in x2a2−y2b2=1 $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ to get the standard form of the equation of the hyperbola

$x 2 16 - y 2 9 = 1 .$ $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1 .$
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 $a = 1$ . The distance between the center (0, 0) and the focus (4, 0) is 4, so c=4 $c = 4$ .

$c 2 42 b 2 = 42 - 12 = 16 - 1 x 2 a 2 - y 2 b 2 x 2 - y 2 15 = = = = = a 2 + b 2 12 + b 2 1511 Relationship between a, b, and c Replace c with 4 and a with 1. Solve for b 2 . Equation for a hyperbola with horizontal transverse axis Replace a 2 with 1 and b 2 with 15 and simplify .$ $\begin{array}{rcll} c^{2} & = & a^{2} + b^{2} & Relationship between a, b, and c \\ 4^{2} & = & 1^{2} + b^{2} & Replace c with 4 and a with 1. \\ b^{2} = 4^{2} - 1^{2} = 16 - 1 & = & 15 & Solve for b^{2} . \\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} & = & 1 & \begin{array}{l} Equation for a hyperbola with \\ horizontal transverse axis \end{array} \\ x^{2} - \frac{y^{2}}{15} & = & 1 & \begin{array}{l} Replace a^{2} with 1 and b^{2} with \\ 15 and simplify . \end{array} \end{array}$
Since the foci (0, −6) $(0, - 6)$ and (0, 6) are on the y-axis, the transverse axis is on the y-axis. The length of the transverse axis is 8, so 2a=8 $2 a = 8$ and a=4 $a = 4$ . The center is midway between the foci (0,±6) $(0, \pm 6)$ , so the center is (0, 0). The distance, c, between the center (0, 0) and the focus (0, 6) is 6, so c=6 $c = 6$ .

$c 2 62 b 2 = 62 - 42 = 36 - 16 y 2 a 2 - x 2 b 2 y 2 16 - x 2 20 = = = = = a 2 + b 2 42 + b 2 2011 Relationship between a, b, and c Replace c with 6 and a with 4. Solve for b 2 . Equation for a hyperbola with vertical transverse axis Replace a 2 with 16 and b 2 with 20.$ $\begin{array}{rcll} c^{2} & = & a^{2} + b^{2} & Relationship between a, b, and c \\ 6^{2} & = & 4^{2} + b^{2} & Replace c with 6 and a with 4. \\ b^{2} = 6^{2} - 4^{2} = 36 - 16 & = & 20 & Solve for b^{2} . \\ \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} & = & 1 & \begin{array}{l} Equation for a hyperbola with \\ vertical transverse axis \end{array} \\ \frac{y^{2}}{16} - \frac{x^{2}}{20} & = & 1 & \begin{array}{l} Replace a^{2} with 16 and b^{2} \\ with 20. \end{array} \end{array}$

Practice Problem 3

Find the standard form of the equation of a hyperbola with
1. vertices at (0, ±3) $(0, \pm 3)$ and foci at (0, ±6) $(0, \pm 6)$ .
2. center (0, 0); vertex: (5, 0); focus: (7, 0).
3. foci: (0,±6) $(0, \pm 6)$ ; the length of the transverse axis is 10.

The Asymptotes of a Hyperbola

2 Find the asymptotes of a hyperbola.

A line y=mx+b $y = m x + b$ is a horizontal asymptote (if m=0 $m = 0$ ) or an oblique asymptote (if m≠0 $m \neq 0$ ) of a graph if the distance from the line to the points on the graph approaches 0 as x→∞ $x \to \infty$ or as x→−∞. $x \to - \infty .$ When horizontal or oblique asymptotes exist, they provide us with information about the end behavior of the graph. This is especially helpful in graphing a hyperbola.

To find the asymptotes of the hyperbola with equation x2a2−y2b2=1, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ we first solve this equation for y.

x 2 a 2 - y 2 b 2 y 2 y 2 y 2 y = = = = = 1 b 2 (x 2 a 2 - 1) b 2 (x 2 a 2 - x 2 a 2 \cdot a 2 x 2) b 2 x 2 a 2 (1 - a 2 x 2) \pm b x a 1 - a 2 x 2 - - - - - - \sqrt Given equation of hyperbola Isolate the y 2 -term . Write 1 = x 2 a 2 \cdot a 2 x 2 . Factor out x 2 a 2 . Square root property

$\begin{array}{rcll} \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} & = & 1 & Given equation of hyperbola \\ y^{2} & = & b^{2} (\frac{x^{2}}{a^{2}} - 1) & Isolate the y^{2} -term . \\ y^{2} & = & b^{2} (\frac{x^{2}}{a^{2}} - \frac{x^{2}}{a^{2}} \cdot \frac{a^{2}}{x^{2}}) & Write 1 = \frac{x^{2}}{a^{2}} \cdot \frac{a^{2}}{x^{2}} . \\ y^{2} & = & \frac{b^{2} x^{2}}{a^{2}} (1 - \frac{a^{2}}{x^{2}}) & Factor out \frac{x^{2}}{a^{2}} . \\ y & = & \pm \frac{b x}{a} \sqrt{1 - \frac{a^{2}}{x^{2}}} & Square root property \end{array}$

As x→∞ $x \to \infty$ or as x→−∞, $x \to - \infty,$ the quantity a2x2 $\frac{a^{2}}{x^{2}}$ approaches 0. Thus, 1−a2x2−−−−−−√ $\sqrt{1 - \frac{a^{2}}{x^{2}}}$ approaches 1 and the value of y approaches ±bxa. $\pm \frac{b x}{a} .$ Therefore, the lines y=bax $y = \frac{b}{a} x$ and y=−bax $y = - \frac{b}{a} x$ are the oblique asymptotes for the graph of the hyperbola x2a2−y2b2=1. $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 .$

Side Note

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, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ replace the 1 on the right side by 0 to obtain x2a2−y2b2= 0. $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 0 .$ Solve this equation for y:

y 2 b 2 y 2 = = x 2 a 2 b 2 a 2 x 2

$\begin{array}{rcl} \frac{y^{2}}{b^{2}} & = & \frac{x^{2}}{a^{2}} \\ y^{2} & = & \frac{b^{2}}{a^{2}} x^{2} \end{array}$

The lines y=±bax $y = \pm \frac{b}{a} x$ are the asymptotes of the hyperbola.

Similarly, you can show that the lines y=abx $y = \frac{a}{b} x$ and y=−abx $y = - \frac{a}{b} x$ are the oblique asymptotes for the graph of the hyperbola y2a2−x2b2=1. $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 .$

Example 4 Finding the Asymptotes of a Hyperbola

Determine the asymptotes of each hyperbola.

x24−y29=1 $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$
y29−x216=1 $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$

Solution

The hyperbola x24−y29=1 $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$ is of the form x2a2−y2b2=1; $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1;$ so $a = 2$ and $b = 3 .$

Substituting these values into $y = \frac{b}{a} x$ and $y = - \frac{b}{a} x,$ we get the asymptotes

$y = \frac{3}{2} x and y = - \frac{3}{2} x .$
The hyperbola $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$ has the form $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1;$ so $a = 3$ and $b = 4 .$

Substituting these values into $y = \frac{a}{b} x$ and $y = - \frac{a}{b} x,$ we get the asymptotes

$y = \frac{3}{4} x and y = - \frac{3}{4} x .$

The hyperbolas of a and b and their asymptotes are shown in Figures 7.29(a) and 7.29(b), respectively.

Figure 7.29

Asymptotes

Practice Problem 4

Determine the asymptotes of the hyperbola $\frac{y^{2}}{4} - \frac{x^{2}}{9} = 1 .$

Example 5 Finding an Equation of a Hyperbola

Find an equation of the hyperbola with center (0, 0) satisfying the given conditions.

Foci: $(\pm 5, 0)$ ; asymptotes $y = \pm 2 x$
Vertices: $(0, \pm 1)$ ; asymptotes $y = \pm \frac{1}{2} x$

Solution

Since the foci are on the x-axis, the transverse axis is on the x-axis and the asymptotes are $y = \pm \frac{b}{a} x$ . We are given that the asymptotes for this hyperbola are $y = \pm 2 x$ .

$\begin{array}{rcll} \frac{b}{a} & = & 2 & \begin{array}{l} Equate the positive values for the \\ asymptotes' slope . \end{array} \\ b & = & 2 a & Multiply both sides by a . \\ c^{2} & = & a^{2} + b^{2} & Relationship between a, b, and c . \\ 5^{2} & = & a^{2} + {(2 a)}^{2} & Foci are (\pm 5, 0), so c = 5. \\ Replace c with 5, and b with 2 a . \\ 25 = a^{2} + 4 a^{2} & = & 5 a^{2} & Simplify . \\ a^{2} & = & 5 & Divide both sides by 5. \\ b^{2} = {(2 a)}^{2} = 4 a^{2} = 4 (5) & = & 20 & Square both sides of b = 2 a \\ and replace a^{2} with 5. \\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} & = & 1 & \begin{array}{l} Equation for a hyperbola with \\ transverse axis on the x -axis \end{array} \\ \frac{x^{2}}{5} - \frac{y^{2}}{20} & = & 1 & Replace a^{2} with 5 and b^{2} with 20. \end{array}$
Since the vertices are $(0, \pm 1), a = 1.$ Because the vertices are on the y-axis, the transverse axis is on the y-axis and the asymptotes are $y = \pm \frac{a}{b} x$ . We are given that the asymptotes for this hyperbola are $y = \pm \frac{1}{2} x$ .

$\begin{array}{rcll} \frac{a}{b} & = & \frac{1}{2} & Equate the positive values for the asymptotes' slope . \\ 2 a & = & b & Multiply both sides by 2 b and simplify . \\ 2 (1) & = & b & Replace a with 1. \\ 4 & = & b^{2} & Square both sides . \\ \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} & = & 1 & \begin{array}{l} Equation for a hyperbola with transverse \\ axis on the y -axis . \end{array} \\ \frac{y^{2}}{1} - \frac{x^{2}}{4} & = & 1 & \begin{array}{l} a = 1 so, a^{2} = 1. Replace a^{2} with 1 \\ and b^{2} with 4. \end{array} \\ y^{2} - \frac{x^{2}}{4} & = & 1 & Simplify . \end{array}$

Practice Problem 5

Find an equation of the hyperbola with center (0, 0) satisfying the given conditions.
1. Foci: $(\pm 2 \sqrt{2}, 0)$ ; asymptotes $y = \pm x$
2. Vertices: $(0, \pm 1)$ ; asymptotes $y = \pm \frac{1}{3} x$

Graphing a Hyperbola with Center (0, 0)

3 Graph a hyperbola.

Consider the hyperbola with equation

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 .$

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 $\frac{b}{a}$ and $- \frac{b}{a}$ . Thus, the extensions of these diagonals are the asymptotes of the hyperbola.

Procedure in Action

Example 6 Graphing a Hyperbola Centered at (0, 0)

OBJECTIVE	EXAMPLE
Sketch the graph of a hyperbola in the form $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ or $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$ .	Sketch the graph of $16 x^{2} - 9 y^{2} = 144 .$	$25 y^{2} - 4 x^{2} = 100 .$
Step 1 Write the equation in standard form. Determine transverse axis and the orientation of the hyperbola.	$\begin{array}{rcl} 16 x^{2} - 9 y^{2} & = & 144 \\ \frac{16 x^{2}}{144} - \frac{9 y^{2}}{144} & = & \frac{144}{144} \\ \frac{x^{2}}{9} - \frac{y^{2}}{16} & = & 1 \end{array}$ Minus sign precedes the $y^{2}$ -term; the transverse axis is on the `x`-axis; the hyperbola opens left and right.	$\begin{array}{rcl} 25 y^{2} - 4 x^{2} & = & 100 \\ \frac{25 y^{2}}{100} - \frac{4 x^{2}}{100} & = & \frac{100}{100} \\ \frac{y^{2}}{4} - \frac{x^{2}}{25} & = & 1 \end{array}$ Minus sign precedes the $x^{2}$ -term; the transverse axis is on the `y`-axis; the hyperbola opens up and down.
Step 2 Locate vertices and the endpoints of the conjugate axis.	Because $a^{2} = 9, a = 3 .$ Also, $b^{2} = 16,$ so $b = 4 .$ The vertices are $(\pm 3, 0),$ and the endpoints of the conjugate axis are $(0, \pm 4) .$	Because $a^{2} = 4, a = 2 .$ Also, $b^{2} = 25,$ so $b = 5 .$ The vertices are $(0, \pm 2),$ and the endpoints of the conjugate axis are $(\pm 5, 0) .$
Step 3 Lightly sketch the fundamental rectangle by drawing dashed lines parallel to the coordinate axes through the points in Step 2.	Draw dashed lines $x = 3,$ $x = - 3, y = 4,$ and $y = - 4$ to form the fundamental rectangle with vertices (3, 4), $(- 3, 4),$ $(- 3, - 4),$ and $(3, - 4) .$	Draw dashed lines $y = 2,$ $y = - 2, x = 5,$ and $x = - 5$ to form the fundamental rectangle with vertices (5, 2), $(- 5, 2),$ $(- 5, - 2),$ and $(5, - 2) .$
Step 4 Sketch the asymptotes. Extend the diagonals of the fundamental rectangle. These are the asymptotes.
Step 5 Sketch the graph. Draw both branches of the hyperbola through the vertices, approaching the asymptotes. See Figures 7.30 and 7.31. The foci are located on the transverse axis, `c` units from the center, where $c^{2} = a^{2} + b^{2} .$	Figure 7.30	Figure 7.31

Practice Problem 6

Sketch the graph of each equation.
1. $25 x^{2} - 4 y^{2} = 100$
2. $9 y^{2} - x^{2} = 1$

Translations of Hyperbolas

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.

Summary of Main Facts

Main properties of hyperbolas centered at (`h`, `k`)
Standard Equation	$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1;$ $a > 0, b > 0$	$\frac{{(y - k)}^{2}}{a^{2}} - \frac{{(x - h)}^{2}}{b^{2}} = 1;$ $a > 0, b > 0$
Transverse axis along the line	$y = k$	$x = h$
Length of transverse axis	2`a`	2`a`
Conjugate axis along the line	$x = h$	$y = k$
Length of conjugate axis	2`b`	2`b`
Center	(`h`, `k`)	(`h`, `k`)
Vertices	$(h - a, k)$ and $(h + a, k)$	$(h, k - a)$ and $(h, k + a)$
Endpoints of conjugate axis	$(h, k - b)$ and $(h, k + b)$	$(h - b, k)$ and $(h + b, k)$
Foci	$(h - c, k)$ and $(h + c, k);$	$(h, k - c)$ and $(h, k + c);$
Equation involving `a`, `b`, and `c`	$c^{2} = a^{2} + b^{2}$	$c^{2} = a^{2} + b^{2}$
Asymptotes	$y - k = \pm \frac{b}{a} (x - h)$	$y - k = \pm \frac{a}{b} (x - h)$

Expanding the binomials in the equation of a hyperbola in standard form and collecting like terms result in an equation of the form

$A x^{2} + C y^{2} + D x + E y + F = 0,$

with A and C of opposite sign (that is, $A C < 0) .$ 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.

Example 7 Graphing a Hyperbola

Show that $9 x^{2} - 16 y^{2} + 18 x + 64 y - 199 = 0$ is an equation of a hyperbola and then graph the hyperbola.

Solution

Since we plan on completing squares, first group the x- and y-terms onto the left side and the constants onto the right side.

$\begin{array}{rcll} 9 x^{2} - 16 y^{2} + 18 x + 64 y - 199 & = & 0 & Given equation \\ (9 x^{2} + 18 x) + (- 16 y^{2} + 64 y) & = & 199 & Group terms . \\ 9 (x^{2} + 2 x) - 16 (y^{2} - 4 y) & = & 199 & Factor out 9 and - 16. \\ 9 (x^{2} + 2 x + 1) - 16 (y^{2} - 4 y + 4) & = & 199 + 9 - 64 & Complete the squares: \\ add 9 \cdot 1 and - 16 \cdot 4. \\ 9 {(x + 1)}^{2} - 16 {(y - 2)}^{2} & = & 144 & Factor and simplify . \\ \frac{9 {(x + 1)}^{2}}{144} - \frac{16 {(y - 2)}^{2}}{144} & = & 1 & \begin{array}{l} Divide both sides by 144 \\ to obtain 1 on the right side . \end{array} \\ \frac{{(x + 1)}^{2}}{16} - \frac{{(y - 2)}^{2}}{9} & = & 1 & Simplify . \end{array}$

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 $(- 1, 2), a^{2} = 16$ and $b^{2} = 9 .$ 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:

$y - 2 = \frac{3}{4} (x + 1) and y - 2 = - \frac{3}{4} (x + 1) .$
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 .

Figure 7.32

The hyperbola $9 x^{2} - 16 y^{2} + 18 x + 64 y - 199 = 0$

Practice Problem 7

Show that $x^{2} - 4 y^{2} - 2 x + 16 y - 20 = 0$ is an equation of a hyperbola and graph the hyperbola.

Example 8 Find an Equation for the Hyperbola Whose Graph Is Shown

Solution

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 $y = 1 .$

$\begin{array}{rcll} c^{2} & = & a^{2} + b^{2} & Relationship between a, b, and c \\ 4^{2} & = & 2^{2} + b^{2} & Replace c with 4 and a with 2. \\ b^{2} = 4^{2} - 2^{2} = 16 - 4 & = & 12 & Solve for b^{2} . Simplify . \\ \frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} & = & 1 & \begin{array}{l} Equation for a hyperbola with horizontal \\ transverse axis \end{array} \\ \frac{{(x - 0)}^{2}}{2^{2}} - \frac{{(y - 1)}^{2}}{12} & = & 1 & \begin{array}{l} Replace h with 0, a with 2, k with 1, \\ and b^{2} with 12. \end{array} \\ \frac{x^{2}}{4} - \frac{{(y - 1)}^{2}}{12} & = & 1 & Simplify . \end{array}$
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 $\frac{3}{2}$ . The transverse axis is on the vertical line $x = 1$ .

$\begin{array}{rcll} \frac{a}{b} & = & \frac{3}{2} & \begin{array}{l} Equate the positive values for \\ the asymptotes' slope, \frac{a}{b} . \end{array} \\ b & = & \frac{2}{3} a & Solve for b . \\ b = \frac{2}{3} (3) & = & 2 & Replace a with 3. Simplify . \\ \frac{{(y - k)}^{2}}{a^{2}} - \frac{{(x - h)}^{2}}{b^{2}} & = & 1 & \begin{array}{l} Equation for a hyperbola with \\ vertical transverse axis \end{array} \\ \frac{{(y - (- 1))}^{2}}{3^{2}} - \frac{{(x - h)}^{2}}{2^{2}} & = & 1 & \begin{array}{l} Replace k with - 1, a with 3, \\ h with 1, and b with 2. \end{array} \\ \frac{{(y + 1)}^{2}}{9} - \frac{{(x - 1)}^{2}}{4} & = & 1 & Simplify . \end{array}$

Practice Problem 8

Find an equation for each hyperbola whose graph is shown.

Applications

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 $P V = c,$ 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 $F_{2} .$
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.

Figure 7.33

How Does LORAN Work?

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).

Example 9 Using LORAN

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.

Solution

With reference to the transmitters A and B, P is located on a hyperbola with $2 a = 240,$ or $a = 120,$ and the hyperbola has foci $(- 130, 0)$ and (130, 0).

$\begin{array}{rcll} \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} & = & 1 & \begin{array}{l} Standard form of the equation of a \\ hyperbola \end{array} \\ a^{2} = {(120)}^{2} & = & 14, 400 & Replace a with 120. \\ b^{2} = c^{2} - a^{2} = {(130)}^{2} - {(120)}^{2} & = & 2500 & Replace c with 130 and a with 120. \\ \frac{x^{2}}{14, 400} - \frac{y^{2}}{2500} & = & 1 & \begin{array}{l} Replace a^{2} with 14, 400 and b^{2} with \\ 2500. \end{array} \end{array}$

The location of the ship therefore lies on a hyperbola (see Figure 7.34) with equation

$\frac{x^{2}}{14, 400} - \frac{y^{2}}{2500} = 1 .$

Practice Problem 9

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.

Answers to Practice Problems

1. y-axis
2. Vertices: $(2 \sqrt{2}, 0)$ and $(- 2 \sqrt{2}, 0);$ foci: $(\sqrt{10}, 0)$ and $(- \sqrt{10}, 0)$
3.
1. $\frac{y^{2}}{9} - \frac{x^{2}}{27} = 1$
2. $\frac{x^{2}}{25} - \frac{y^{2}}{24} = 1$
3. $\frac{y^{2}}{25} - \frac{x^{2}}{11} = 1$
4. $y = \frac{2}{3} x$ and $y = - \frac{2}{3} x$
5.
1. $\frac{x^{2}}{4} - \frac{y^{2}}{4} = 1$
2. $y^{2} - \frac{x^{2}}{9} = 1$
6.
1. The standard form of the equation is $\frac{x^{2}}{4} - \frac{y^{2}}{25} = 1 .$ Vertices: (2, 0) and $(- 2, 0);$ endpoints of the conjugate axis: (0, 5) and $(0, - 5);$ asymptotes: $y = \frac{5}{2} x$ and $y = - \frac{5}{2} x$
2. The standard form of the equation is $\frac{y^{2}}{\frac{1}{9}} - \frac{x^{2}}{1} = 1 .$ Vertices: $(0, \frac{1}{3})$ and $(0, - \frac{1}{3});$ endpoints of the conjugate axis: (1, 0) and $(- 1, 0);$ asymptotes: $y = \frac{1}{3} x$ and $y = - \frac{1}{3} x$
7. $\frac{{(x - 1)}^{2}}{5} - \frac{{(y - 2)}^{2}}{\frac{5}{4}} = 1 .$ Center: (1, 2); vertices: $(1 + \sqrt{5}, 2)$ and $(1 - \sqrt{5}, 2);$ endpoints of the conjugate axis: $(1, 2 + \frac{\sqrt{5}}{2})$ and $(1, 2 - \frac{\sqrt{5}}{2});$ asymptotes:

$y - 2 = \pm \frac{1}{2} (x - 1)$ foci: $(\frac{7}{2}, 2)$ and $(- \frac{3}{2}, 2)$
8.
1. $x^{2} - \frac{{(y + 2)}^{2}}{8} = 1$
2. $y^{2} - \frac{{(x + 1)}^{2}}{4} = 1$
9. $\frac{x^{2}}{16, 900} - \frac{y^{2}}{5, 600} = 1$

Section 7.4 Exercises

Concepts and Vocabulary

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 $(\pm a, 0),$ and foci $(\pm c, 0)$ is

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, where b^{2} = \underline{} .$
For the hyperbola $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1,$ the vertices are and and the foci are $(0, \pm c),$ where $c^{2} =$ .
True or False. The graph of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1$ is a hyperbola.
True or False. The graph of $A x^{2} + C y^{2} + D x + E y + F = 0$ is a degenerate conic or a hyperbola if $A C < 0 .$
True or False. If one asymptote of a hyperbola has equation $y = - 2 x$ , the other must have equation $y = \frac{1}{2} x$ .
True or False. If one focus point of a hyperbola is (0, 5), the other must be $(0, - 5) .$

Building Skills

In Exercises 9–16, the equation of a hyperbola is given. Match each equation with its graph shown in (a)–(h).

1. $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$
2. $\frac{x^{2}}{4} - \frac{y^{2}}{25} = 1$
3. $\frac{y^{2}}{25} - \frac{x^{2}}{4} = 1$
4. $\frac{y^{2}}{4} - \frac{x^{2}}{9} = 1$
5. $4 x^{2} - 25 y^{2} = 100$
6. $16 x^{2} - 49 y^{2} = 196$
7. $16 y^{2} - 9 x^{2} = 100$
8. $9 y^{2} - 16 x^{2} = 144$

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.

$x^{2} - \frac{y^{2}}{4} = 1$
$y^{2} - \frac{x^{2}}{4} = 1$
$x^{2} - y^{2} = - 1$
$y^{2} - x^{2} = - 1$
$9 y^{2} - x^{2} = 36$
$9 x^{2} - y^{2} = 36$
$4 x^{2} - 9 y^{2} - 36 = 0$
$4 x^{2} - 9 y^{2} + 36 = 0$
$y = \pm \sqrt{4 x^{2} + 1}$
$y = \pm 2 \sqrt{x^{2} + 1}$
$y = \pm \sqrt{9 x^{2} - 1}$
$y = \pm 3 \sqrt{x^{2} - 4}$

In Exercises 29–42, find an equation of the hyperbola satisfying the given conditions. Graph the hyperbola.

Vertices: $(\pm 2, 0);$ foci: $(\pm 3, 0)$
Vertices: $(\pm 3, 0);$ foci: $(\pm 5, 0)$
Vertices: $(0, \pm 4);$ foci: $(0, \pm 6)$
Vertices: $(0, \pm 5);$ foci: $(0, \pm 8)$
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: $(0, \pm 5);$ the length of the transverse axis is 6.
Foci: $(\pm 2, 0);$ the length of the transverse axis is 2.
Foci: $(\pm \sqrt{5}, 0);$ asymptotes: $y = \pm 2 x$
Foci: $(0, \pm 10);$ asymptotes: $y = \pm 3 x$
Vertices: $(0, \pm 4);$ asymptotes: $y = \pm x$
Vertices: $(\pm 3, 0);$ asymptotes: $y = \pm 2 x$

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.

$\frac{{(x - 1)}^{2}}{9} - \frac{{(y + 1)}^{2}}{16} = 1$
$\frac{{(y - 1)}^{2}}{16} - \frac{{(x + 1)}^{2}}{4} = 1$
$\frac{{(x + 2)}^{2}}{25} - \frac{y^{2}}{49} = 1$
$\frac{{(x + 1)}^{2}}{9} - \frac{{(y + 2)}^{2}}{36} = 1$
$\frac{{(x + 4)}^{2}}{25} - \frac{{(y + 3)}^{2}}{49} = 1$
$\frac{{(x - 3)}^{2}}{9} - \frac{{(y + 1)}^{2}}{9} = 1$
$4 x^{2} - {(y + 1)}^{2} = 25$
$9 {(x - 1)}^{2} - y^{2} = 144$
${(y + 1)}^{2} - 9 {(x - 2)}^{2} = 25$
$6 {(x - 4)}^{2} - 3 {(y + 3)}^{2} = 4$
$x^{2} - y^{2} + 6 x = 36$
$x^{2} - 4 y^{2} - 4 x = 0$
$x^{2} + 4 x - 4 y^{2} = 12$
$x^{2} - 2 y^{2} - 8 y = 12$
$2 x^{2} - y^{2} + 12 x - 8 y + 3 = 0$
$y^{2} - 9 x^{2} - 4 y - 30 x = 33$
$3 x^{2} - 18 x - 2 y^{2} - 8 y + 1 = 0$
$4 y^{2} - 9 x^{2} - 8 y - 36 x = 68$
$y^{2} + 2 \sqrt{2} - x^{2} + 2 \sqrt{2} x = 1$
$x^{2} + x = \frac{y^{2} - 1}{4}$

In Exercises 63–72, identify the conic section represented by each equation and sketch the graph.

$x^{2} - 6 x + 12 y + 33 = 0$
$9 y^{2} = x^{2} - 4 x$
$y^{2} - 9 x^{2} = - 1$
$4 x^{2} + 9 y^{2} - 8 x + 36 y + 4 = 0$
$x^{2} + y^{2} - 4 x + 8 y = 16$
$2 y^{2} + 3 x - 4 y - 7 = 0$
$2 x^{2} - 4 x + 3 y + 8 = 0$
$y^{2} - 9 x^{2} + 18 x - 4 y = 14$
$4 x^{2} + 9 y^{2} + 8 x - 54 y + 49 = 0$
$x^{2} - 4 y^{2} + 6 x + 12 y = 0$

In Exercises 73–78, find an equation (in standard form) for the hyperbola whose graph is shown.

Applying the Concepts

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.

Beyond the Basics

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 $F_{1}$ and $F_{2}$ is two units, where
1. $F_{1}$ is (0, 6) and $F_{2}$ is $(0, - 6);$
2. $F_{1}$ is (0, 4) and $F_{2}$ is $(0, - 4);$
3. $F_{1}$ is (0, 3) and $F_{2}$ is $(0, - 3) .$
Sketch the graphs of all three hyperbolas in Exercise 90 on the same coordinate plane.
Rewrite equation

$\sqrt{{(x + c)}^{2} + y^{2}} - \sqrt{{(x - c)}^{2} + y^{2}} = \pm 2 a$

from page 691 as

$\sqrt{{(x + c)}^{2} + y^{2}} = \pm 2 a + \sqrt{{(x - c)}^{2} + y^{2}} .$

Square both sides to obtain

${(x + c)}^{2} + y^{2} = 4 a^{2} \pm 4 a \sqrt{{(x - c)}^{2} + y^{2}} + {(x - c)}^{2} + y^{2} .$

Simplify and then isolate the radical; then square both sides again to show that the equation can be simplified to

$(c^{2} - a^{2}) x^{2} - a^{2} y^{2} = a^{2} (c^{2} - a^{2}) .$
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

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 is \frac{2 b^{2}}{a} .$
The eccentricity of a hyperbola, denoted by e, is defined by

$e = \frac{Distance between the foci}{Distance between the vertices} = \frac{2 c}{2 a} = \frac{c}{a} .$

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.

$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
$36 x^{2} - 25 y^{2} = 900$
$x^{2} - y^{2} = 49$
$8 {(x - 1)}^{2} - {(y + 2)}^{2} = 2$
$5 x^{2} - 4 y^{2} - 10 x - 8 y - 19 = 0$
For the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ with eccentricity e, show that $b^{2} = a^{2} (e^{2} - 1) .$
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 $(\pm 6, 0)$ ; length of its latus rectum is 10.
Foci are $(\pm 3 \sqrt{5}, 0)$ ; length of its lactus rectum is 8.
Foci are $(0, \pm 12)$ ; length of its latus rectum is 36.
Foci are $(0, \pm 2 \sqrt{3})$ ; length of its lactus rectum is 8.
Foci are $(0, \pm \sqrt{10)}$ , and the graph passes through the point (2, 3).
Foci are $(\pm 4, 0)$ , and the graph passes through the point $(- 2 \sqrt{2}, 2 \sqrt{3})$ .

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.

$y - 2 x - 20 = 0$ and $y^{2} - 4 x^{2} = 36$
$x - 2 y^{2} = 0$ and $x^{2} = 8 y^{2} + 5$
$x^{2} + y^{2} = 15$ and $x^{2} - y^{2} = 1$
$x^{2} + 9 y^{2} = 9$ and $4 x^{2} - 25 y^{2} = 36$

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 $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ with $a = \sqrt{\frac{5}{3}}$ and $b = \sqrt{\frac{5}{7}}$ 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 $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ at the point $(x_{1}, y_{1}),$ is $y = \frac{b^{2} x_{1}}{a^{2} y_{1}} x - \frac{b^{2} x_{1}^{2}}{a^{2} y_{1}} + y_{1} .$

[Hint: See the steps for Exercise 92 in Section 7.2 and show that ${(a^{2} y_{1} m - b^{2} x_{1})}^{2} = 0 .$ ]

Critical Thinking/Discussion/Writing

Explain why each of the following represents a degenerate conic section. Where possible, sketch the graph.
1. $4 x^{2} - 9 y^{2} = 0$
2. $x^{2} + 4 y^{2} + 6 = 0$
3. $8 x^{2} + 5 y^{2} = 0$
4. $x^{2} + y^{2} - 4 x + 8 y = - 20$
5. $y^{2} - 2 x^{2} = 0$
Discuss the graph of a quadratic equation in x and y of the form

$A x^{2} + C y^{2} + D x + E y + F = 0, A C \neq 0 .$

Rewrite this equation in the form

$\begin{array}{l} A (x^{2} + \frac{D}{A} x + \frac{D^{2}}{4 A^{2}}) + C (y^{2} + \frac{E}{C} y + \frac{E^{2}}{4 C^{2}}) \\ = \frac{D^{2}}{4 A} + \frac{E^{2}}{4 C} - F . \end{array}$

Include in your discussion the types of graphs you will obtain in each of the following cases:
1. $A > 0, C > 0$
2. $A > 0, C < 0$
3. $A < 0, C > 0$
4. $A < 0, C < 0$
In each case, discuss how the classification is affected by the sign of

$\frac{D^{2}}{4 A} + \frac{E^{2}}{4 C} - F .$

Getting Ready for the Next Section

In Exercises 117–120, find the value of each function for the given number.

$f (x) = \frac{1}{1 + x}; x = 3$
$g (x) = 2 x + 1; x = 10$
$h (x) = {(- 1)}^{3} 3^{x - 1}; x = 5$
$f (x) = \frac{16 x}{x^{2} + 12}; x = 6$

In Exercises 121–124, find the values of the given expression for the specified integer, n.

${(- 1)}^{n}; n = 17$
${(- 1)}^{n} \frac{1}{n + 1}; n = 8$
${(- 1)}^{n + 1} 2^{n}; n = 5$
${(- 1)}^{n - 2} (2 n + 1) + {(- 1)}^{n}; n = 7$

In Exercises 125 and 126, specify the integer that is the result of the computation, where a represents a nonzero real number.

$\frac{3 \cdot π \cdot e^{2} \cdot 7 \cdot 11 \cdot 19 \cdot a}{a \cdot 3 \cdot π \cdot e^{2} \cdot 19}$
$\frac{a \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10}{3 \cdot 5 \cdot 7 \cdot 8 \cdot 9 \cdot 10 \cdot a}$

In Exercises 127 and 128, specify whether the given statement is true or false.

$c (1 - 5 a + 3 b) = c - 5 a c + 3 b c$
$3 a - 2 b + 6 c - 5 d = (3 a + 6 c) - (2 b + 5 d)$

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Table of Contents for Section 7.4 The Hyperbola

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Table of Contents for
Section 7.4 The Hyperbola