8.5 Hyperbolas

In the definition of an ellipse, Definition 8.4, we fixed two points called foci and looked at points whose distances to the foci always added to a constant distance $d$ . Those prone to syntactical tinkering may wonder what, if any, curve we'd generate if we replaced added with subtracted. The answer is a hyperbola.

Definition

Given two distinct points $F sub 1$ and $F sub 2$ in the plane and a fixed distance $d$ , a hyperbola is the set of all points $(x, y)$ in the plane such that the absolute value of the difference of each of the distances from $F sub 1$ and $F sub 2$ to $(x, y)$ is $d$ . The points $F sub 1$ and $F sub 2$ are called the foci of the hyperbola.

Coordinate-plane figure. — **Figure 8.96**

In the figure above:

$\begin{matrix} the distance from F_{1} to (x_{1}, y_{1}) - the distance from F_{2} to (x_{1}, y_{1}) & = & d \end{matrix}$

and

$\begin{matrix} the distance from F_{2} to (x_{2}, y_{2}) - the distance from F_{1} to (x_{2}, y_{2}) & = & d \end{matrix}$

Note that the hyperbola has two parts, called branches. The center of the hyperbola is the midpoint of the line segment connecting the two foci. The transverse axis of the hyperbola is the line segment connecting two opposite ends of the hyperbola which also contains the center and foci. The vertices of a hyperbola are the points of the hyperbola which lie on the transverse axis.

In addition, we will show momentarily that the hyperbola has a pair of asymptotes which the branches of the hyperbola approach for large $x$ and $y$ values. They serve as guides to the graph. Schematically:

A hyperbola with center $C$ ; foci $F sub 1$ , $F sub 2$ ; and vertices $V sub 1$ , $V sub 2$ and asymptotes (dashed)

Before we derive the standard equation of the hyperbola, we need to discuss one further parameter, the conjugate axis of the hyperbola. The conjugate axis of a hyperbola is the line segment through the center which is perpendicular to the transverse axis and has the same length as the line segment through a vertex which connects the asymptotes. Schematically:

Note that in the diagram, we can construct a rectangle using line segments with lengths equal to the lengths of the transverse and conjugate axes whose center is the center of the hyperbola and whose diagonals are contained in the asymptotes. This guide rectangle, much akin to the one we saw Section 8.4 to help us graph ellipses, will aid us in graphing hyperbolas.

Suppose we wish to derive the equation of a hyperbola. For simplicity, we shall assume that the center is $open parenthesis 0, 0 close parenthesis$ , the vertices are $open parenthesis a, 0 close parenthesis$ and $open parenthesis minus a, 0 close parenthesis$ and the foci are $open parenthesis c, 0 close parenthesis$ and $open parenthesis minus c, 0 close parenthesis$ . We label the endpoints of the conjugate axis $open parenthesis 0, b close parenthesis$ and $open parenthesis 0, minus b close parenthesis$ . (Although $b$ does not enter into our derivation, we will have to justify this choice as you shall see later.) As before, we assume $a$ , $b$ , and $c$ are all positive numbers. Schematically:

Since $open parenthesis a, 0 close parenthesis$ is on the hyperbola, it must satisfy the conditions of Definition 8.6. That is, the distance from $open parenthesis minus c, 0 close parenthesis$ to $open parenthesis a, 0 close parenthesis$ minus the distance from $open parenthesis c, 0 close parenthesis$ to $open parenthesis a, 0 close parenthesis$ must equal the fixed distance $d$ . Since all these points lie on the $x$ -axis, we get

$\begin{matrix} distance from (- c, 0) to (a, 0) - distance from (c, 0) to (a, 0) & = & d \\ (a + c) - (c - a) & = & d \\ 2 a & = & d \end{matrix}$

In other words, the fixed distance $d$ from the definition of the hyperbola is actually the length of the transverse axis! (Where have we seen that type of coincidence before?) Now consider a point $(x, y)$ on the hyperbola. Applying Definition 8.6, we get

$\begin{matrix} distance from (- c, 0) to (x, y) - distance from (c, 0) to (x, y) & = & 2 a \\ \sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} - \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} & = & 2 a \\ \sqrt{{(x + c)}^{2} + y^{2}} - \sqrt{{(x - c)}^{2} + y^{2}} & = & 2 a \end{matrix}$

Using the same arsenal of Intermediate Algebra weaponry we used in deriving the standard formula of an ellipse, Equation 8.5, we arrive at the following.¹

$\begin{matrix} (a^{2} - c^{2}) x^{2} + a^{2} y^{2} & = & a^{2} (a^{2} - c^{2}) \end{matrix}$

What remains is to determine the relationship between $a$ , $b$ and $c$ . To that end, we note that since $a$ and $c$ are both positive numbers with $a < c$ , we get $a^{2} < c^{2}$ so that $a^{2} - c^{2}$ is a negative number. Hence, $c^{2} - a^{2}$ is a positive number. For reasons which will become clear soon, we solve the equation for $\frac{y^{2}}{x^{2}}$ :

$\begin{matrix} (a^{2} - c^{2}) x^{2} + a^{2} y^{2} & = & a^{2} (a^{2} - c^{2}) \\ - (c^{2} - a^{2}) x^{2} + a^{2} y^{2} & = & - a^{2} (c^{2} - a^{2}) \\ a^{2} y^{2} & = & (c^{2} - a^{2}) x^{2} - a^{2} (c^{2} - a^{2}) \\ \frac{y^{2}}{x^{2}} & = & \frac{(c^{2} - a^{2})}{a^{2}} - \frac{(c^{2} - a^{2})}{x^{2}} \end{matrix}$

As $vertical bar x vertical bar → ∞$ ,² the quantity $\frac{(c^{2} - a^{2})}{x^{2}} \to 0$ so that $\frac{y^{2}}{x^{2}} \approx \frac{(c^{2} - a^{2})}{a^{2}}$ . By setting $b^{2} = c^{2} - a^{2}$ we get $\frac{y^{2}}{x^{2}} \approx \frac{b^{2}}{a^{2}}$ . This shows that $y \approx \pm \frac{b}{a} x$ , so that $y = \pm \frac{b}{a} x$ are the asymptotes to the graph as predicted and our choice of labels for the endpoints of the conjugate axis is justified. In our equation of the hyperbola we can substitute $a^{2} - c^{2} = - b^{2}$ which yields

$\begin{matrix} (a^{2} - c^{2}) x^{2} + a^{2} y^{2} & = & a^{2} (a^{2} - c^{2}) \\ - b^{2} x^{2} + a^{2} y^{2} & = & - a^{2} b^{2} \\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} & = & 1 \end{matrix}$

The equation above is for a hyperbola whose center is the origin and which opens to the left and right. If the hyperbola were centered at a point $(h, k)$ , we would get the following.

Equation

The Standard Equation of a Horizontal³ Hyperbola

For positive numbers $a$ and $b$ , the equation of a horizontal hyperbola with center $(h, k)$ is

$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$

If the roles of $x$ and $y$ were interchanged, then the hyperbola's branches would open upwards and downwards and we would get a `vertical' hyperbola.

Equation

The Standard Equation of a Vertical Hyperbola

For positive numbers $a$ and $b$ , the equation of a vertical hyperbola with center $(h, k)$ is:

$\frac{{(y - k)}^{2}}{b^{2}} - \frac{{(x - h)}^{2}}{a^{2}} = 1$

The values of $a$ and $b$ determine how far in the $x$ and $y$ directions, respectively, one counts from the center to determine the guide rectangle. In both cases, the distance from the center to the foci, $c$ , as seen in the derivation, can be found by the formula $c = \sqrt{a^{2} + b^{2}}$ . Lastly, note that we can quickly distinguish the equation of a hyperbola from that of a circle or ellipse because the hyperbola formula involves a difference of squares where the circle and ellipse formulas both involve the sum of squares.

Example 1

Graph each of the following equations below in the $x y$ -plane. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes.
1. $25 {(x - 2)}^{2} - 4 y^{2} = 100$ .
2. $9 y^{2} - x^{2} - 6 x = 10$ .
Graph $f (x) = \sqrt{x^{2} - 2 x - 3}$ .
Find the standard form of the equation of a hyperbola which satisfies the following characteristics:
1. the asymptotes are $y = \pm 2 x$ and the vertices are $(\pm 5, 0)$ .
2. the hyperbola graphed below:
  Figure 8.100

1. Owing to the difference of squares in $25 {(x - 2)}^{2} - 4 y^{2} = 100$ , we work towards putting this equation into the form of Equation 8.6 or Equation 8.7. To that end, we rewrite $y^{2}$ as $open parenthesis y minus 0 close parenthesis squared$ and divide through by $100$ :
  $25 {(x - 2)}^{2} - 4 y^{2} = 100 \to \frac{{(x - 2)}^{2}}{4} - \frac{{(y - 0)}^{2}}{25} = 1 \leftrightarrow \frac{{(x - 2)}^{2}}{{(2)}^{2}} - \frac{{(y - 0)}^{2}}{{(5)}^{2}} = 1 .$
  We identify $h equals 2$ and $k equals 0$ , so the hyperbola is centered at $open parenthesis 2, 0 close parenthesis$ . We also see $a equals 2$ , and $b equals 5$ , which means we move $2$ units to the left and to the right of the center and $5$ units up and down from the center to arrive at points on the guide rectangle: $(2 - 2, 0) = (0, 0)$ , $(2 + 2, 0) = (4, 0)$ , $(2, 0 + 5) = (2, 5)$ , and $(2, 0 - 5) = (2, - 5)$ . Since slant asymptotes pass through the center of the hyperbola as well as the corners of the rectangle, we get the set-up as drawn below on the left.
  Since the $y^{2}$ term is being subtracted from the $x^{2}$ term, we are in the situation of Equation 8.6. Hence, the branches of the hyperbola open to the left and right so the transverse axis lies along the $x$ -axis and the conjugate axis lies along the vertical line $x equals 2$ .
  Since the vertices of the hyperbola are where the hyperbola intersects the transverse axis, we get that the vertices are $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 4, 0 close parenthesis$ . To find the foci, we need $c = \sqrt{a^{2} + b^{2}} = \sqrt{4 + 25} = \sqrt{29}$ . Since the foci lie on the transverse axis, we move $the square root of 29$ units to the left and right of $open parenthesis 2, 0 close parenthesis$ to arrive at $(2 - \sqrt{29}, 0)$ (approximately $open parenthesis minus 3.39, 0 close parenthesis$ ) and $(2 + \sqrt{29}, 0)$ (approximately $open parenthesis 7.39, 0 close parenthesis$ ).
  Lastly, to determine the equations of the asymptotes, recall that the asymptotes pass through the center of the hyperbola, $open parenthesis 2, 0 close parenthesis$ , as well as the corners of guide rectangle. As such, they have slopes of $\pm \frac{b}{a} = \pm \frac{5}{2}$ . Feeding this information into the point-slope equation of a line, Equation A.5, we get $y - 0 = \pm \frac{5}{2} (x - 2)$ , so the asymptotes are $y = \frac{5}{2} x - 5$ and $y = - \frac{5}{2} x + 5$ . Putting it all together, we get our final graph below on the right.
  Figure 8.101
  Figure 8.102 The graph of $25 {(x - 2)}^{2} - 4 y^{2} = 100$ .
2. Since we have a difference of squares in $9 y^{2} - x^{2} - 6 x = 10$ , we aim to transform our given equation into Equation 8.6 or Equation 8.7. As we've seen with the other conic sections, we begin with completing the square.
  
  Now that this equation is in the standard form of Equation 8.7, we identify $h = - 3$ and $k equals 0$ so the center is $open parenthesis minus 3, 0 close parenthesis$ . We also see so $a equals 1$ , and $b = \frac{1}{3}$ which means that we move $1$ unit to the left and to the right of the center and $\frac{1}{3}$ units up and down from the center to arrive at points on the guide rectangle: $(- 3 - 1, 0) = (- 4, 0)$ , $(- 3 + 1, 0) = (- 2, 0)$ , $(- 3, 0 + \frac{1}{3}) = (- 3, \frac{1}{3})$ and $(- 3, 0 - \frac{1}{3}) = (- 3, - \frac{1}{3})$ .
  Since the $x^{2}$ term is being subtracted from the $y^{2}$ term, we know the branches of the hyperbola open upwards and downwards. This means the transverse axis lies along the vertical line $x = - 3$ and the conjugate axis lies along the $x$ -axis. As a result, we get the vertices are $(- 3, \frac{1}{3})$ and $(- 3, - \frac{1}{3})$ .
  To find the foci, we use $c = \sqrt{a^{2} + b^{2}} = \sqrt{\frac{1}{9} + 1} = \frac{\sqrt{10}}{3}$ . Since the foci lie on the transverse axis, we move $\frac{\sqrt{10}}{3}$ units above and below $open parenthesis minus 3, 0 close parenthesis$ to arrive at $(- 3, \frac{\sqrt{10}}{3})$ and $(- 3, - \frac{\sqrt{10}}{3})$ .
  To determine the asymptotes, we use the fact the asymptotes pass through the center of the hyperbola, $open parenthesis minus 3, 0 close parenthesis$ , as well as the corners of guide rectangle, so they have slopes of $\pm \frac{b}{a} = \pm \frac{1}{3}$ . Once again we use the point-slope equation of a line, Equation A.5, to get the two asymptotes $y = \frac{1}{3} x + 1$ and $y = - \frac{1}{3} x - 1$ . Our final graph is below.
  Figure 8.103 The graph of $9 y^{2} - x^{2} - 6 x = 10$ .
Graphing $f (x) = \sqrt{x^{2} - 2 x - 5}$ amounts to graphing the equation $y = \sqrt{x^{2} - 2 x - 5}$ . In order to use the tools we've learned in this chapter, we first square both sides to get a quadratic equation in two variables: $y^{2} = {(\sqrt{x^{2} - 2 x - 3})}^{2}$ . We get $y^{2} = x^{2} - 2 x - 3$ or $y^{2} - x^{2} + 2 x = - 3$ . We now set about transforming this equation into the form stated in Equation 8.6 or Equation 8.7.

We get the equation into the form of Equation 8.6 and identify $h equals 1$ and $k equals 0$ so the center is $open parenthesis 1, 0 close parenthesis$ . We have $a = b = 2$ , which means we move $2$ units to the left, to the right, up and down from the center to find points on the guide rectangle: $(1 - 2, 0) = (- 1, 0)$ , $(1 + 2, 0) = (3, 0)$ , $(1, 0 - 2) = (1, - 2)$ and $(1, 0 + 2) = (1, 2)$ . Of these four points, the vertices are $open parenthesis minus 1, 0 close parenthesis$ and $open parenthesis 3, 0 close parenthesis$ since the hyperbola opens to the left and to the right. As usual, we the guide rectangle helps us sketch the hyperbola along with its slant asymptotes, which we find are $y - 0 = \pm (x - 1)$ or $y = x - 1$ and $y = - x + 1$ .
We know since $f$ is a function, the graph of $f$ cannot be the entire hyperbola, otherwise the graph would fail the vertical line test. Since, by definition, $\sqrt{x^{2} - 2 x - 5} \geq 0$ , we know $f ⁢ open parenthesis x close parenthesis greater than or equal to 0$ . Hence the graph of $f$ is the upper half of the hyperbola, as shown below.
Figure 8.104 The graph of $f (x) = \sqrt{x^{2} - 2 x - 5}$ .
1. Plotting the data given to us below, we know the branches of the hyperbola open to the left and to the right. This means the our answer will take the form of Equation 8.6.
  Since the center is the midpoint of the vertices, we see the center is $open parenthesis 0, 0 close parenthesis$ , so $h = k = 0$ . Moreover, since the vertices are exactly $5$ units from the center, we know $a equals 5$ so $a squared equals 25$ . All that remains to find is the value of $b^{2}$ .
  Recall that the slopes of the asymptotes are $\pm \frac{b}{a}$ . Since $a equals 5$ and the slope of the line $y equals 2 ⁢ x$ is $2$ , we have that $\frac{b}{5} = 2$ , so $b equals 10$ . Hence, $b squared equals 100$ . Our final answer is $\frac{x^{2}}{25} - \frac{y^{2}}{100} = 1$ .
  Figure 8.105
2. From what we are given on the graph, the equation of the hyperbola takes the form of Equation 8.7. The vertices appear to be $open parenthesis 3, 1 close parenthesis$ and $open parenthesis 3, minus 3 close parenthesis$ whose midpoint gives us the center as $open parenthesis 3, minus 1 close parenthesis$ . Hence, $h equals 3$ and $k = - 1$ . Moreover, since the vertices are $2$ units above and below the center, we know $b equals 2$ so $b squared equals 4$ . All that remains is for us to find the value of $a^{2}$ .
  Since we are given two additional points, $open parenthesis 0, 3 close parenthesis$ and $open parenthesis 0, minus 5 close parenthesis$ , we choose one of them, $open parenthesis 0, 3 close parenthesis$ to find $a^{2}$ and use the other, $open parenthesis 0, minus 5 close parenthesis$ to partially check our answer.
  At this stage, we know the equation of the hyperbola is
  $\frac{{(y + 1)}^{2}}{4} - \frac{{(x - 3)}^{2}}{a^{2}} = 1,$
  so substituting $x equals 0$ and $y equals 3$ into this equation, we get $\frac{16}{4} - \frac{9}{a^{2}} = 1$ so $a squared equals 3$ . Hence, our final answer is
  $\frac{{(y + 1)}^{2}}{4} - \frac{{(x - 3)}^{2}}{3} = 1 .$
  We leave it to the reader to check.

As seen in Example 8.5.1, it is often the case we need to transform a given equation into the form specified by Equations 8.6 or 8.7. We summarize one method below.

To Write the Equation of a Hyperbola in Standard Form

Group common variables together on one side of the equation and put the constant on the other.
Complete the square on both variables as needed.
Divide both sides, if needed, to obtain $1$ on one side of the equation.

Hyperbolas can be used in so-called ` trilateration ,' or `positioning' problems. The procedure outlined in the next example is the basis of the (now defunct) LOng Range Aid to Navigation ( LORAN for short) system.⁴

Example 2

Jeff is stationed $10$ miles due west of Carl in an otherwise empty forest in an attempt to locate an elusive Sasquatch. At the stroke of midnight, Jeff records a Sasquatch call $9$ seconds earlier than Carl. If the speed of sound that night is $760$ miles per hour, determine a hyperbolic path along which Sasquatch must be located.
By a stroke of luck, Kai is also camping in the woods at this time. He is $6$ miles due north of Jeff and heard the Sasquatch call $18$ seconds after Jeff did. Use this added information to locate Sasquatch.

Since Jeff hears Sasquatch sooner, it is closer to Jeff than it is to Carl. Since the speed of sound is $760$ miles per hour, we can determine how much closer Sasquatch is to Jeff by multiplying
$760 \frac{miles}{hour} \times \frac{1 hour}{3600 seconds} \times 9 seconds = 1.9 miles$
This means that Sasquatch is $1.9$ miles closer to Jeff than it is to Carl. In other words, Sasquatch must lie on a path where
$(the distance to Carl) - (the distance to Jeff) = 1.9$
This is exactly the situation in the definition of a hyperbola, Definition 8.6. In this case, Jeff and Carl are located at the foci,⁵ and our fixed distance $d$ is 1.9. For simplicity, we assume the hyperbola is centered at $open parenthesis 0, 0 close parenthesis$ with its foci at $open parenthesis minus 5, 0 close parenthesis$ and $open parenthesis 5, 0 close parenthesis$ . Schematically:
Figure 8.106
We are seeking a curve of the form $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ in which the distance from the center to each focus is $c equals 5$ . As we saw in the derivation of the standard equation of the hyperbola, Equation 8.6, $d equals 2 ⁢ a$ , so that $2 ⁢ a equals 1.9$ , or $a equals 0.95$ and $a squared equals 0.9025$ . All that remains is to find $b^{2}$ . To that end, we recall that $a^{2} + b^{2} = c^{2}$ so $b^{2} = c^{2} - a^{2} = 25 - 0.9025 = 24.0975$ . Since Sasquatch is closer to Jeff than it is to Carl, it must be on the western (left hand) branch of
$\frac{x^{2}}{0.9025} - \frac{y^{2}}{24.0975} = 1 .$
Kai and Jeff are at the foci of a second hyperbola where the fixed distance $d$ is:
$760 \frac{miles}{hour} \times \frac{1 hour}{3600 seconds} \times 18 seconds = 3.8 miles$
Since Jeff is positioned at $open parenthesis minus 5, 0 close parenthesis$ , we place Kai at $open parenthesis minus 5, 6 close parenthesis$ . This puts the center of the new hyperbola at $open parenthesis minus 5, 3 close parenthesis$ . Plotting Kai's position and the new center gives us the diagram below on the left.
The second hyperbola is vertical, so it must be of the form $\frac{{(y - 3)}^{2}}{b^{2}} - \frac{{(x + 5)}^{2}}{a^{2}} = 1$ . As before, the distance $d$ is the length of the major axis, which in this case is $2 ⁢ b$ . We get $2 ⁢ b equals 3.8$ so that $b equals 1.9$ and $b squared equals 3.61$ . With Kai $6$ miles due North of Jeff, we have that the distance from the center to the focus is $c equals 3$ . Since $a^{2} + b^{2} = c^{2}$ , we get $a^{2} = c^{2} - b^{2} = 9 - 3.61 = 5.39$ .
Kai heard the Sasquatch call after Jeff, so Kai is farther from Sasquatch than Jeff. Thus Sasquatch must lie on the southern branch of the hyperbola
$\frac{{(y - 3)}^{2}}{3.61} - \frac{{(x + 5)}^{2}}{5.39} = 1 .$
Looking at the western branch of the hyperbola determined by Jeff and Carl along with the southern branch of the hyperbola determined by Kai and Jeff, we see that there is exactly one point in common, and this is where Sasquatch must have been when it called.
Figure 8.107
Figure 8.108
To determine the coordinates of this point of intersection exactly, we would need techniques for solving systems of non-linear equations (which we won't see until Section 9.7), so we use a graphing utility. Doing so, we get Sasquatch is approximately at $(- 0.9629, - 0.8113)$ .

Each of the conic sections we have studied in this chapter result from graphing equations of the form $A x^{2} + C y^{2} + D x + E y + F = 0$ for different choices of $A$ , $C$ , $D$ , $E$ , and⁶ $F$ . While we've seen examples demonstrate how to convert an equation from this general form to one of the standard forms, we close this chapter with some advice about which standard form to choose.⁷

Strategies for Identifying Conic Sections

Suppose the graph of equation $A x^{2} + C y^{2} + D x + E y + F = 0$ is a non-degenerate conic section.⁸

If just one variable is squared, the graph is a parabola. Rewrite the equation in the standard form given in Equation 8.1 (if $x$ is squared) or Equation 8.2 (if $y$ is squared).

If both variables are squared, look at the coefficients of $x^{2}$ and $y^{2}$ , $A$ and $C$ .

If $A = C$ , the graph is a circle. Rewrite the equation in the standard form given in Equation 8.3.
If $A \neq C$ but $A$ and $C$ have the same sign, the graph is an ellipse. Rewrite the equation in the standard form given in Equation 8.5.
If $A$ and $C$ have the different signs, the graph is a hyperbola. Rewrite the equation in the standard form given in either Equation 8.6 or Equation 8.7.

Exercises

In Exercises -, graph the hyperbola in the $x y$ -plane. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes.

$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
$\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$
$\frac{{(x - 2)}^{2}}{4} - \frac{{(y + 3)}^{2}}{9} = 1$
$\frac{{(y - 3)}^{2}}{11} - \frac{{(x - 1)}^{2}}{10} = 1$
$\frac{{(x + 4)}^{2}}{16} - {(y - 4)}^{2} = 1$
$\frac{{(x + 1)}^{2}}{9} - \frac{{(y - 3)}^{2}}{4} = 1$
$\frac{{(y + 2)}^{2}}{16} - \frac{{(x - 5)}^{2}}{20} = 1$
$\frac{{(x - 4)}^{2}}{8} - \frac{{(y - 2)}^{2}}{18} = 1$

In Exercises -, put the equation in standard form. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes.⁹

$12 x^{2} - 3 y^{2} + 30 y - 111 = 0$
$18 y^{2} - 5 x^{2} + 72 y + 30 x - 63 = 0$
$9 x^{2} - 25 y^{2} - 54 x - 50 y - 169 = 0$
$- 6 x^{2} + 5 y^{2} - 24 x + 40 y + 26 = 0$
For each of the odd numbered equations given in Exercises -, find two or more explicit functions of $x$ represented by each of the equations. (See Example 8.2.2 in Section 8.2.)

In Exercises -, graph each function by recognizing it as a portion of a hyperbola.

$f (x) = \sqrt{x^{2} - 4}$
$g (x) = - \sqrt{x^{2} - 4 x}$
$f (x) = - 2 \sqrt{x^{2} + 2 x - 3}$
$g (x) = - 2 + 2 \sqrt{x^{2} - 9}$

In Exercises -, find an equation for the hyperbola whose graph is given.

Figure 8.109
Figure 8.110

In Exercises -, find the standard form of the equation of the hyperbola which has the given properties.

Center $open parenthesis 3, 7 close parenthesis$ , Vertex $open parenthesis 3, 3 close parenthesis$ , Focus $open parenthesis 3, 2 close parenthesis$
Vertex $open parenthesis 0, 1 close parenthesis$ , Vertex $open parenthesis 8, 1 close parenthesis$ , Focus $open parenthesis minus 3, 1 close parenthesis$
Foci $(0, \pm 8)$ , Vertices $(0, \pm 5)$ .
Foci $(\pm 5, 0)$ , length of the Conjugate Axis $6$
Vertices $open parenthesis 3, 2 close parenthesis$ , $open parenthesis 13, 2 close parenthesis$ ; Endpoints of the Conjugate Axis $open parenthesis 8, 4 close parenthesis$ , $open parenthesis 8, 0 close parenthesis$
Vertex $open parenthesis minus 10, 5 close parenthesis$ , Asymptotes $y = \pm \frac{1}{2} (x - 6) + 5$

In Exercises -, find the standard form of the equation using the guidelines on page Section 8.5 and then graph the conic section.

$x^{2} - 2 x - 4 y - 11 = 0$
$x^{2} + y^{2} - 8 x + 4 y + 11 = 0$
$9 x^{2} + 4 y^{2} - 36 x + 24 y + 36 = 0$
$9 x^{2} - 4 y^{2} - 36 x - 24 y - 36 = 0$
$y^{2} + 8 y - 4 x + 16 = 0$
$4 x^{2} + y^{2} - 8 x + 4 = 0$
$4 x^{2} + 9 y^{2} - 8 x + 54 y + 49 = 0$
$x^{2} + y^{2} - 6 x + 4 y + 14 = 0$
$2 x^{2} + 4 y^{2} + 12 x - 8 y + 25 = 0$
$4 x^{2} - 5 y^{2} - 40 x - 20 y + 160 = 0$
The location of an earthquake's epicenter the point on the surface of the Earth directly above where the earthquake actually occurred can be determined by a process similar to how we located Sasquatch in Example 8.5.2. (As we said back in Exercise in Section 7.2, earthquakes are complicated events and it is not our intent to provide a complete discussion of the science involved in them. Instead, we refer the interested reader to a course in Geology or the U.S. Geological Survey's Earthquake Hazards Program found here .) Our technique works only for relatively small distances because we need to assume that the Earth is flat in order to use hyperbolas in the plane. The P-waves (“P” stands for Primary) of an earthquake in Sasquatchia travel at 6 kilometers per second.¹⁰ Station A records the waves first. Then Station B, which is 100 kilometers due north of Station A, records the waves 2 seconds later. Station C, which is 150 kilometers due west of Station A records the waves 3 seconds after that (a total of 5 seconds after Station A). Where is the epicenter?
The notion of eccentricity introduced for ellipses in Definition 8.5 in Section 8.4 is the same for hyperbolas in that we can define the eccentricity $e$ of a hyperbola as
$e = \frac{distance from the center to a focus}{distance from the center to a vertex}$
1. With the help of your classmates, explain why $e greater than 1$ for any hyperbola.
2. Find the equation of the hyperbola with vertices $(\pm 3, 0)$ and eccentricity $e equals 2$ .
3. With the help of your classmates, find the eccentricity of each of the hyperbolas in Exercises -. What role does eccentricity play in the shape of the graphs?
On page in Section 8.2, we discussed paraboloids of revolution when studying the design of satellite dishes and parabolic mirrors. In much the same way, `natural draft' cooling towers are often shaped as hyperboloids of revolution. Each vertical cross section of these towers is a hyperbola. Suppose the a natural draft cooling tower has the cross section below. Suppose the tower is 450 feet wide at the base, 275 feet wide at the top, and 220 feet at its narrowest point (which occurs 330 feet above the ground.) Determine the height of the tower to the nearest foot.
Figure 8.111
With the help of your classmates, research the Cassegrain Telescope. It uses the reflective property of the hyperbola as well as that of the parabola to make an ingenious telescope.
With the help of your classmates show that if $A x^{2} + C y^{2} + D x + E y + F = 0$ determines a non-degenerate conic¹¹ then
- $A ⁢ C less than 0$ means that the graph is a hyperbola
- $A ⁢ C equals 0$ means that the graph is a parabola
- $A ⁢ C greater than 0$ means that the graph is an ellipse or circle
NOTE: This result will be generalized in Theorem 14.9 in Section 14.4.1.

Answers

$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
Center $open parenthesis 0, 0 close parenthesis$ Transverse axis on $y equals 0$ Conjugate axis on $x equals 0$ Vertices $open parenthesis 4, 0 close parenthesis, open parenthesis minus 4, 0 close parenthesis$ Foci $open parenthesis 5, 0 close parenthesis, open parenthesis minus 5, 0 close parenthesis$ Asymptotes $y = \pm \frac{3}{4} x$
Figure 8.112
$\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$
Center $open parenthesis 0, 0 close parenthesis$ Transverse axis on $x equals 0$ Conjugate axis on $y equals 0$ Vertices $open parenthesis 0, 3 close parenthesis, open parenthesis 0, minus 3 close parenthesis$ Foci $open parenthesis 0, 5 close parenthesis, open parenthesis 0, minus 5 close parenthesis$ Asymptotes $y = \pm \frac{3}{4} x$
Figure 8.113
$\frac{{(x - 2)}^{2}}{4} - \frac{{(y + 3)}^{2}}{9} = 1$
Center $open parenthesis 2, minus 3 close parenthesis$ Transverse axis on $y = - 3$ Conjugate axis on $x equals 2$ Vertices $(0, - 3), (4, - 3)$ Foci $(2 + \sqrt{13}, - 3), (2 - \sqrt{13}, - 3)$ Asymptotes $y = \pm \frac{3}{2} (x - 2) - 3$
Figure 8.114
$\frac{{(y - 3)}^{2}}{11} - \frac{{(x - 1)}^{2}}{10} = 1$
Center $open parenthesis 1, 3 close parenthesis$ Transverse axis on $x equals 1$ Conjugate axis on $y equals 3$ Vertices $(1, 3 + \sqrt{11}), (1, 3 - \sqrt{11})$ Foci $(1, 3 + \sqrt{21}), (1, 3 - \sqrt{21})$ Asymptotes $y = \pm \frac{\sqrt{110}}{10} (x - 1) + 3$
Figure 8.115
$\frac{{(x + 4)}^{2}}{16} - \frac{{(y - 4)}^{2}}{1} = 1$
Center $open parenthesis minus 4, 4 close parenthesis$ Transverse axis on $y equals 4$ Conjugate axis on $x = - 4$ Vertices $open parenthesis minus 8, 4 close parenthesis, open parenthesis 0, 4 close parenthesis$ Foci $(- 4 + \sqrt{17}, 4), (- 4 - \sqrt{17}, 4)$ Asymptotes $y = \pm \frac{1}{4} (x + 4) + 4$
Figure 8.116
$\frac{{(x + 1)}^{2}}{9} - \frac{{(y - 3)}^{2}}{4} = 1$
Center $open parenthesis minus 1, 3 close parenthesis$ Transverse axis on $y equals 3$ Conjugate axis on $x = - 1$ Vertices $open parenthesis 2, 3 close parenthesis, open parenthesis minus 4, 3 close parenthesis$ Foci $(- 1 + \sqrt{13}, 3), (- 1 - \sqrt{13}, 3)$ Asymptotes $y = \pm \frac{2}{3} (x + 1) + 3$
Figure 8.117
$\frac{{(y + 2)}^{2}}{16} - \frac{{(x - 5)}^{2}}{20} = 1$
Center $open parenthesis 5, minus 2 close parenthesis$ Transverse axis on $x equals 5$ Conjugate axis on $y = - 2$ Vertices $open parenthesis 5, 2 close parenthesis, open parenthesis 5, minus 6 close parenthesis$ Foci $open parenthesis 5, 4 close parenthesis, open parenthesis 5, minus 8 close parenthesis$ Asymptotes $y = \pm \frac{2 \sqrt{5}}{5} (x - 5) - 2$
Figure 8.118
$\frac{{(x - 4)}^{2}}{8} - \frac{{(y - 2)}^{2}}{18} = 1$
Center $open parenthesis 4, 2 close parenthesis$ Transverse axis on $y equals 2$ Conjugate axis on $x equals 4$ Vertices $(4 + 2 \sqrt{2}, 2), (4 - 2 \sqrt{2}, 2)$ Foci $(4 + \sqrt{26}, 2), (4 - \sqrt{26}, 2)$ Asymptotes $y = \pm \frac{3}{2} (x - 4) + 2$
Figure 8.119
$\frac{x^{2}}{3} - \frac{{(y - 5)}^{2}}{12} = 1$
Center $open parenthesis 0, 5 close parenthesis$ Transverse axis on $y equals 5$ Conjugate axis on $x equals 0$ Vertices $(\sqrt{3}, 5), (- \sqrt{3}, 5)$ Foci $(\sqrt{15}, 5), (- \sqrt{15}, 5)$ Asymptotes $y = \pm 2 x + 5$
$\frac{{(y + 2)}^{2}}{5} - \frac{{(x - 3)}^{2}}{18} = 1$
Center $open parenthesis 3, minus 2 close parenthesis$ Transverse axis on $x equals 3$ Conjugate axis on $y = - 2$ Vertices $(3, - 2 + \sqrt{5}), (3, - 2 - \sqrt{5})$ Foci $(3, - 2 + \sqrt{23}), (3, - 2 - \sqrt{23})$ Asymptotes $y = \pm \frac{\sqrt{10}}{6} (x - 3) - 2$
$\frac{{(x - 3)}^{2}}{25} - \frac{{(y + 1)}^{2}}{9} = 1$
Center $open parenthesis 3, minus 1 close parenthesis$ Transverse axis on $y = - 1$ Conjugate axis on $x equals 3$ Vertices $(8, - 1), (- 2, - 1)$ Foci $(3 + \sqrt{34}, - 1), (3 - \sqrt{34}, - 1)$ Asymptotes $y = \pm \frac{3}{5} (x - 3) - 1$
$\frac{{(y + 4)}^{2}}{6} - \frac{{(x + 2)}^{2}}{5} = 1$
Center $(- 2, - 4)$ Transverse axis on $x = - 2$ Conjugate axis on $y = - 4$ Vertices $(- 2, - 4 + \sqrt{6}), (- 2, - 4 - \sqrt{6})$ Foci $(- 2, - 4 + \sqrt{11}), (- 2, - 4 - \sqrt{11})$ Asymptotes $y = \pm \frac{\sqrt{30}}{5} (x + 2) - 4$
For number:
- $f (x) = \frac{3}{4} \sqrt{x^{2} - 16}$ represents the upper half of the hyperbola.
- $g (x) = - \frac{3}{4} \sqrt{x^{2} - 16}$ represents the lower half of the hyperbola.
For number:
- $f (x) = - 3 + \frac{3}{2} \sqrt{x^{2} - 4 x}$ represents the upper half of the hyperbola.
- $g (x) = - 3 - \frac{3}{2} \sqrt{x^{2} - 4 x}$ represents the lower half of the hyperbola.
For number:
- $f (x) = 4 + \frac{1}{4} \sqrt{x^{2} + 8 x}$ represents the upper half of the hyperbola.
- $g (x) = 4 - \frac{1}{4} \sqrt{x^{2} + 8 x}$ represents the lower half of the hyperbola.
For number:
- $f (x) = - 2 + \frac{2}{5} \sqrt{5 x^{2} - 50 x + 225}$ represents the upper half of the hyperbola.
- $g (x) = - 2 - \frac{2}{5} \sqrt{5 x^{2} - 50 x + 225}$ represents the lower half of the hyperbola.
For number:
- $f (x) = 5 + 2 \sqrt{x^{2} - 3}$ represents the upper half of the hyperbola.
- $g (x) = 5 - 2 \sqrt{x^{2} - 3}$ represents the lower half of the hyperbola.
For number:
- $f (x) = - 1 + \frac{3}{5} \sqrt{x^{2} - 6 x - 16}$ represents the upper half of the hyperbola.
- $g (x) = - 1 - \frac{3}{5} \sqrt{x^{2} - 6 x - 16}$ represents the lower half of the hyperbola.
$f (x) = \sqrt{x^{2} - 4}$
Figure 8.120
$g (x) = - \sqrt{x^{2} - 4 x}$
Figure 8.121
$f (x) = - 2 \sqrt{x^{2} + 2 x - 3}$
Figure 8.122
$g (x) = - 2 + 2 \sqrt{x^{2} - 9}$
Figure 8.123
$\frac{x^{2}}{16} - \frac{y^{2}}{16} = 1$
$\frac{{(y - 4)}^{2}}{4} - \frac{{(x - 4)}^{2}}{3} = 1$
$\frac{{(y - 7)}^{2}}{16} - \frac{{(x - 3)}^{2}}{9} = 1$
$\frac{{(x - 4)}^{2}}{16} - \frac{{(y - 1)}^{2}}{33} = 1$
$\frac{y^{2}}{25} - \frac{x^{2}}{39} = 1$
$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
$\frac{{(x - 8)}^{2}}{25} - \frac{{(y - 2)}^{2}}{4} = 1$
$\frac{{(x - 6)}^{2}}{256} - \frac{{(y - 5)}^{2}}{64} = 1$
${(x - 1)}^{2} = 4 (y + 3)$
Figure 8.124
${(x - 4)}^{2} + {(y + 2)}^{2} = 9$
Figure 8.125
$\frac{{(x - 2)}^{2}}{4} + \frac{{(y + 3)}^{2}}{9} = 1$
Figure 8.126
$\frac{{(x - 2)}^{2}}{4} - \frac{{(y + 3)}^{2}}{9} = 1$
Figure 8.127
${(y + 4)}^{2} = 4 x$
Figure 8.128
$\frac{{(x - 1)}^{2}}{1} + \frac{y^{2}}{4} = 0$ The graph is the point $open parenthesis 1, 0 close parenthesis$ only.
$\frac{{(x - 1)}^{2}}{9} + \frac{{(y + 3)}^{2}}{4} = 1$
Figure 8.129
${(x - 3)}^{2} + {(y + 2)}^{2} = - 1$ There is no graph.
$\frac{{(x + 3)}^{2}}{2} + \frac{{(y - 1)}^{2}}{1} = - \frac{3}{4}$ There is no graph.
$\frac{{(y + 2)}^{2}}{16} - \frac{{(x - 5)}^{2}}{20} = 1$
Figure 8.130
By placing Station A at $open parenthesis 0, minus 50 close parenthesis$ and Station B at $open parenthesis 0, 50 close parenthesis$ , the two second time difference yields the hyperbola $\frac{y^{2}}{36} - \frac{x^{2}}{2464} = 1$ with foci A and B and center $open parenthesis 0, 0 close parenthesis$ . Placing Station C at $(- 150, - 50)$ and using foci A and C gives us a center of $(- 75, - 50)$ and the hyperbola $\frac{{(x + 75)}^{2}}{225} - \frac{{(y + 50)}^{2}}{5400} = 1$ . The point of intersection of these two hyperbolas which is closer to A than B and closer to A than C is $(- 57.8444, - 9.21336)$ so that is the epicenter.
1. $\frac{x^{2}}{9} - \frac{y^{2}}{27} = 1$ .
The tower may be modeled (approximately)¹² by $\frac{x^{2}}{12100} - \frac{{(y - 330)}^{2}}{34203} = 1$ . To find the height, we plug in $x equals 137.5$ which yields $y approximately equals 191$ or $y approximately equals 469$ . Since the top of the tower is above the narrowest point, we get the tower is approximately 469 feet tall.

It is a good exercise to actually work this out. ↩
Recall this means we are analyzing the behavior as $x → ∞$ and as $x → minus ∞$ . ↩
That is, a hyperbola whose branches open to the left and right ↩
GPS now rules the positioning kingdom. Is there still a place for LORAN? Do satellites ever malfunction? ↩
We usually like to be the center of attention, but being the focus of attention works equally well. ↩
See Section 14.4 to see why we skip $B$ . ↩
We formalize this in Exercise . ↩
That is, a parabola, circle, ellipse, or hyperbola – see Section 8.1. ↩
…assuming the equation were graphed in the $x y$ -plane … ↩
Depending on the composition of the crust at a specific location, P-waves can travel between 5 kps and 8 kps. ↩
Recall that this means its graph is either a circle, parabola, ellipse or hyperbola. ↩
The exact value underneath $open parenthesis y minus 330 close parenthesis squared$ is $\frac{52707600}{1541}$ in case you need more precision. ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.