Contents

Preface
1Foundations
- Introduction
- 1.1Use the Language of Algebra
- 1.2Integers
- 1.3Fractions
- 1.4Decimals
- 1.5Properties of Real Numbers
- Chapter Review
  Key Terms
  Key Concepts
- Exercises
  Review Exercises
  Practice Test
2Solving Linear Equations
3Graphs and Functions
4Systems of Linear Equations
5Polynomials and Polynomial Functions
6Factoring
7Rational Expressions and Functions
8Roots and Radicals
9Quadratic Equations and Functions
10Exponential and Logarithmic Functions
11Conics
12Sequences, Series and Binomial Theorem
- Introduction
- 12.1Sequences
- 12.2Arithmetic Sequences
- 12.3Geometric Sequences and Series
- 12.4Binomial Theorem
- Chapter Review
  Key Terms
Answer Key
Index

Hyperbolas

Learning Objectives

By the end of this section, you will be able to:

Graph a hyperbola with center at $(0, 0) (0, 0)$
Graph a hyperbola with center at $(h, k) (h, k)$
Identify conic sections by their equations

Be Prepared 11.4

Before you get started, take this readiness quiz.

Solve: $x^{2} = 12 . x^{2} = 12 .$
If you missed this problem, review Example 9.1.
Expand: ${(x - 4)}^{2} . {(x - 4)}^{2} .$
If you missed this problem, review Example 5.32.
Graph $y = - \frac{2}{3} x . y = - \frac{2}{3} x .$
If you missed this problem, review Example 3.4.

Graph a Hyperbola with Center at (0, 0)

The last conic section we will look at is called a hyperbola. We will see that the equation of a hyperbola looks the same as the equation of an ellipse, except it is a difference rather than a sum. While the equations of an ellipse and a hyperbola are very similar, their graphs are very different.

We define a hyperbola as all points in a plane where the difference of their distances from two fixed points is constant. Each of the fixed points is called a focus of the hyperbola.

Hyperbola

A hyperbola is all points in a plane where the difference of their distances from two fixed points is constant. Each of the fixed points is called a focus of the hyperbola.

The line through the foci, is called the transverse axis. The two points where the transverse axis intersects the hyperbola are each a vertex of the hyperbola. The midpoint of the segment joining the foci is called the center of the hyperbola. The line perpendicular to the transverse axis that passes through the center is called the conjugate axis. Each piece of the graph is called a branch of the hyperbola.

The figure shows two graphs of a hyperbola. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci are shown with points that lie on the transverse axis, which is the x-axis. The branches pass through the vertices and open left and right. The y-axis is the conjugate axis. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices and foci lie are shown with points that lie on the transverse axis, which is the y-axis. The branches pass through the vertices and open up and down. The x-axis is the conjugate axis.

Again our goal is to connect the geometry of a conic with algebra. Placing the hyperbola on a rectangular coordinate system gives us that opportunity. In the figure, we placed the hyperbola so the foci $((− c, 0), (c, 0)) ((− c, 0), (c, 0))$ are on the x-axis and the center is the origin.

The figure shows the graph of a hyperbola. The graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The foci (negative c, 0) and (c, 0) are marked with a point and lie on the x-axis. The vertices are marked with a point and lie on the x-axis. The branches pass through the vertices and open left and right. The distance from (negative c, 0) to a point on the branch (x, y) is marked d sub 1. The distance from (x, y) on the branch to (c, 0) is marked d sub 2.

The definition states the difference of the distance from the foci to a point $(x, y) (x, y)$ is constant. So $| d_{1} - d_{2} | | d_{1} - d_{2} |$ is a constant that we will call $2 a 2 a$ so $| d_{1} - d_{2} | = 2 a . | d_{1} - d_{2} | = 2 a .$ We will use the distance formula to lead us to an algebraic formula for an ellipse.

$\begin{matrix} | d_{1} - d_{2} | = 2 a \\ Use the distance formula to find d_{1}, d_{2} & | \sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} - \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} | = 2 a \\ Eliminate the radicals. \\ \begin{matrix} To simplify the equation of the ellipse, we \\ let c^{2} - a^{2} = b^{2} . \end{matrix} & \frac{x^{2}}{a^{2}} + \frac{y^{2}}{c^{2} - a^{2}} = 1 \\ \begin{matrix} So, the equation of a hyperbola centered at \\ the origin in standard form is: \end{matrix} & \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \end{matrix} \begin{matrix} Use the distance formula to find \\ Eliminate the radicals. \\ \begin{matrix} To simplify the equation of the ellipse, we \\ let \end{matrix} \\ \begin{matrix} So, the equation of a hyperbola centered at \\ the origin in standard form is: \end{matrix} \end{matrix}$

To graph the hyperbola, it will be helpful to know about the intercepts. We will find the x-intercepts and y-intercepts using the formula.

$\begin{matrix} \begin{matrix} x -intercepts & y -intercepts \end{matrix} \\ \begin{matrix} \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} & = & 1 & \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} & = & 1 \\ Let y = 0 . & \frac{x^{2}}{a^{2}} - \frac{0^{2}}{b^{2}} & = & 1 & Let x = 0 . & \frac{0^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} & = & 1 \\ \frac{x^{2}}{a^{2}} & = & 1 & - \frac{y^{2}}{b^{2}} & = & 1 \\ x^{2} & = & a^{2} & y^{2} & = & − b^{2} \\ x & = & ± a & y & = & ± \sqrt{− b^{2}} \end{matrix} \\ \begin{matrix} The x -intercepts are (a, 0) and (− a, 0) . & There are no y -intercepts. \end{matrix} \end{matrix} \begin{matrix} \begin{matrix} \end{matrix} \\ \begin{matrix} \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \\ Let & \frac{x^{2}}{a^{2}} - \frac{0^{2}}{b^{2}} = 1 \frac{0^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \\ \frac{x^{2}}{a^{2}} = 1 - \frac{y^{2}}{b^{2}} = 1 \\ x^{2} = a^{2} y^{2} = − b^{2} \\ x = ± a y = ± \sqrt{− b^{2}} \end{matrix} \\ \begin{matrix} \end{matrix} \end{matrix}$

The a, b values in the equation also help us find the asymptotes of the hyperbola. The asymptotes are intersecting straight lines that the branches of the graph approach but never intersect as the x, y values get larger and larger.

To find the asymptotes, we sketch a rectangle whose sides intersect the x-axis at the vertices $(− a, 0), (− a, 0),$ $(a, 0) (a, 0)$ and intersect the y-axis at $(0, − b), (0, − b),$ $(0, b) . (0, b) .$ The lines containing the diagonals of this rectangle are the asymptotes of the hyperbola. The rectangle and asymptotes are not part of the hyperbola, but they help us graph the hyperbola.

The asymptotes pass through the origin and we can evaluate their slope using the rectangle we sketched. They have equations $y = \frac{b}{a} x y = \frac{b}{a} x$ and $y = - \frac{b}{a} x . y = - \frac{b}{a} x .$

There are two equations for hyperbolas, depending whether the transverse axis is vertical or horizontal. We can tell whether the transverse axis is horizontal by looking at the equation. When the equation is in standard form, if the x²-term is positive, the transverse axis is horizontal. When the equation is in standard form, if the y²-term is positive, the transverse axis is vertical.

The second equations could be derived similarly to what we have done. We will summarize the results here.

Standard Form of the Equation a Hyperbola with Center $(0, 0) (0, 0)$

The standard form of the equation of a hyperbola with center $(0, 0), (0, 0),$ is

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

The figure shows the graph of two hyperbolas. The first graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (negative a, 0) and (a, 0) and are marked with a point and lie on the x-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the x-axis at the vertices (negative a, 0) and (a, 0) and intersect the y-axis at (0, b) and (0, negative b). The asymptotes are given by y is equal to b divided by a times x and y is equal to negative b divided by a times x and are drawn as the diagonals of the central rectangle. The branches of the hyperbola pass through the vertices, open left and right, and approach the asymptotes. The second graph shows the x-axis and y-axis that both run in the negative and positive directions, but at unlabeled intervals. The center of the hyperbola is the origin. The vertices are (0, a) and (0, negative a) and are marked with a point and lie on the y-axis. The points (0, b) and (0, negative) lie on the on the y-axis. There is a central rectangle who sides intersect the y-axis at the vertices (0, a) and (0, negative a) and intersect the y-axis at (negative b, 0) and (b, 0). The branches of the hyperbola pass through the vertices, open up and down, and approach the asymptotes.

Notice that, unlike the equation of an ellipse, the denominator of $x^{2} x^{2}$ is not always $a^{2} a^{2}$ and the denominator of $y^{2} y^{2}$ is not always $b^{2} . b^{2} .$

Notice that when the $x^{2} x^{2}$ -term is positive, the transverse axis is on the x-axis. When the $y^{2} y^{2}$ -term is positive, the transverse axis is on the y-axis.

Standard Forms of the Equation a Hyperbola with Center $(0, 0) (0, 0)$
	$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$	$\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$
Orientation	Transverse axis on the x-axis. Opens left and right	Transverse axis on the y-axis. Opens up and down
Vertices	$(− a, 0), (− a, 0),$ $(a, 0) (a, 0)$	$(0, − a), (0, − a),$ $(0, a) (0, a)$
x-intercepts	$(− a, 0), (− a, 0),$ $(a, 0) (a, 0)$	none
y-intercepts	none	$(0, − a), (0, − a),$ $(0, a) (0, a)$
Rectangle	Use $(± a, 0) (± a, 0)$ $(0, ± b) (0, ± b)$	Use $(0, ± a) (0, ± a)$ $(± b, 0) (± b, 0)$
asymptotes	$y = \frac{b}{a} x, y = \frac{b}{a} x,$ $y = - \frac{b}{a} x y = - \frac{b}{a} x$	$y = \frac{a}{b} x, y = \frac{a}{b} x,$ $y = - \frac{a}{b} x y = - \frac{a}{b} x$

We will use these properties to graph hyperbolas.

Example 11.27

How to Graph a Hyperbola with Center $(0, 0) (0, 0)$

Graph $\frac{x^{2}}{25} - \frac{y^{2}}{4} = 1 . \frac{x^{2}}{25} - \frac{y^{2}}{4} = 1 .$

Show solution

Solution

Step 2 is to determine whether the transverse axis is horizontal or vertical. Since the x squared term is positive, the transverse axis is horizontal.

Step 3 is to find the vertices. Since a squared is equal to 25, then a is equal to plus or minus 5. The vertices lie on the x-axis and are (negative 5, 0) and (5, 0).

Step 4 is to sketch the rectangle centered at the origin, intersecting one axis at plus or minus a and the other at plus or minus b. Since a is equal to plus or minus 5, the rectangle will intersect the x-axis at the vertices. Since b is equal to plus or minus 2, the rectangle will intersect the y-axis at (0, negative 2) and (0, 2). The rectangle is shown on a coordinate plane with the points (0, 2), (0, negative 2), (negative 5, 0), and (5, 0) labeled.

Step 5 is to sketch the asymptotes, the lines through the diagonals of the rectangle. The asymptotes have the equations y is equal to five-halves times x and y is equal to negative five-halves x. The coordinate plane shows the rectangle with the points (0, 2), (0, negative 2), (negative 5, 0), and (5, 0) labeled and the lines that represent the asymptotes.

Step 6 is to draw the two branches of the hyperbola. Start at each vertex and use the asymptotes as a guide. The coordinate plane shows the rectangle with the points (0, 2), (0, negative 2), (negative 5, 0), and (5, 0) labeled, the lines that represent the asymptotes, y is equal to plus or minus five-halves times x, and the branches that pass through (plus or minus 5, 0) and open left and right.

Try It 11.53

Graph $\frac{x^{2}}{16} - \frac{y^{2}}{4} = 1 . \frac{x^{2}}{16} - \frac{y^{2}}{4} = 1 .$

Try It 11.54

Graph $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 . \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 .$

We summarize the steps for reference.

How To

Graph a hyperbola centered at $(0, 0) . (0, 0) .$

Step 1. Write the equation in standard form.
Step 2. Determine whether the transverse axis is horizontal or vertical.
Step 3. Find the vertices.
Step 4. Sketch the rectangle centered at the origin intersecting one axis at $± a ± a$ and the other at $± b . ± b .$
Step 5. Sketch the asymptotes—the lines through the diagonals of the rectangle.
Step 6. Draw the two branches of the hyperbola.

Sometimes the equation for a hyperbola needs to be first placed in standard form before we graph it.

Example 11.28

Graph $4 y^{2} - 16 x^{2} = 64 . 4 y^{2} - 16 x^{2} = 64 .$

Show solution

Solution

	$4 y^{2} - 16 x^{2} = 64 4 y^{2} - 16 x^{2} = 64$
To write the equation in standard form, divide each term by 64 to make the equation equal to 1.	$\frac{4 y^{2}}{64} - \frac{16 x^{2}}{64} = \frac{64}{64} \frac{4 y^{2}}{64} - \frac{16 x^{2}}{64} = \frac{64}{64}$
Simplify.	$\frac{y^{2}}{16} - \frac{x^{2}}{4} = 1$
Since the y²-term is positive, the transverse axis is vertical. Since $a^{2} = 16 a^{2} = 16$ then $a = ± 4 . a = ± 4 .$
The vertices are on the y-axis, $(0, − a), (0, − a),$ $(0, a) . (0, a) .$ Since $b^{2} = 4 b^{2} = 4$ then $b = ± 2 . b = ± 2 .$	$(0, −4), (0, −4),$ $(0, 4) (0, 4)$
Sketch the rectangle intersecting the x-axis at $(−2, 0), (−2, 0),$ $(2, 0) (2, 0)$ and the y-axis at the vertices. Sketch the asymptotes through the diagonals of the rectangle. Draw the two branches of the hyperbola.

Try It 11.55

Graph $4 y^{2} - 25 x^{2} = 100 . 4 y^{2} - 25 x^{2} = 100 .$

Try It 11.56

Graph $25 y^{2} - 9 x^{2} = 225 . 25 y^{2} - 9 x^{2} = 225 .$

Graph a Hyperbola with Center at $(h, k) (h, k)$

Hyperbolas are not always centered at the origin. When a hyperbola is centered at $(h, k) (h, k)$ the equations changes a bit as reflected in the table.

Standard Forms of the Equation a Hyperbola with Center $(h, k) (h, k)$
	$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1 \frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$	$\frac{{(y - k)}^{2}}{a^{2}} - \frac{{(x - h)}^{2}}{b^{2}} = 1 \frac{{(y - k)}^{2}}{a^{2}} - \frac{{(x - h)}^{2}}{b^{2}} = 1$
Orientation	Transverse axis is horizontal. Opens left and right	Transverse axis is vertical. Opens up and down
Center	$(h, k) (h, k)$	$(h, k) (h, k)$
Vertices	a units to the left and right of the center	a units above and below the center
Rectangle	Use a units left/right of center b units above/ below the center	Use a units above/below the center b units left/right of center

Example 11.29

How to Graph a Hyperbola with Center $(h, k) (h, k)$

Graph $\frac{{(x - 1)}^{2}}{9} - \frac{{(y - 2)}^{2}}{16} = 1 \frac{{(x - 1)}^{2}}{9} - \frac{{(y - 2)}^{2}}{16} = 1$

Show solution

Solution

Step 2 is to deteremine whether the transverse axis is horizonal or vertical. Since the x squared term is positive, the hyperbola opens left and right. The transverse axis is horizontal. The hyperbola opens left and right.

Step 3 is to find the center and a and b. h is equal to 1 and k is equal 2. a squared is equal to 9 and b squared is equal to 16. You can see tha x minus h is x minus 1, and that y minus k is y minus 2. So, the center is (1, 2) and a is equal to 3 and b is equal to 4.

Step 4 is to sketch the rectangle centered at (h, k) using a and b. Mark the center (1, 2) on a coordinate plane. Sketch the rectangle that goes through the points 3 units to the left and right of the center and 4 units above and below the center.

Step 5 is to sketch the asymptotes on the coordinate plane. They are the lines through the diagonals of the retcangle. Mark the vertices which lie on the rectangle 3 units to the left and right of the center. The vertices are (negative 2, 2) and (4, 2).

Step 6 is to draw the branches of the hyperbola. Start the vertices, (negative 2, 2) and (4, 2) and use the asymptotes as a guide. The branches should open left and right.

Try It 11.57

Graph $\frac{{(x - 3)}^{2}}{25} - \frac{{(y - 1)}^{2}}{9} = 1 . \frac{{(x - 3)}^{2}}{25} - \frac{{(y - 1)}^{2}}{9} = 1 .$

Try It 11.58

Graph $\frac{{(x - 2)}^{2}}{4} - \frac{{(y - 2)}^{2}}{9} = 1 . \frac{{(x - 2)}^{2}}{4} - \frac{{(y - 2)}^{2}}{9} = 1 .$

We summarize the steps for easy reference.

How To

Graph a hyperbola centered at $(h, k) . (h, k) .$

Step 1. Write the equation in standard form.
Step 2. Determine whether the transverse axis is horizontal or vertical.
Step 3. Find the center and a, b.
Step 4. Sketch the rectangle centered at $(h, k) (h, k)$ using a, b.
Step 5. Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices.
Step 6. Draw the two branches of the hyperbola.

Be careful as you identify the center. The standard equation has $x - h x - h$ and $y - k y - k$ with the center as $(h, k) . (h, k) .$

Example 11.30

Graph $\frac{{(y + 2)}^{2}}{9} - \frac{{(x + 1)}^{2}}{4} = 1 . \frac{{(y + 2)}^{2}}{9} - \frac{{(x + 1)}^{2}}{4} = 1 .$

Show solution

Solution


Since the $y^{2} - y^{2} -$ term is positive, the hyperbola opens up and down.
Find the center, $(h, k) . (h, k) .$	Center: $(−1, −2) (−1, −2)$
Find a, b.	$a = 3 a = 3$ $b = 2 b = 2$
Sketch the rectangle that goes through the points 3 units above and below the center and 2 units to the left/right of the center. Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices. Graph the branches.

Try It 11.59

Graph $\frac{{(y + 3)}^{2}}{16} - \frac{{(x + 2)}^{2}}{9} = 1 . \frac{{(y + 3)}^{2}}{16} - \frac{{(x + 2)}^{2}}{9} = 1 .$

Try It 11.60

Graph $\frac{{(y + 2)}^{2}}{9} - \frac{{(x + 2)}^{2}}{9} = 1 . \frac{{(y + 2)}^{2}}{9} - \frac{{(x + 2)}^{2}}{9} = 1 .$

Again, sometimes we have to put the equation in standard form as our first step.

Example 11.31

Write the equation in standard form and graph $4 x^{2} - 9 y^{2} - 24 x - 36 y - 36 = 0 . 4 x^{2} - 9 y^{2} - 24 x - 36 y - 36 = 0 .$

Show solution

Solution


To get to standard form, complete the squares.


Divide each term by 36 to get the constant to be 1.

Since the $x^{2} - x^{2} -$ term is positive, the hyperbola opens left and right.
Find the center, $(h, k) . (h, k) .$	Center: $(3, −2) (3, −2)$
Find a, b.	$\begin{matrix} a = 3 \\ b = 4 \end{matrix} \begin{matrix} a = 3 \\ b = 4 \end{matrix}$
Sketch the rectangle that goes through the points 3 units to the left/right of the center and 2 units above and below the center. Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices. Graph the branches.

Try It 11.61

ⓐ Write the equation in standard form and ⓑ graph $9 x^{2} - 16 y^{2} + 18 x + 64 y - 199 = 0 . 9 x^{2} - 16 y^{2} + 18 x + 64 y - 199 = 0 .$

Try It 11.62

ⓐ Write the equation in standard form and ⓑ graph $16 x^{2} - 25 y^{2} + 96 x - 50 y - 281 = 0 . 16 x^{2} - 25 y^{2} + 96 x - 50 y - 281 = 0 .$

Identify Conic Sections by their Equations

Now that we have completed our study of the conic sections, we will take a look at the different equations and recognize some ways to identify a conic by its equation. When we are given an equation to graph, it is helpful to identify the conic so we know what next steps to take.

To identify a conic from its equation, it is easier if we put the variable terms on one side of the equation and the constants on the other.

Conic	Characteristics of $x^{2} - x^{2} -$ and $y^{2} - y^{2} -$ terms	Example
Parabola	Either $x^{2} x^{2}$ OR $y^{2} . y^{2} .$ Only one variable is squared.	$x = 3 y^{2} - 2 y + 1 x = 3 y^{2} - 2 y + 1$
Circle	$x^{2} - x^{2} -$ and $y^{2} - y^{2} -$ terms have the same coefficients	$x^{2} + y^{2} = 49 x^{2} + y^{2} = 49$
Ellipse	$x^{2} - x^{2} -$ and $y^{2} - y^{2} -$ terms have the same sign, different coefficients	$4 x^{2} + 25 y^{2} = 100 4 x^{2} + 25 y^{2} = 100$
Hyperbola	$x^{2} - x^{2} -$ and $y^{2} - y^{2} -$ terms have different signs, different coefficients	$25 y^{2} - 4 x^{2} = 100 25 y^{2} - 4 x^{2} = 100$

Example 11.32

Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

ⓐ $9 x^{2} + 4 y^{2} + 56 y + 160 = 0 9 x^{2} + 4 y^{2} + 56 y + 160 = 0$ ⓑ $9 x^{2} - 16 y^{2} + 18 x + 64 y - 199 = 0 9 x^{2} - 16 y^{2} + 18 x + 64 y - 199 = 0$ ⓒ $x^{2} + y^{2} - 6 x - 8 y = 0 x^{2} + y^{2} - 6 x - 8 y = 0$ ⓓ $y = −2 x^{2} - 4 x - 5 y = −2 x^{2} - 4 x - 5$

Show solution

Solution

ⓐ
$\begin{matrix} 9 x^{2} + 4 y^{2} + 56 y + 160 = 0 \\ \begin{matrix} The x^{2} - and y^{2} -terms have the same sign and \\ different coefficients. \end{matrix} & Ellipse \end{matrix} \begin{matrix} \begin{matrix} The \\ different coefficients. \end{matrix} \end{matrix}$

ⓑ
$\begin{matrix} 9 x^{2} - 16 y^{2} + 18 x + 64 y - 199 = 0 \\ \begin{matrix} The x^{2} - and y^{2} -terms have different signs and \\ different coefficients. \end{matrix} & Hyperbola \end{matrix} \begin{matrix} \begin{matrix} The \\ different coefficients. \end{matrix} \end{matrix}$

ⓒ
$\begin{matrix} x^{2} + y^{2} - 6 x - 8 y = 0 \\ The x^{2} - and y^{2} -terms have the same coefficients. & Circle \end{matrix} \begin{matrix} The \end{matrix}$

ⓓ
$\begin{matrix} y = −2 x^{2} - 4 x - 5 \\ Only one variable, x, is squared. & Parabola \end{matrix} \begin{matrix} Only one variable, \end{matrix}$

Try It 11.63

Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

ⓐ $x^{2} + y^{2} - 8 x - 6 y = 0 x^{2} + y^{2} - 8 x - 6 y = 0$ ⓑ $4 x^{2} + 25 y^{2} = 100 4 x^{2} + 25 y^{2} = 100$ ⓒ $y = 6 x^{2} + 2 x - 1 y = 6 x^{2} + 2 x - 1$ ⓓ $16 y^{2} - 9 x^{2} = 144 16 y^{2} - 9 x^{2} = 144$

Try It 11.64

Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

ⓐ $16 x^{2} + 9 y^{2} = 144 16 x^{2} + 9 y^{2} = 144$ ⓑ $y = 2 x^{2} + 4 x + 6 y = 2 x^{2} + 4 x + 6$ ⓒ $x^{2} + y^{2} + 2 x + 6 y + 9 = 0 x^{2} + y^{2} + 2 x + 6 y + 9 = 0$ ⓓ $4 x^{2} - 16 y^{2} = 64 4 x^{2} - 16 y^{2} = 64$

Media

Access these online resources for additional instructions and practice with hyperbolas.

Practice Makes Perfect

Graph a Hyperbola with Center at $(0, 0) (0, 0)$

In the following exercises, graph.

147.

$\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1 \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$

148.

$\frac{x^{2}}{25} - \frac{y^{2}}{9} = 1 \frac{x^{2}}{25} - \frac{y^{2}}{9} = 1$

149.

$\frac{x^{2}}{16} - \frac{y^{2}}{25} = 1 \frac{x^{2}}{16} - \frac{y^{2}}{25} = 1$

150.

$\frac{x^{2}}{9} - \frac{y^{2}}{36} = 1 \frac{x^{2}}{9} - \frac{y^{2}}{36} = 1$

151.

$\frac{y^{2}}{25} - \frac{x^{2}}{4} = 1 \frac{y^{2}}{25} - \frac{x^{2}}{4} = 1$

152.

$\frac{y^{2}}{36} - \frac{x^{2}}{16} = 1 \frac{y^{2}}{36} - \frac{x^{2}}{16} = 1$

153.

$16 y^{2} - 9 x^{2} = 144 16 y^{2} - 9 x^{2} = 144$

154.

$25 y^{2} - 9 x^{2} = 225 25 y^{2} - 9 x^{2} = 225$

155.

$4 y^{2} - 9 x^{2} = 36 4 y^{2} - 9 x^{2} = 36$

156.

$16 y^{2} - 25 x^{2} = 400 16 y^{2} - 25 x^{2} = 400$

157.

$4 x^{2} - 16 y^{2} = 64 4 x^{2} - 16 y^{2} = 64$

158.

$9 x^{2} - 4 y^{2} = 36 9 x^{2} - 4 y^{2} = 36$

Graph a Hyperbola with Center at $(h, k) (h, k)$

In the following exercises, graph.

159.

$\frac{{(x - 1)}^{2}}{16} - \frac{{(y - 3)}^{2}}{4} = 1 \frac{{(x - 1)}^{2}}{16} - \frac{{(y - 3)}^{2}}{4} = 1$

160.

$\frac{{(x - 2)}^{2}}{4} - \frac{{(y - 3)}^{2}}{16} = 1 \frac{{(x - 2)}^{2}}{4} - \frac{{(y - 3)}^{2}}{16} = 1$

161.

$\frac{{(y - 4)}^{2}}{9} - \frac{{(x - 2)}^{2}}{25} = 1 \frac{{(y - 4)}^{2}}{9} - \frac{{(x - 2)}^{2}}{25} = 1$

162.

$\frac{{(y - 1)}^{2}}{25} - \frac{{(x - 4)}^{2}}{16} = 1 \frac{{(y - 1)}^{2}}{25} - \frac{{(x - 4)}^{2}}{16} = 1$

163.

$\frac{{(y + 4)}^{2}}{25} - \frac{{(x + 1)}^{2}}{36} = 1 \frac{{(y + 4)}^{2}}{25} - \frac{{(x + 1)}^{2}}{36} = 1$

164.

$\frac{{(y + 1)}^{2}}{16} - \frac{{(x + 1)}^{2}}{4} = 1 \frac{{(y + 1)}^{2}}{16} - \frac{{(x + 1)}^{2}}{4} = 1$

165.

$\frac{{(y - 4)}^{2}}{16} - \frac{{(x + 1)}^{2}}{25} = 1 \frac{{(y - 4)}^{2}}{16} - \frac{{(x + 1)}^{2}}{25} = 1$

166.

$\frac{{(y + 3)}^{2}}{16} - \frac{{(x - 3)}^{2}}{36} = 1 \frac{{(y + 3)}^{2}}{16} - \frac{{(x - 3)}^{2}}{36} = 1$

167.

$\frac{{(x - 3)}^{2}}{25} - \frac{{(y + 2)}^{2}}{9} = 1 \frac{{(x - 3)}^{2}}{25} - \frac{{(y + 2)}^{2}}{9} = 1$

168.

$\frac{{(x + 2)}^{2}}{4} - \frac{{(y - 1)}^{2}}{9} = 1 \frac{{(x + 2)}^{2}}{4} - \frac{{(y - 1)}^{2}}{9} = 1$

In the following exercises, ⓐ write the equation in standard form and ⓑ graph.

169.

$9 x^{2} - 4 y^{2} - 18 x + 8 y - 31 = 0 9 x^{2} - 4 y^{2} - 18 x + 8 y - 31 = 0$

170.

$16 x^{2} - 4 y^{2} + 64 x - 24 y - 36 = 0 16 x^{2} - 4 y^{2} + 64 x - 24 y - 36 = 0$

171.

$y^{2} - x^{2} - 4 y + 2 x - 6 = 0 y^{2} - x^{2} - 4 y + 2 x - 6 = 0$

172.

$4 y^{2} - 16 x^{2} - 24 y + 96 x - 172 = 0 4 y^{2} - 16 x^{2} - 24 y + 96 x - 172 = 0$

173.

$9 y^{2} - x^{2} + 18 y - 4 x - 4 = 0 9 y^{2} - x^{2} + 18 y - 4 x - 4 = 0$

Identify the Graph of each Equation as a Circle, Parabola, Ellipse, or Hyperbola

In the following exercises, identify the type of graph.

174.

ⓐ $x = − y^{2} - 2 y + 3 x = − y^{2} - 2 y + 3$ ⓑ $9 y^{2} - x^{2} + 18 y - 4 x - 4 = 0 9 y^{2} - x^{2} + 18 y - 4 x - 4 = 0$ ⓒ $9 x^{2} + 25 y^{2} = 225 9 x^{2} + 25 y^{2} = 225$ ⓓ $x^{2} + y^{2} - 4 x + 10 y - 7 = 0 x^{2} + y^{2} - 4 x + 10 y - 7 = 0$

175.

ⓐ $x = −2 y^{2} - 12 y - 16 x = −2 y^{2} - 12 y - 16$ ⓑ $x^{2} + y^{2} = 9 x^{2} + y^{2} = 9$ ⓒ $16 x^{2} - 4 y^{2} + 64 x - 24 y - 36 = 0 16 x^{2} - 4 y^{2} + 64 x - 24 y - 36 = 0$ ⓓ $16 x^{2} + 36 y^{2} = 576 16 x^{2} + 36 y^{2} = 576$

Mixed Practice

In the following exercises, graph each equation.

176.

$\frac{{(y - 3)}^{2}}{9} - \frac{{(x + 2)}^{2}}{16} = 1 \frac{{(y - 3)}^{2}}{9} - \frac{{(x + 2)}^{2}}{16} = 1$

177.

$x^{2} + y^{2} - 4 x + 10 y - 7 = 0 x^{2} + y^{2} - 4 x + 10 y - 7 = 0$

178.

$y = {(x - 1)}^{2} + 2 y = {(x - 1)}^{2} + 2$

179.

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1 \frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

180.

${(x + 2)}^{2} + {(y - 5)}^{2} = 4 {(x + 2)}^{2} + {(y - 5)}^{2} = 4$

181.

$9 x^{2} - 4 y^{2} + 54 x + 8 y + 41 = 0 9 x^{2} - 4 y^{2} + 54 x + 8 y + 41 = 0$

182.

$x = − y^{2} - 2 y + 3 x = − y^{2} - 2 y + 3$

183.

$16 x^{2} + 9 y^{2} = 144 16 x^{2} + 9 y^{2} = 144$

Writing Exercises

184.

In your own words, define a hyperbola and write the equation of a hyperbola centered at the origin in standard form. Draw a sketch of the hyperbola labeling the center, vertices, and asymptotes.

185.

Explain in your own words how to create and use the rectangle that helps graph a hyperbola.

186.

Compare and contrast the graphs of the equations $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1 \frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$ and $\frac{y^{2}}{9} - \frac{x^{2}}{4} = 1 . \frac{y^{2}}{9} - \frac{x^{2}}{4} = 1 .$

187.

Explain in your own words, how to distinguish the equation of an ellipse with the equation of a hyperbola.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and four rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was graph a hyperbola with center at (0, 0). In row 3, the I can was graph a hyperbola with a center at (h, k). In row 4, the I can was identify conic sections by their equations.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Learning Objectives

Be Prepared 11.4

Graph a Hyperbola with Center at (0, 0)

Hyperbola

Standard Form of the Equation a Hyperbola with Center ( 0 , 0 ) ( 0 , 0 )

Example 11.27

How to Graph a Hyperbola with Center (0,0)(0,0)

Solution

Try It 11.53

Try It 11.54

How To

Graph a hyperbola centered at (0,0).(0,0).

Example 11.28

Solution

Try It 11.55

Try It 11.56

Graph a Hyperbola with Center at (h,k)(h,k)

Example 11.29

How to Graph a Hyperbola with Center (h,k)(h,k)

Solution

Try It 11.57

Try It 11.58

How To

Graph a hyperbola centered at (h,k).(h,k).

Example 11.30

Solution

Try It 11.59

Try It 11.60

Example 11.31

Solution

Try It 11.61

Try It 11.62

Identify Conic Sections by their Equations

Example 11.32

Solution

Try It 11.63

Try It 11.64

Media

Practice Makes Perfect

Writing Exercises

Self Check

Standard Form of the Equation a Hyperbola with Center $(0, 0) (0, 0)$

How to Graph a Hyperbola with Center $(0, 0) (0, 0)$

Graph a hyperbola centered at $(0, 0) . (0, 0) .$

Graph a Hyperbola with Center at $(h, k) (h, k)$

How to Graph a Hyperbola with Center $(h, k) (h, k)$

Graph a hyperbola centered at $(h, k) . (h, k) .$