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The Hyperbola

Author: Sophia
PDF Version

what's covered
In this lesson, you will analyze the equation and graph of a hyperbola. Specifically, this lesson will cover:

Table of Contents

1. Determining Graphical Aspects of a Hyperbola

1a. Graphical Aspects of a Hyperbola With Center at the Origin

A hyperbola is obtained by cutting a plane through a double cone, as shown in the figure.

A 3D representation of a double cone with a vertical plane slicing through both the upper and lower cones. The intersection of the plane with the cones creates two symmetrical curves, called a hyperbola.

The graph of a hyperbola with center open parentheses 0 comma space 0 close parentheses and vertices along the horizontal axis is shown below.

A graph with an x-axis and a y-axis intersecting at the origin and representing a hyperbola. The x-axis has the marked points (–c, 0), (–a, 0), (a, 0), and (c, 0) from left to right. The left portion of a hyperbola opens toward the left with vertex at (-a, 0). The right portion of a hyperbola has vertex at (a, 0) and opens toward the right. The marked point (x, y) is contained on the right portion of the hyberbola. A line labeled ‘d1’ slants upward from the marked point (–c, 0) to the marked point (x, y). Another line labeled ‘d2’ slants upward from the marked point (c, 0) to the marked point (x, y).

A hyperbola is defined as the set of all points open parentheses x comma space y close parentheses such that the absolute difference between the distances from open parentheses x comma space y close parentheses to each focus is constant. That is, open vertical bar d subscript 1 minus d subscript 2 close vertical bar is constant.

Considering this definition with both horizontal and vertical axes, there are two possible basic graphs and equations of hyperbolas when centered at the origin, as shown below.

Graph 1 Graph 2
A general representation of a hyperbola that opens leftward and rightward. The center is located at the point (0, 0), the vertices are a units to the left and right of the center, and the foci are located c units to the left and right of the center. The asymptotes are represented by dashed lines, passing through the center and having slopes (b over a) and (-b over a), showing how the hyperbola opens in both directions.

Vertices: open parentheses plus-or-minus a comma space 0 close parentheses
Foci: open parentheses plus-or-minus c comma space 0 close parentheses
Asymptotes: y equals plus-or-minus b over a x 		A general representation of a hyperbola that opens upward and downward. The center is located at the point (0, 0), the vertices are a units above and below the center, and the foci are located c units above and below the center. The asymptotes are represented by dashed lines, passing through the center and having slopes (a over b) and (-a over b), showing how the hyperbola opens in both directions.

Vertices: open parentheses 0 comma space plus-or-minus a close parentheses
Foci: open parentheses 0 comma space plus-or-minus c close parentheses
Asymptotes: y equals plus-or-minus a over b x

Note the following:

  • In both cases, c squared equals a squared plus b squared.
  • Each orientation of the hyperbola has slant asymptotes that intersect at the origin.
  • When the x squared term has a positive coefficient, the vertices and foci are along the x-axis.
  • When the y squared term has a positive coefficient, the vertices and foci are along the y-axis.
  • The line segment joining the vertices is called the transverse axis.

EXAMPLE

Consider the hyperbola whose equation is x squared over 9 minus y squared over 16 equals 1.

  • Since the x squared term has a positive coefficient, this matches to the equation x squared over a squared minus y squared over b squared equals 1.
  • Thus, a squared equals 9 and b squared equals 16 comma which means a equals 3 and b equals 4.
  • Furthermore, c squared equals a squared plus b squared equals 9 plus 16 equals 25 comma which gives c equals 5.
  • The vertices are located at open parentheses short dash 3 comma space 0 close parentheses and open parentheses 3 comma space 0 close parentheses.
  • The foci are located at open parentheses short dash 5 comma space 0 close parentheses and open parentheses 5 comma space 0 close parentheses.
  • The slopes of the asymptotes are plus-or-minus b over a equals plus-or-minus 4 over 3. Therefore, the equations of the asymptotes are y equals plus-or-minus 4 over 3 x.
The graph of the hyperbola and its asymptotes are pictured below.

A graph with an x-axis and a y-axis ranging from −8 to 8 represents a hyperbola with two branches. The left branch has vertex at the point (-3, 0) and opens leftward toward two dashed lines with equations y equals (-4 over 3)x below and y equals (4 over 3)x above. The right branch has vertex at the point (3, 0) and opens rightward toward the lines with equations y equals (-4 over 3)x below and y equals (4 over 3)x above. There are marked points at (-5, 0) and (5, 0), which are the foci, indicating that the hyperbola opens to the left of (-3, 0) and to the right of (3, 0).

try it
Consider the hyperbola whose equation is y squared over 16 minus x squared over 4 equals 1.
What are the coordinates of the center of the hyperbola?
The center is located at open parentheses 0 comma space 0 close parentheses.
What are the coordinates of the vertices of the hyperbola?
The vertices correspond to the positive term in the standard form, so we focus on the term y squared over 16. This means a squared equals 16 comma or a equals 4. Since this corresponds to the y squared term, the vertices are 4 units above and below the center, which places them at the points open parentheses 0 comma space 4 close parentheses and open parentheses 0 comma space short dash 4 close parentheses.
What are the coordinates of the foci of the hyperbola?
The foci are c units from the center, where c squared equals a squared plus b squared.

From the equation, a squared equals 16 and b squared equals 4 comma so c squared equals 16 plus 4 equals 20.

Then, c equals plus-or-minus square root of 20 equals plus-or-minus 2 square root of 5.

Thus, the foci are 2 square root of 5 units above and below the center, which places them at open parentheses 0 comma space plus-or-minus 2 square root of 5 close parentheses.
What are the equations of the asymptotes?
The asymptotes both pass through the center open parentheses 0 comma space 0 close parentheses.

From the equation, we have a equals 4 and b equals 2. Since a corresponds to the y-term and b corresponds to the x-term, the slopes of the asymptotes are plus-or-minus 4 over 2 equals plus-or-minus 2.

Thus, the equations of the asymptotes are y equals plus-or-minus 2 x.

Similar to ellipses, hyperbolas can look very different from one another in terms of how wide or narrow they open. The following figures below show hyperbolas, each with a horizontal major axis. Note that the vertices of each hyperbola are open parentheses short dash 3 comma space 0 close parentheses and open parentheses 3 comma space 0 close parentheses.

Equation Figure
x squared over 3 squared minus y squared over 2 squared equals 1 Figure 1
A graph containing a hyperbola that opens leftward and rightward. The leftward branch has vertex at the point (-3, 0) and opens toward two slanted lines (represented as dashed lines), one with equation y equals (-2 over 3)x above and the other y equals (2 over 3)x below. The branch on the right has vertex at (3, 0) and opens to the right toward the lines y equals (2 over 3)x above and y equals (-2 over 3)x below. Two other points, each marked with 'Focus', have coordinates (3.5, 0) and (-3.5, 0), indicating the direction that the hyperbola opens.
x squared over 3 squared minus y squared over 6 squared equals 1 Figure 2
A graph containing a hyperbola that opens leftward and rightward. The leftward branch has vertex at the point (-3, 0) and opens toward two slanted lines (represented as dashed lines), one with equation y equals -2x above and the other y equals 2x below. The branch on the right has vertex at (3, 0) and opens to the right toward the lines y equals 2x above and y equals -2x below. Two other points, each marked with 'Focus', have coordinates (6.75, 0) and (-6.75, 0), indicating the direction that the hyperbola opens.
x squared over 3 squared minus y squared over 9 squared equals 1 Figure 3
A graph containing a hyperbola that opens leftward and rightward. The leftward branch has vertex at the point (-3, 0) and opens toward two slanted lines (represented as dashed lines), one with equation y equals -3x and the other y equals 3x. The branch on the right has vertex at (3, 0) and opens to the right toward the lines y equals 3x above and y equals -3x below. Two other points, each marked with 'Focus', are located at (9.5, 0) and (-9.5, 0), indicating the direction that the hyperbola opens.

Even though each is classified as a hyperbola, they look very different. Figure 1 has a very narrow opening in the horizontal direction compared to the others, while Figure 3 has the widest opening in the horizontal direction compared to the others. Notice that the wider the opening, the further the foci are from the center.

Eccentricity can also be used to distinguish hyperbolas from each other in terms of how narrow or wide their openings are. The formula for eccentricity as it applies to hyperbolas is the same as before: e equals c over a

For any hyperbola, c greater than a since the foci are further away from the center than the vertices are. Also, since a greater than 0 and c greater than 0 comma it follows that every hyperbola has eccentricity e greater than 1.

EXAMPLE

From the three hyperbolas given above, we find their eccentricities as follows:

Equation Find the Value of c Eccentricity bold italic e bold equals bold c over bold a
Figure 1
x squared over 3 squared minus y squared over 2 squared equals 1 		table attributes columnalign left end attributes row cell c squared equals a squared plus b squared end cell row cell c squared equals 3 squared plus 2 squared end cell row cell c squared equals 13 end cell row cell space space c equals square root of 13 end cell end table e equals c over a equals fraction numerator square root of 13 over denominator 3 end fraction almost equal to 1.20
Figure 2
x squared over 3 squared minus y squared over 6 squared equals 1 		table attributes columnalign left end attributes row cell c squared equals a squared plus b squared end cell row cell c squared equals 3 squared plus 6 squared end cell row cell c squared equals 45 end cell row cell space space c equals 3 square root of 5 end cell end table e equals c over a equals fraction numerator 3 square root of 5 over denominator 3 end fraction almost equal to 2.24
Figure 3
x squared over 3 squared minus y squared over 9 squared equals 1 		table attributes columnalign left end attributes row cell c squared equals a squared plus b squared end cell row cell c squared equals 3 squared plus 9 squared end cell row cell c squared equals 90 end cell row cell space space c equals 3 square root of 10 end cell end table e equals c over a equals fraction numerator 3 square root of 10 over denominator 3 end fraction almost equal to 3.16

In summary:

  • The hyperbola in Figure 1 has the narrowest opening of the three, and its eccentricity is closest to 1.
  • The hyperbola in Figure 3 has the widest opening and has the largest eccentricity.

try it
Consider the hyperbola whose equation is y squared over 20 minus x squared over 10 equals 1.
Identify the values of a, b, and c. Give exact values.
The value of a corresponds to the positive term, so a equals square root of 20 equals 2 square root of 5.

The value of b corresponds to the negative term, so b equals square root of 10.

Using the equation c squared equals a squared plus b squared comma we have c squared equals 20 plus 10 equals 30 comma so c equals square root of 30.
Find the eccentricity of the hyperbola.
The eccentricity is e equals c over a equals fraction numerator square root of 30 over denominator 2 square root of 5 end fraction equals fraction numerator square root of 6 over denominator 2 end fraction comma or approximately 1.22.

term to know
Transverse Axis
The line segment that connects the vertices of a hyperbola.

1b. Graphical Aspects of a Hyperbola With Center Not at the Origin

When the center is located at open parentheses h comma space k close parentheses comma the graphs and equations are as follows:

Graph 1 Graph 2
A general representation of a hyperbola that opens upward and downward. The center is located at the point (h, k), the vertices are a units above and below the center, and the foci are located c units above and below the center. The asymptotes are represented by dashed lines, passing through the center and having slopes (a over b) and (-a over b), showing how the hyperbola opens in both directions.

open parentheses x minus h close parentheses squared over a squared minus fraction numerator begin display style open parentheses y minus k close parentheses squared end style over denominator b squared end fraction equals 1
Vertices: open parentheses h plus-or-minus a comma space k close parentheses
Foci: open parentheses h plus-or-minus c comma space k close parentheses
Asymptotes: y minus k equals plus-or-minus b over a open parentheses x minus h close parentheses 		A general representation of a hyperbola that opens leftward and rightward. The center is located at the point (h, k), the vertices are a units to the left and right of the center, and the foci are located c units to the left and right of the center. The asymptotes are represented by dashed lines, passing through the center and having slopes (b over a) and (-b over a), showing how the hyperbola opens in both directions.

open parentheses y minus k close parentheses squared over a squared minus fraction numerator begin display style open parentheses x minus h close parentheses squared end style over denominator b squared end fraction equals 1
Vertices: open parentheses h comma space k plus-or-minus a close parentheses
Foci: open parentheses h comma space k plus-or-minus c close parentheses
Asymptotes: y minus k equals plus-or-minus a over b open parentheses x minus h close parentheses

Note the following:

  • In both cases, c squared equals a squared plus b squared.
  • Each orientation of the hyperbola has slant asymptotes that intersect at the center.
  • When the x squared term has a positive coefficient, the transverse axis is parallel to the x-axis.
  • When the y squared term has a positive coefficient, the transverse axis is parallel to the y-axis.
formula to know
Standard Form of the Equation of a Hyperbola
open parentheses x minus h close parentheses squared over a squared minus fraction numerator begin display style open parentheses y minus k close parentheses squared end style over denominator b squared end fraction equals 1
open parentheses y minus k close parentheses squared over a squared minus fraction numerator begin display style open parentheses x minus h close parentheses squared end style over denominator b squared end fraction equals 1 comma where open parentheses h comma space k close parentheses is the center.
The foci are c units from the center, where c squared equals a squared plus b squared.

watch
In this video, we will explore the graphical aspects of the hyperbola whose equation is open parentheses x minus 1 close parentheses squared over 25 minus y squared over 11 equals 1.

try it
Consider the hyperbola having the equation open parentheses y minus 2 close parentheses squared over 4 minus open parentheses x plus 1 close parentheses squared over 9 equals 1.
What are the coordinates of the center of the hyperbola?
Comparing x plus 1 to x minus h in the standard form, we have h equals short dash 1 semicolon and comparing y minus 2 to y minus k in the standard form, we have k equals 2. This means the center is located at open parentheses short dash 1 comma space 2 close parentheses.
What are the coordinates of the vertices of the hyperbola?
Since the y-term is positive, the vertices are determined by this term.

This means a squared equals 4 comma or a equals 2.

This means that the vertices are 2 units above and below the center, which means they are located at open parentheses short dash 1 comma space 2 plus-or-minus 2 close parentheses comma or written out, open parentheses short dash 1 comma space 0 close parentheses and open parentheses short dash 1 comma space 4 close parentheses.
What are the coordinates of the foci of the hyperbola?
From the equation, a squared equals 4 and b squared equals 9.

Using c squared equals a squared plus b squared comma we have c squared equals 4 plus 9 equals 13 comma which means c equals square root of 13.

This means the foci are square root of 13 units above and below the center. Therefore, they are located at open parentheses short dash 1 comma space 2 plus-or-minus square root of 13 close parentheses.
What are the equations of the asymptotes?
The asymptotes both pass through the center, open parentheses short dash 1 comma space 2 close parentheses.

Since a equals 2 corresponds to the y-term and b equals 3 corresponds to the x-term, the slopes of the asymptotes are plus-or-minus 2 over 3.

Using point-slope form, the equations of the asymptotes are y minus 2 equals plus-or-minus 2 over 3 open parentheses x plus 1 close parentheses.

Written in expanded form, the equations are y equals 2 over 3 x plus 8 over 3 and y equals short dash 2 over 3 x plus 4 over 3.
What is the eccentricity of the hyperbola?
The eccentricity is e equals c over a equals fraction numerator square root of 13 over denominator 2 end fraction comma or approximately 1.80.

EXAMPLE

Consider the equation short dash 9 x squared plus 72 x plus 16 y squared plus 16 y plus 4 equals 0.

To make the equation more recognizable, complete the square.

short dash 9 x squared plus 72 x plus 16 y squared plus 16 y plus 4 equals 0 This is the original equation.
short dash 9 x squared plus 72 x plus 16 y squared plus 16 y equals short dash 4 Subtract 4 from both sides so all the variable terms are on one side and the constant is on the other side.
short dash 9 open parentheses x squared minus 8 x close parentheses plus 16 open parentheses y squared plus y close parentheses equals short dash 4 Factor out -9 from the x-terms and 16 from the y-terms.
short dash 9 open parentheses x squared minus 8 x plus 16 close parentheses plus 16 open parentheses y squared plus y plus 1 fourth close parentheses equals short dash 4 minus 9 open parentheses 16 close parentheses plus 16 open parentheses 1 fourth close parentheses Add 16 to complete the square for x squared minus 8 x. This means that short dash 9 open parentheses 16 close parentheses is added to the left side; therefore, add short dash 9 open parentheses 16 close parentheses to the right side as well.

Add 1 fourth to complete the square for y squared plus y. This means that 16 open parentheses 1 fourth close parentheses is added to the left side; therefore, add 16 open parentheses 1 fourth close parentheses to the right side as well.
short dash 9 open parentheses x minus 4 close parentheses squared plus 16 open parentheses y plus 1 half close parentheses squared equals short dash 144 Rewrite the quadratic terms as perfect squares on the left side, and simplify the right side.
open parentheses x minus 4 close parentheses squared over 16 minus open parentheses y plus 1 half close parentheses squared over 9 equals 1 Divide both sides by -144.

Thus, the standard form of the original equation is open parentheses x minus 4 close parentheses squared over 16 minus open parentheses y plus 1 half close parentheses squared over 9 equals 1 comma which makes it a hyperbola with center at open parentheses 4 comma space short dash 1 half close parentheses. Now that this equation is in standard form, other characteristics like the vertices, foci, and asymptotes can also be identified.

try it
Consider the hyperbola with equation x squared plus 2 x minus 100 y squared minus 1000 y minus 2399 equals 0.
Write the equation in standard form.
To write the equation in standard form, complete the square, then write the equation so that two square terms are on one side and 1 is on the other side.

x squared plus 2 x minus 100 y squared minus 1000 y equals 2399 Add the constant term, 2399, to both sides.
x squared plus 2 x minus 100 open parentheses y squared plus 10 y close parentheses equals 2399 Factor out -100 from the y-terms.
x squared plus 2 x plus 1 minus 100 open parentheses y squared plus 10 y plus 25 close parentheses equals 2399 plus 1 minus 100 open parentheses 25 close parentheses Add 1 to both sides to complete the square on the x terms.
Add 25 inside the parentheses to complete the square on the y terms, which means we really added short dash 100 open parentheses 25 close parentheses.
open parentheses x plus 1 close parentheses squared minus 100 open parentheses y plus 5 close parentheses squared equals short dash 100 Write all trinomials as perfect squares, and simplify the right-hand side.
fraction numerator open parentheses x plus 1 close parentheses squared over denominator short dash 100 end fraction plus open parentheses y plus 5 close parentheses squared equals 1 Divide both sides by -100.
open parentheses y plus 5 close parentheses squared minus open parentheses x plus 1 close parentheses squared over 100 equals 1 Rewrite so that the positive term appears first.

The equation in standard form is open parentheses y plus 5 close parentheses squared minus open parentheses x plus 1 close parentheses squared over 100 equals 1.
What is the center of the hyperbola?
Comparing y plus 5 with y minus k in the standard form, we have k equals short dash 5 semicolon and comparing x plus 1 to x minus h in the standard form, we have h equals short dash 1. The center is located at open parentheses short dash 1 comma space short dash 5 close parentheses.
What are the coordinates of the vertices?
Since the y-term is positive, this term determines the vertices. If it helps, write open parentheses y plus 5 close parentheses squared as open parentheses y plus 5 close parentheses squared over 1. Now it is clear to see that a equals 1. This means the vertices are 1 unit above and below the center, which places the vertices at the points open parentheses short dash 1 comma space short dash 5 plus-or-minus 1 close parentheses comma or written out, open parentheses short dash 1 comma space short dash 4 close parentheses and open parentheses short dash 1 comma space short dash 6 close parentheses.
What are the coordinates of the foci?
Note that c squared equals a squared plus b squared. From the equation, a squared equals 1 and b squared equals 100 comma which means c squared equals 1 plus 100 equals 101 comma and c equals square root of 101.

The foci are square root of 101 units above and below the center, placing them at open parentheses short dash 1 comma space short dash 5 plus-or-minus square root of 101 close parentheses.
What are the equations of the asymptotes?
The asymptotes both contain the center, open parentheses short dash 1 comma space short dash 5 close parentheses.

Note that a equals 1 and b equals 10. Since a equals 1 corresponds to the y-term, it follows that the slopes of the asymptotes are plus-or-minus 1 over 10.

Therefore, the equations of the asymptotes are y plus 5 equals plus-or-minus 1 over 10 open parentheses x plus 1 close parentheses. If written in expanded form, the equations are y equals 1 over 10 x minus 49 over 10 and y equals short dash 1 over 10 x minus 51 over 10.

2. Writing the Equation of a Hyperbola From Given Information

Given information about the various graphical aspects of a hyperbola, we can write the equation of said hyperbola.

EXAMPLE

A hyperbola has foci open parentheses plus-or-minus 10 comma space 1 close parentheses and vertices open parentheses plus-or-minus 5 comma space 1 close parentheses. What is the equation of the hyperbola?

To write the equation of a hyperbola, we need to know the coordinates of its center, its transverse axis, and the values of a squared and b squared.

The center is the midpoint of the line segment joining either the vertices or the foci. Using either, the center is open parentheses 0 comma space 1 close parentheses.

This picture shows the center along with the foci and vertices.

There are five marked points in the graph at coordinates (−10, 1), (−5, 1), (0, 1), (5, 1), and (10, 1).

Since these all lie on a horizontal line, the transverse axis is parallel to the x-axis, which means the equation open parentheses x minus h close parentheses squared over a squared minus fraction numerator begin display style open parentheses y minus k close parentheses squared end style over denominator b squared end fraction equals 1 is used.

Values of a and b  :

  • The vertices are each 5 units away from the center, therefore a equals 5.
  • The foci are each 10 units from the center, therefore c equals 10.
  • Then, use the equation a squared plus b squared equals c squared to find b squared.
a squared plus b squared equals c squared This is the equation that relates a comma b, and c.
5 squared plus b squared equals 10 squared Substitute a equals 5 and c equals 10.
25 plus b squared equals 100 Simplify.
b squared equals 75 Solve for b squared.

Thus, b squared equals 75. Then, substituting a squared equals 25 comma b squared equals 75 comma and open parentheses h comma space k close parentheses equals open parentheses 0 comma space 1 close parentheses comma the equation of the hyperbola is x squared over 25 minus fraction numerator begin display style open parentheses y minus 1 close parentheses squared end style over denominator 75 end fraction equals 1.

try it
A hyperbola has its center at open parentheses short dash 2 comma space 3 close parentheses comma a vertex at open parentheses short dash 2 comma space 7 close parentheses comma and a focus at open parentheses short dash 2 comma space 10 close parentheses.
Write the equation of the hyperbola in standard form.
To write the equation, we need a comma b, and the center.

Since the center, one of the vertices, and one of the foci all fall in a vertical line, the standard form is open parentheses y minus k close parentheses squared over a squared minus open parentheses x minus h close parentheses squared over b squared equals 1.

The center is open parentheses short dash 2 comma space 3 close parentheses comma which means h equals short dash 2 and k equals 3.

A vertex is located at open parentheses short dash 2 comma space 7 close parentheses comma which is 4 units above the center. This means a equals 4.

A focus is located at open parentheses short dash 2 comma space 10 close parentheses comma which is 7 units above the center. This means c equals 7.

Now, use c squared equals a squared plus b squared to find b:

7 squared equals 4 squared plus b squared Replace a equals 4 and c equals 7.
49 equals 16 plus b squared Simplify.
33 equals b squared Solve for b squared.

There is no need to solve for b since b squared appears in the standard form.

Now, with the center at open parentheses short dash 2 comma space 3 close parentheses comma a squared equals 16 comma and b squared equals 33 comma the standard form equation of the hyperbola is open parentheses y minus 3 close parentheses squared over 16 minus fraction numerator begin display style open parentheses x plus 2 close parentheses squared end style over denominator 33 end fraction equals 1.

EXAMPLE

A hyperbola has its center at open parentheses 0 comma space 0 close parentheses comma a focus at open parentheses 0 comma space 7 close parentheses comma and eccentricity e equals 1.4. Write the equation of the hyperbola in standard form.

First, notice that the focus is 7 units above the center, which indicates two things:

  • c equals 7
  • The major axis of the hyperbola is vertical.
Since the center is at open parentheses 0 comma space 0 close parentheses comma this also means that the equation of the hyperbola has the form y squared over a squared minus x squared over b squared equals 1.

We know c equals 7 comma so we will need to use the given eccentricity to find a squared and b squared.

1.4 equals 7 over a Use the eccentricity formula e equals c over a with e equals 1.4 and c equals 7.
a equals fraction numerator 7 over denominator 1.4 end fraction equals 5 Solve for a.
7 squared equals 5 squared plus b squared Next, use the formula c squared equals a squared plus b squared with c equals 7 and a equals 5 to find b squared.
49 equals 25 plus b squared Simplify.
b squared equals 24 Solve for b squared.

Replacing a squared with 25 and b squared with 24, the equation of the ellipse is y squared over 25 minus x squared over 24 equals 1.

try it
A hyperbola has center open parentheses 0 comma space 0 close parentheses comma a vertex at open parentheses 18 comma space 0 close parentheses comma and eccentricity e equals 3 over 2.
Find the value of c.
The vertex is 18 units to the right of the center, so a equals 18.

Next, use the eccentricity formula to find c.

e equals c over a The eccentricity formula.
3 over 2 equals c over 18 Substitute e equals 3 over 2 and a equals 18.
c equals 3 over 2 open parentheses 18 close parentheses equals 27 Multiply both sides by 18 to find c.

Therefore, c equals 27.
Write the equation of the hyperbola in standard form.
Since the center is at open parentheses 0 comma space 0 close parentheses and the vertex is to the right of the center, the equation has standard form x squared over a squared minus y squared over b squared equals 1.

We already know a equals 18 comma which means a squared equals 324. Also, c squared equals 27 squared equals 729.

To find b squared comma use the formula c squared equals a squared plus b squared.

729 equals 324 plus b squared Replace a squared equals 324 and c squared equals 729.
405 equals b squared Solve for b squared.

Replacing a squared with 324 and b squared with 405 into the standard form, the equation of the hyperbola is x squared over 324 minus y squared over 405 equals 1.

3. Solving Applied Problems Involving Hyperbolas

Hyperbolas are used to model several real-world applications.

  • Paths of some comets are hyperbolic.
  • The cooling towers at a nuclear power plant are hyperbolic in shape.
  • When an object travels faster than the speed of sound, a shock wave in the form of a cone is created. When the wave intersects the ground, a portion of a hyperbola is formed (see figure below).

watch
In this video, we will find the equation of the hyperbola that models the sides of a cooling tower.

watch
A hyperbolic hedge is to be designed around a fountain at its center. In this video, we will find the equation of a hyperbolic hedge that follows the asymptotes y equals plus-or-minus 2 x and at its closest points is 5 feet from the fountain.

try it
A hyperbolic hedge is to be designed around a fountain at its center and follows the asymptotes y equals plus-or-minus 2 over 3 x and is 12 feet from the fountain at its closest points.
Write the equation of the hyperbolic hedge. Assume a horizontal transverse axis.
Note that the asymptotes intersect at open parentheses 0 comma space 0 close parentheses comma so this is where the center is placed.

Assuming a horizontal transverse axis, one vertex is at the point open parentheses 12 comma space 0 close parentheses. Therefore, a equals 12.

The standard form of the equation of this hyperbola is x squared over 12 squared minus y squared over b squared equals 1.

The slopes of the asymptotes are plus-or-minus b over 12 comma but are also given as plus-or-minus 2 over 3.

Thus, b over 12 equals 2 over 3 comma which means b equals 8.

Substituting into the standard equation, the equation for this hyperbola is x squared over 12 squared minus y squared over 8 squared equals 1 comma or in simpler form, x squared over 144 minus y squared over 64 equals 1.

summary
In this lesson, you determined the graphical aspects of a hyperbola, which is a collection of points whose distances from two fixed points have a constant difference. The key characteristics of a hyperbola are its center, vertices, foci, asymptotes, eccentricity, the relationship c squared equals a squared plus b squared, and transverse axis, and you explored these graphical aspects of a hyperbola with center at the origin and with center not at the origin. You also learned that given a hyperbola's equation, you can analyze these characteristics. In addition, given information about a hyperbola, notably the coordinates of its center, its transverse axis, and the values of a squared and b squared comma you can write the equation of a hyperbola, which is very useful in solving applied problems that involve hyperbolas.

SOURCE: THIS TUTORIAL HAS BEEN ADAPTED FROM OPENSTAX "PRECALCULUS” BY JAY ABRAMSON. ACCESS FOR FREE AT OPENSTAX.ORG/DETAILS/BOOKS/PRECALCULUS-2E. LICENSE: CREATIVE COMMONS ATTRIBUTION 4.0 INTERNATIONAL.

Terms to Know
Transverse Axis

The line segment that connects the vertices of a hyperbola.

Formulas to Know
Standard Form of the Equation of a Hyperbola

open parentheses x minus h close parentheses squared over a squared minus fraction numerator begin display style open parentheses y minus k close parentheses squared end style over denominator b squared end fraction equals 1

open parentheses y minus k close parentheses squared over a squared minus fraction numerator begin display style open parentheses x minus h close parentheses squared end style over denominator b squared end fraction equals 1 comma where open parentheses h comma space k close parentheses is the center.

The foci are c units from the center, where c squared equals a squared plus b squared.

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