Your homework assignment is to graph the hyperbola

### Graphing Hyperbolas

We know that the resulting graph of a rational function is a hyperbola with two **branches**. A hyperbola is also a conic section. To create a hyperbola, you would slice a plane through two inverted cones, such that the plane is perpendicular to the bases of the cones.

By the conic definition, a **hyperbola** is the set of all points such that the of the differences of the distances from the **foci** is constant.

Using the picture, any point,

Comparing this to the ellipse, where

For a hyperbola, then, the equation will be **vertices**, on the hyperbola. Here, they are the two points that are closest to each other on the graph. The line through the vertices and foci is called the **transverse axis**. Its midpoint is the **center** of the hyperbola. In this concept, the center will be the origin. There will always be two branches for any hyperbola and two asymptotes.

Let's graph

First, this hyperbola has a horizontal transverse axis because the ** is not** always greater than

Therefore,

Draw the hyperbola branches with the vertices on the transverse axis and the rectangle. Sketch the branches to get close to the asymptotes, but not touch them.

The vertices are *further away* from the center than the vertices.

The foci are

Now, let's graph

This equation is not in standard form. To rewrite it in standard form, the right side of the equation must be 1. Divide everything by 324.

Now, we can see that this is a vertical hyperbola, where

To find the foci, use

The foci are

Finally, let's graph

This will be a horizontal hyperbola, because the

The asymptotes are

**Important Note:** The asymptotes and square are *not* a part of the function. They are included in graphing a hyperbola because it makes it easier to do so.

Also, when graphing hyperbolas, we are *sketching* each branch. We did not make a table of values to find certain points and then connect. You can do this, but using the square or rectangle with the asymptotes produces a pretty accurate graph and is much simpler.

### Examples

#### Example 1

Earlier, you were asked to find the asymptotes and foci of your graph.

First, we need to get the equation in the form

Now we can see that *y*-term comes first, the hyperbola is vertically oriented. Therefore, the asymptotes are

Substituting for *a* and *b*, we get \begin{align*}y = -\frac{2}{3}x\end{align*} and \begin{align*}y = \frac{2}{3}x\end{align*}.

Finally, to find the foci, use \begin{align*}c^2=a^2+b^2\end{align*}.

\begin{align*}c^2 &= 4+9=13 \\ c &= \sqrt{13}\end{align*}

The foci are \begin{align*}\left(0, \sqrt{13}\right)\end{align*} and \begin{align*}\left(0, -\sqrt{13}\right)\end{align*}.

#### Example 2

Find the vertices, foci, and asymptotes of \begin{align*}y^2-\frac{x^2}{25}=1\end{align*}.

First, let’s rewrite the equation like this: \begin{align*}\frac{y^2}{1} - \frac{x^2}{25}=1\end{align*}. We know that the transverse axis is vertical because the \begin{align*}y\end{align*}-term is first, making \begin{align*}a=1\end{align*} and \begin{align*}b=5\end{align*}. Therefore, the vertices are \begin{align*}(0, -1)\end{align*} and \begin{align*}(0, 1)\end{align*}. The asymptotes are \begin{align*}y= \frac{1}{5}x\end{align*} and \begin{align*}y= -\frac{1}{5}x\end{align*}. Lastly, let’s find the foci using \begin{align*}c^2=a^2+b^2\end{align*}.

\begin{align*}c^2 &= 1+25=26 \\ c &= \sqrt{26}\end{align*}

The foci are \begin{align*}\left(0, -\sqrt{26}\right)\end{align*} and \begin{align*}\left(0, \sqrt{26}\right)\end{align*}.

#### Example 3

Graph Example 2.

#### Example 4

Graph \begin{align*}9x^2-49y^2=411\end{align*}.

Rewrite the equation so that the right side is equal to 1. Divide everything by 441.

\begin{align*}\frac{9x^2}{441} - \frac{49y^2}{441} &= \frac{441}{441} \\ \frac{x^2}{49} - \frac{y^2}{9} &= 1\end{align*}

\begin{align*}a=9\end{align*} and \begin{align*}b=6\end{align*} with a horizontal transverse axis.

### Review

Find the vertices, asymptotes, and foci of each hyperbola below.

- \begin{align*}\frac{x^2}{9} - \frac{y^2}{16} =1\end{align*}
- \begin{align*}4y^2-25x^2=100\end{align*}
- \begin{align*}\frac{x^2}{81} - \frac{y^2}{64} =1\end{align*}
- \begin{align*}x^2-y^2=16\end{align*}
- \begin{align*}\frac{y^2}{49} - \frac{x^2}{25} =1\end{align*}
- \begin{align*}121y^2-9x^2=1089\end{align*}
- \begin{align*}y^2-x^2=1\end{align*}
- \begin{align*}\frac{x^2}{64} - \frac{y^2}{4} =1\end{align*}
- \begin{align*}\frac{y^2}{4} - \frac{x^2}{64} =1\end{align*}
- Graph #1.
- Graph #2.
- Graph #8.
- Graph #9.
**Writing**Compare the hyperbolas from #8 and #9. How are they the same? How are they different? What do you know about the asymptotes and foci?**Critical Thinking**Compare the equations \begin{align*}\frac{x^2}{25} - \frac{y^2}{9} =1\end{align*} and \begin{align*}\frac{x^2}{25} + \frac{y^2}{9} =1\end{align*}. Graph them on the same axes and find their foci.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 10.7.