# Graphing Hyperbolas (Part 2)

(This page is Part 2. Click here for Part 1.)

## Asymptotes for Hyperbolas

### Asymptotes for $\,\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

The lines $\,y = \pm\frac bax\,$ are asymptotes for the hyperbola $\,\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\,.$

As $\,x\,$ gets big (positive or negative), the graph of $\,\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\,$ gets closer and closer to these lines, as shown above.

The asymptotes form an ‘envelope’ inside which the hyperbola lives. The envelope gets tighter and tighter as $\,x\,$ gets bigger and bigger!

The green box is called the
*central box* of the hyperbola,
and is a useful graphing aid:

- Its sides are parallel to the axes
- It passes through the vertices of the hyperbola
- It crosses the other axis at $\,\pm b$

The *diagonals* of the central box
are the asymptotes for the hyperbola!

### Asymptotes for $\,\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

Similarly, the lines $\,y = \pm\frac abx\,$ are asymptotes for the hyperbola $\,\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\,.$

Compare $\,\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\,$ with the previously-discussed equation $\,\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\,$: the variables $\,x\,$ and $\,y\,$ have just been switched.

Thus, the exact same derivation as before (with variables switched) yields the asymptotes $\,x = \pm \frac ba y\,.$ Solve for $\,y\,$ to get the asymptotes $\,y = \pm\frac ab x\,.$

Alternatively, solve for $\,y\,$ in $\,\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\,$ to get:

$$ \cssId{s18}{y = \pm \frac ab\sqrt{\strut x^2\bigl(1 + \frac{b^2}{x^2}\bigr)}} $$For large $\,x\,,$ the graph looks like $\,y = \pm \frac ab x\,.$

## Getting the Central Box Graphically

If you have a hyperbola graphed with the foci already located, then it's easy to get the central box graphically:

- Make sure that ‘$\,1\,$’ on the $x$-axis is the same as ‘$\,1\,$’ on the $y$-axis (so that a circle actually looks like a circle).
- Sweep out (part of) the circle with center at the origin and radius $\,c\,,$ as shown above.
- Mark where the circle intersects the vertical line through the vertex.
- The green triangle has hypotenuse $\,c\,$ and bottom leg $\,a\,.$
- Since $\,c^2 = a^2 + b^2\,,$ the remaining leg must have length $\,b\,.$

Of course, you can use this information in a different way: if you already have $\,a\,$ and $\,b\,$ marked, then just rotate the hypotenuse to locate the focus!

## Summary: Graphing Hyperbolas

Vertices:

Set $\,y = 0\,,$ solve for $\,x\,,$ to get $\,x = \pm a\,.$

The *positive* term
($\color{green}{\frac{x^2}{a^2}}\,$)
determines the vertices!

Vertices:

Set $\,x = 0\,,$ solve for $\,y\,,$ to get $\,y = \pm a\,.$

The *positive* term
($\color{green}{\frac{y^2}{a^2}}\,$)
determines the vertices!

Use the other term to find $\,b\,$; draw in the central box.

The diagonals of the central box are the asymptotes for the hyperbola.

## Steps to Graph a Hyperbola

### Vertices

Find the vertices.

### Central Box

Use the other term to find $\,b\,$; draw in the central box. (See the examples below.)

### Diagonals

The diagonals of the central box are the asymptotes of the hyperbola.

### Draw Hyperbola

Draw in the hyperbola, using the asymptotes as an ‘envelope’.

### Foci

If needed, use $\,c^2 = a^2 + b^2\,$ to locate the foci.

## Examples: Graphing Hyperbolas

In both examples below, there are only $\,x^2\,,$ $\,y^2\,,$ and constant terms. When the variable terms are on the same side, they have different signs. So, we know we're dealing with hyperbolas!

## Example #1: Put the Equation in Standard Form

Graph: $\,9x^2 - 4y^2 = 36$

Find the vertices, central box, asymptotes, and foci.

When all the numbers ‘work out nicely’, it may be easiest to put the equation in standard form. Here, both $\,9\,$ and $\,4\,$ go into $\,36\,$ evenly (resulting in perfect square denominators).

### Put the Equation in Standard Form

$$ \cssId{s67}{\frac{9x^2}{36} - \frac{4y^2}{36} = \frac{36}{36}} $$Get a $\,1\,$ on the right-hand side

$$ \cssId{s69}{\frac{x^2}{4} - \color{red}{\frac{y^2}{9}} = 1} $$
Simplify

(This is an equation of the form
$\,\frac{x^2}{a^2} - \color{red}{\frac{y^2}{b^2}} = 1\,.$ )

### Find the Vertices

Set $\,y = 0\,$: $\, \frac{x^2}{4} = 1\,,$ $\, x^2 = 4\,,$ $\, x = \pm 2$

### Use the Other Term to Find $\,b\,.$ Draw the Central Box, its Diagonals, and the Hyperbola

Compare $\,\frac{y^2}{b^2}\,$ with $\,\frac{y^2}{9}\,$ to see that $\,b = 3\,.$ Mark $\,\pm 3\,$ on the $y$-axis.

Sketch in the central box and its diagonals (which are the asymptotes). Sketch the hyperbola inside the ‘envelope’ formed by the diagonals.

### Foci

We have $\,a = 2\,$ and $\,b = 3\,$:

$$ \begin{gather} \cssId{s85}{c^2 = a^2 + b^2 = 4 + 9 = 13}\cr\cr \cssId{s86}{c = \sqrt{13} \approx 3.6} \end{gather} $$### Report Results

Vertices: $\,(\pm 2,0)$

Central Box: has corners $\,(\pm 2,\pm 3)$

Asymptotes: $\,y = \pm \frac 32x$

Foci: $\,(\pm\sqrt{13},0)$

You can also head up to WolframAlpha and type in:

graph 9x^2 - 4y^2 = 36

You'll get everything: graph, vertices, foci, asymptotes, and more!

## Example #2: Don't Put the Equation (Completely) in Standard Form

If the numbers don't work out nicely,
then you don't have to write it
*completely* in standard form.
As long as you get the variable terms on the left,
and the
‘$\,1\,$’
on the right, you're good to go!

Graph: $\,9y^2 = 7 + 14x^2$

Find the vertices, central box, asymptotes,
and foci.

### Get the Variable Terms on the Left, ‘$\,1\,$’ on the Right

$$ \cssId{s100}{9y^2 - 14x^2 = 7} $$Variable terms on left, constant on right

$$ \cssId{s102}{\frac{9y^2}{7} - \frac{14x^2}{7} = \frac{7}{7}} $$Get a $\,1\,$ on the right-hand side

$$ \cssId{s104}{\frac{9y^2}{7} - \color{red}{2x^2} = 1} $$(This is an equation of the form $\,\frac{y^2}{a^2} - \color{red}{\frac{x^2}{b^2}} = 1\,.$ )

To go from here to ‘complete’ standard form would be a bit ugly, resulting in:

$$ \cssId{s107}{\frac{y^2}{7/9} - \frac{x^2}{1/2} = 1} $$As you'll see, this isn't needed!

### Find the Vertices

Set $\,x = 0\,$: $\,\frac{9y^2}{7} = 1\,,$ $\, y^2 = \frac{7}{9}\,,$ $\, y = \pm \sqrt{\frac79} = \pm\frac{\sqrt 7}3 \approx \pm 0.88$

So: $\,a = \frac{\sqrt 7}{3}$

### Use the Other Term to Find $\,b\,.$ Draw the Central Box, its Diagonals, and the Hyperbola

Compare $\,\frac{x^2}{b^2}\,$ with $\,2x^2\,$ to find $\,b\,$:

$$ \cssId{s118}{\frac{x^2}{b^2} = 2x^2} $$Compare the term to the desired form

$$ \cssId{s120}{\frac{1}{b^2} = 2} $$Cancel $\,x^2$

$$ \cssId{s122}{b^2 = \frac 12} $$Solve for $\,b^2$

$$ \cssId{s124}{b = \pm \frac{1}{\sqrt 2} \approx \pm 0.71} $$Mark $\,\pm \frac{1}{\sqrt 2}\,$ on the $x$-axis.

Sketch in the central box and its diagonals (which are the asymptotes). Sketch the hyperbola inside the ‘envelope’ formed by the diagonals.

### Foci

We have $\,a = \frac{\sqrt 7}{3}\,$ and $\,b = \frac{1}{\sqrt 2}\,$:

$$ \begin{align} &\cssId{s130}{c^2 = a^2 + b^2}\cr\cr &\quad\, \cssId{s131}{= \bigl(\frac{\sqrt 7}{3}\bigr)^2 + \bigl(\frac{1}{\sqrt 2}\bigr)^2}\cr\cr &\quad\, \cssId{s132}{= \frac 79 + \frac 12}\cr\cr &\quad\, \cssId{s133}{= \frac{14}{18} + \frac{9}{18}}\cr\cr &\quad\, \cssId{s134}{= \frac{23}{18}}\cr\cr\cr &\cssId{s135}{c = \frac{\sqrt{23}}{3\sqrt{2}} \approx 1.13} \end{align} $$### Report Results

Note that:

$$ \cssId{s138}{\frac ab \ =\ \frac{\frac{\sqrt 7}{3}}{\frac{1}{\sqrt 2}}} \cssId{s139}{\ =\ \frac{\sqrt 7}{3}\cdot \frac{\sqrt 2}{1}} \cssId{s140}{\ =\ \frac{\sqrt{14}}{3}} $$
Vertices: $\,\left(0 \ ,\ \pm \frac{\sqrt{7}}3\right)$

Central Box: has corners $\,\left(\pm \frac{1}{\sqrt{2}} \ ,\ \pm \frac{\sqrt{7}}{3}\right)$

Asymptotes: $\,y = \pm \frac{\sqrt{14}}{3}x$

Foci: $\,\left(0 \ ,\ \pm\frac{\sqrt{23}}{3\sqrt 2}\right)$