# What does the equation 9y^2-4x^2=36 tell me about its hyperbola?

Sep 13, 2014

Before we start interpreting our hyperbola, we want to set it in standard form first. Meaning, we want it to be in ${y}^{2} / {a}^{2} - {x}^{2} / {b}^{2} = 1$ form. To do this, we start by dividing both sides by 36, to get 1 on the left side. Once that's done, you should have:

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

Once you have this, we can make a few observations:

1. There is no h and k
2. It is a ${y}^{2} / {a}^{2}$ hyperbola (which means that it has a vertical transverse axis.

Now we can begin to find some things. I will guide you through how to find some of the things most teachers will ask you to find on tests or quizzes:

1. Center
2. Vertices
3.Foci
3. Asymptotes

Look at the illustration below to get a good idea of what goes where and how the picture looks: Since there is no h or k, we know that it is a hyperbola with a center at the origin (0,0).

The vertices are simply the points at which the hyperbola's branches begin to curve either way. As displayed in the diagram, we know that they are simply $\left(0 , \pm a\right)$.

So once we find $a$ from our equation ($\sqrt{4} =$ 2), we can plug it in and get the coordinates of our vertices: (0,2) and (0,-2).

The foci are points that are the same distance from the vertices as the vertices are from the center. We usually label them with the variable $c$.They can be found using the following formula: ${c}^{2} = {a}^{2} + {b}^{2}$.

So now we plug in our ${a}^{2}$ and ${b}^{2}$. Keep in mind that what we have in the equation is already squared, so we don't need to square it again.

$4 + 9 = {c}^{2}$
$c = \pm \sqrt{13}$

Our foci are always on the same vertical line as the vertices. So we know that our foci will be (0,$\sqrt{13}$) and (0, $- \sqrt{13}$).

Lastly, we have our asymptotes. Asymptotes are simply "barriers" that prevent the branches from simply carrying straight on into space, and forcing them to curve.

As indicated by the picture, our asymptotes are simply the lines $y = \pm \frac{a}{b} x$

So all we need to do is to plug in our stuff, and our asymptotes are $y = \frac{2}{3} x$ and $y = - \frac{2}{3} x$

Hope that helps :)