# How do you find the foci and sketch the hyperbola x^2/9-y^2/4=1?

Jul 4, 2017

We know that the standard Cartesian form for the equation of a hyperbola with a transverse horizontal axis,

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ [1]}$

has foci at $\left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right)$ and $\left(h + \sqrt{{a}^{2} + {b}^{2}} , k\right)$

If we write the given equation in the same form as equation [1], then it is a simple matter to find the foci:

${\left(x - 0\right)}^{2} / {3}^{2} - {\left(y - 0\right)}^{2} / {2}^{2} = 1 \text{ [2]}$

Now that we have the given equation in the same form, the computation for the foci is trivial:

$\left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right)$ and $\left(h + \sqrt{{a}^{2} + {b}^{2}} , k\right)$

$\left(0 - \sqrt{9 + 4} , 0\right)$ and $\left(0 + \sqrt{9 + 4} , 0\right)$

$\left(- \sqrt{13} , 0\right)$ and $\left(\sqrt{13} , 0\right) \leftarrow$ these are the foci.

To graph the equation you will need the vertices:

$\left(h - a , k\right)$ and $\left(h + a , k\right)$

$\left(0 - 3 , 0\right)$ and $\left(0 + 3 , 0\right)$

$\left(- 3 , 0\right)$ and $\left(3 , 0\right) \leftarrow$ these are the vertices.

And you will need the equations of the asymptotes:

$y = - \frac{b}{a} \left(x - h\right) + k$ and $y = \frac{b}{a} \left(x - h\right) + k$

$y = - \frac{2}{3} \left(x - 0\right) + 0$ and $y = \frac{2}{3} \left(x - 0\right) + 0$

$y = - \frac{2}{3} x$ and $y = \frac{2}{3} x \leftarrow$ there are the equations of the asymptotes.

Here is a graph of, the equation, the foci, the vertices, and the asymptotes.