# How do you find all the critical points to graph (x/2)^2 - (y/3)^2 = 1 including vertices, foci and asymptotes?

Oct 19, 2016

The center is $\left(0 , 0\right)$ at the origin
The vertices are $\left(2 , 0\right)$ and $\left(- 2 , 0\right)$
The slopes of the asymptotes are $\left(\frac{3}{2}\right)$ and $\left(- \frac{3}{2}\right)$
The equations of the asymptotes are $y = \frac{3 x}{2}$ and $y = \frac{- 3 x}{2}$
To determine the foci, we calculate, $c = \pm \sqrt{{2}^{2} + {3}^{2}} = \pm \sqrt{13}$
So the foci are $\left(\sqrt{13} , 0\right)$ and $\left(- \sqrt{13} , 0\right)$