# How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola (y-3)^2/25-(x-2)^2/16=1?

Oct 24, 2016

The vertices are $\left(2 , 8\right)$ and $\left(2 , - 2\right)$
The foci are $\left(2 , 3 + \sqrt{41}\right)$ and $\left(2 , 3 - \sqrt{41}\right)$
The equations of the asymptotes are $\left(y = 3 + \frac{5}{4} \left(x - 3\right)\right)$
and $\left(3 - \frac{5}{4} \left(x - 3\right)\right)$

#### Explanation:

Looking at the equation it's an up-down hyperbola

The center of the hyperbola is $\left(2 , 3\right)$

The vertices are $\left(2 , 8\right)$ and $\left(2 , - 2\right)$

The slope of the asymptotes ares $\left(\frac{5}{4}\right)$ and $\left(- \frac{5}{4}\right)$

The equations of the asymptotes are $\left(y = 3 + \frac{5}{4} \left(x - 3\right)\right)$
and $\left(3 - \frac{5}{4} \left(x - 3\right)\right)$

To calculate the foci, we need $c = \pm \sqrt{16 + 25} = \sqrt{41}$

The foci are $\left(2 , 3 + \sqrt{41}\right)$ and $\left(2 , 3 - \sqrt{41}\right)$