# How do you identify the vertices, foci, and direction of (y+8)^2/36-(x+2)^2/25=1?

Dec 24, 2016

Because the x term is negative, we know that the hyperbola opens vertically; the type of hyperbola is called a vertical transverse axis type; its standard form is:

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

The center of this type is the point $\left(h , k\right)$
The vertices are the points $\left(h , k - a\right) \mathmr{and} \left(h , k + a\right)$
The foci of this are are the points $\left(h , k - \sqrt{{a}^{2} + {b}^{2}}\right) \mathmr{and} \left(h , k + \sqrt{{a}^{2} + {b}^{2}}\right)$

Rewrite the given equation in the form of equation [1]:

${\left(y - - 8\right)}^{2} / {6}^{2} - \frac{x - - 2}{5} ^ 2 = 1 \text{ [2]}$

he center of this type is the point $\left(- 8 , - 2\right)$
The vertices are the points $\left(- 8 , - 8\right) \mathmr{and} \left(- 8 , 4\right)$
The foci of this are are the points $\left(- 8 , - 2 - \sqrt{61}\right) \mathmr{and} \left(- 8 , - 2 + \sqrt{61}\right)$