# How do you find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola (y+6)^2/20-(x-1)^2/25=1?

Jan 11, 2017

Given: ${\left(y - k\right)}^{2} / {a}^{2} - {\left(x - h\right)}^{2} / {b}^{2} = 1 \text{ [1]}$
vertices: $\left(h , k \pm a\right)$
foci:$\left(h , k \pm \sqrt{{a}^{2} + {b}^{2}}\right)$
asymptotes: $y = \pm \frac{a}{b} \left(x - h\right) + k$

#### Explanation:

This is a good reference for Hyperbolas

Write the given equation in the same form as equation [1]:

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

The vertices are:

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

$\left(1 , - 6 - 2 \sqrt{5}\right) \mathmr{and} \left(1 , - 6 + 2 \sqrt{5}\right)$

The foci are:

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

$\left(1 , - 6 - \sqrt{20 + 25}\right) \mathmr{and} \left(1 , - 6 + \sqrt{20 + 25}\right)$

$\left(1 , - 6 - \sqrt{45}\right) \mathmr{and} \left(1 , - 6 + \sqrt{45}\right)$

The equations of the asymptotes are:

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

$y = - 2 \frac{\sqrt{5}}{5} \left(x - 1\right) - 6$ and $y = 2 \frac{\sqrt{5}}{5} \left(x - 1\right) - 6$