# How do you find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola y^2/18-x^2/20=1?

Oct 26, 2016

The vertices are $\left(0 , 3 \sqrt{2}\right)$ and $\left(0 , - 3 \sqrt{2}\right)$
The foci are $\left(0 , \sqrt{38}\right)$ and $\left(0 , - \sqrt{38}\right)$
the asymptotes are $y = \frac{3 x}{\sqrt{10}}$ and $\frac{- 3 x}{\sqrt{10}}$

#### Explanation:

Looking at the equation, this is an up-down hyperbola

and, we compare to the standard equation
${\left(y - k\right)}^{2} / {b}^{2} - {\left(x - h\right)}^{2} / {a}^{2} = 1$

The center is $\left(h , k\right) = \left(0 , 0\right)$

The vertices are $\left(h , k + b\right) = \left(0 , \sqrt{18}\right)$ and $\left(h . k - b\right) = \left(0 , - \sqrt{18}\right)$

The slopes of the asymptotes are $\frac{b}{a} = \frac{\sqrt{18}}{\sqrt{20}} = \frac{3}{\sqrt{10}}$ and $- \frac{b}{a} = - \frac{\sqrt{18}}{\sqrt{20}} = - \frac{3}{\sqrt{10}}$
The equations of the asymptotes are $y = k \pm \frac{b}{a} \left(x - h\right)$

The equations of the asymptotes are $y = 0 + 3 \frac{x}{\sqrt{10}}$ and $y = 0 - 3 \frac{x}{\sqrt{10}}$

To calculate the foci, we need $c = \pm \sqrt{{a}^{2} + {b}^{2}} = \pm \sqrt{18 + 20} = \pm \sqrt{38}$
The foci are $h , k \pm c$

And the foci are $\left(0 , \sqrt{38}\right)$ and $\left(0 , - \sqrt{38}\right)$