# How do you graph y^2/16-x^2/4=1 and identify the foci and asympototes?

Oct 22, 2016

The foci are $\left(0 , + 2 \sqrt{5}\right)$ and $\left(0 , - 2 \sqrt{5}\right)$
And the asymptotes are $y = 2 x$ and $y = - 2 x$

#### Explanation:

The graph is a hyperbola "opens up and down"

The center is $\left(0 , 0\right)$

The vertices are $\left(0 , 4\right)$ and $\left(0 , - 4\right)$

The slopes of the asymptotes are $2$ and$- 2$

The equations of the asymptotes are $y = 2 x$ and $y = - 2 x$

To determine the foci, we need $c = \pm \sqrt{16 + 4} = \pm \sqrt{20} = \pm 2 \sqrt{5}$

The foci are $\left(0 , + 2 \sqrt{5}\right)$ and $\left(0 , - 2 \sqrt{5}\right)$

graph{(y/4)^2-(x/2)^2=1 [-40, 40, -20, 20]}