# How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola y^2/36-x^2/4=1?

Apr 1, 2018

$\text{Vertices : "(0,+-6)," Asymptotes : } x \pm 3 y = 0 , \mathmr{and} ,$

$F o c i i : \left(0 , \pm 2 \sqrt{10}\right)$.

#### Explanation:

We know that, for the $\text{Hyperbola S : } {y}^{2} / {b}^{2} - {x}^{2} / {a}^{2} = 1$,

$\left(1\right) : \text{ Vertices are "(0,+-b), (2)" Focii are } \left(0 , \pm b e\right)$

$\left(3\right) : \text{ the Asymptotes are } y = \pm \frac{b}{a} \cdot x$.

The Eccentricity $e$ is given by, ${a}^{2} = {b}^{2} \left({e}^{2} - 1\right)$.

We have, $b = 6 , a = 2$.

$\therefore$ Vertices are $\left(0 , \pm 6\right)$, and the eqns. of the Asymptotes are

$x \pm 3 y = 0$.

For $e$, we have, $4 = 36 \left({e}^{2} - 1\right) , \mathmr{and} , e = \frac{\sqrt{10}}{3}$.

$\therefore$ Focii are $\left(0 , \pm 2 \sqrt{10}\right)$.