# How do you find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola x^2-36y^2=36?

Dec 21, 2017

Vertices are $\left(6 , 0\right)$ and $\left(- 6 , 0\right)$, focii are $\left(\sqrt{37} , 0\right)$ and $\left(- \sqrt{37} , 0\right)$ and asymptotes are $x \pm 6 y = 0$.

#### Explanation:

If the general equation of hyperbola is

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$, then

vertices are $\left(h \pm a , k\right)$, focii are $\left(h \pm c , k\right)$, where ${c}^{2} = {a}^{2} + {b}^{2}$, directrix are $x = \pm {a}^{2} / c$

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

Here equation is ${x}^{2} - 36 {y}^{2} = 36$ or ${x}^{2} / {6}^{2} - {y}^{2} / {1}^{2} = 1$

as $a = 6$ and $b = 1$, vertices are $\left(6 , 0\right)$ and $\left(- 6 , 0\right)$,

and $c = \sqrt{36 + 1} = \sqrt{37}$,

focii are $\left(\sqrt{37} , 0\right)$ and $\left(- \sqrt{37} , 0\right)$

directrix are $x = \pm \frac{36}{\sqrt{37}}$

Asymptotes are $y = \pm \frac{x}{6}$ or $x \pm 6 y = 0$

graph{(x^2-36y^2-36)(x+6y)(x-6y)=0 [-10, 10, -5, 5]}