# How do you find the vertices, asymptote, foci and graph x^2/36-y^2=1?

May 17, 2017

The General form for a hyperbola of this type is:${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ }$
Vertices: $\left(h - a , k\right)$ and $\left(h + a , k\right)$
Foci: $\left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right)$ and $\left(h + \sqrt{{a}^{2} + {b}^{2}} , k\right)$
Asymptotes: $y = \pm \frac{b}{a} \left(x - h\right) + k$

#### Explanation:

Here is a reference for Hyperbola equations provided in the answer section.

Let's convert the given equation, ${x}^{2} / 36 - {y}^{2} = 1$, to the form of equation :

${\left(x - 0\right)}^{2} / {6}^{2} - {\left(y - 0\right)}^{2} / {1}^{2} = 1 \text{ }$

By inspection of equation  we observe that $h = k = 0 , a = 6 , \mathmr{and} b = 1$

Vertices: $\left(- 6 , 0\right)$ and $\left(6 , 0\right)$
Foci: $\left(- \sqrt{37} , 0\right)$ and $\left(\sqrt{37} , 0\right)$

Asymptotes: $y = \frac{1}{6} x \mathmr{and} y = - \frac{1}{6} x$

Here is a graph:

graph{x^2/36-y^2=1 [-10, 10, -5, 5]}