# How do you find the vertices, asymptote, foci and graph x^2/144-y^2/121=1?

Jul 7, 2018

#### Explanation:

This is the equation of a hyperbola

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

Here, we have

${x}^{2} / {12}^{2} - {y}^{2} / {11}^{2} = 1$

The center is $C = \left(h , k\right) = \left(0 , 0\right)$ at the origin.

The vertices are

$V = \left(h + a , k\right) = \left(12 , 0\right)$

and

$V ' = \left(h - a , k\right) = \left(- 12 , 0\right)$

The equations of the asymptotes are

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

$y = \pm \frac{11}{12} x$

That is

$y = - \frac{11}{12} x$ and $y = \frac{11}{12} x$

Calculate

$c = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{144 + 121} = \sqrt{265}$

The foci are $F = \left(\sqrt{265} , 0\right)$ and $F ' = \left(- \sqrt{265} , 0\right)$

And the graph is

graph{(x^2/144-y^2/121-1)(y-11/12x)(y+11/12x)=0 [-65.8, 65.84, -32.95, 32.9]}