# How do you find the vertices, asymptote, foci and graph 100x^2-81y^2=8100?

Nov 1, 2016

#### Explanation:

The standard form for this kind of hyperbola is:

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

where:

$\left(h , k\right)$ is the center

$\left(h - a , k\right)$ and $\left(h + a , k\right)$ are the vertices

$\left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right)$ and $\left(h + \sqrt{{a}^{2} + {b}^{2}} , k\right)$ are the foci

and the asymptotes are

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

Given:

$100 {x}^{2} - 81 {y}^{2} = 8100$

Divide both sides by 8100:

${x}^{2} / 81 - {y}^{2} / 100 = 1$

Write the denominators as squares and add -0 inside the numerator squares:

${\left(x - 0\right)}^{2} / {9}^{2} - {\left(y - 0\right)}^{2} / {10}^{2} = 1$

Now, the vertices can be written by observation $\left(- 9 , 0\right)$ and $\left(9 , 0\right)$

The foci are written with at trivial computation of $\sqrt{{10}^{2} + {9}^{2}} = \sqrt{181}$:

$\left(- \sqrt{181} , 0\right) \mathmr{and} \left(\sqrt{181} , 0\right)$

The asymptotes are:

$y = \frac{10}{9} x$ and $y = - \frac{10}{9} x$