# How do you find the vertices, asymptote, foci and graph x^2-9y^2=25?

Nov 17, 2016

#### Explanation:

The standard form for the Horizontal Transverse Axis type is:

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

center: $\left(h , k\right)$

vertices: $\left(h - a , k\right) \mathmr{and} \left(h + a , k\right)$

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

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

Given: ${x}^{2} - 9 {y}^{2} = 25$

Put the given equation in standard form. Divide both sides by $25$:

${x}^{2} / 25 - 9 {y}^{2} / 25 = 1$

Flip the 9 down under the 25:

${x}^{2} / 25 - {y}^{2} / \left(\frac{25}{9}\right) = 1$

Write the denominators as squares:

${x}^{2} / {5}^{2} - {y}^{2} / {\left(\frac{5}{3}\right)}^{2} = 1$

Insert - 0s in the numerators:

${\left(x - 0\right)}^{2} / {5}^{2} - {\left(y - 0\right)}^{2} / {\left(\frac{5}{3}\right)}^{2} = 1$

center: $\left(0 , 0\right)$

vertices: $\left(- 5 , 0\right) \mathmr{and} \left(5 , 0\right)$

foci: $\left(- 5 \frac{\sqrt{10}}{3} , 0\right) \mathmr{and} \left(5 \frac{\sqrt{10}}{3} , 0\right)$

Asymptotes:

$y = - \frac{1}{3} x \mathmr{and} y = \frac{1}{3} x$