# How do you find the vertices, asymptote, foci and graph y^2/144-x^2/100=1?

##### 1 Answer
Oct 22, 2016

Asymptotes

$y = \frac{6}{5} x$
$y = - \frac{6}{5} x$
Foci

$\left(0 , 2 \sqrt{61}\right)$
$\left(0 , - 2 \sqrt{61}\right)$

vertices

$\left(0 , 12\right)$
$\left(0 , - 12\right)$

#### Explanation:

Given -

${y}^{2} / 144 - {x}^{2} / 100 = 1$

It is in the form

${y}^{2} / {b}^{2} - {x}^{2} / {a}^{2} = 1$

In this case

Asymptotes are

$y = \frac{b}{a} x$

$y = - \frac{b}{a} x$

When ${x}^{2} / {a}^{2}$is negative hyperbola opens up and down along the Y-axis

Vertices are

$\left(0 , b\right)$

$\left(0 , - b\right)$

Foci are -

$\left(0 , c\right)$

$\left(0 , - c\right)$

Let us apply this in our problem -

${a}^{2} = 100$
$a = \pm \sqrt{100} = \pm 10$
${b}^{2} = 144$
$b = \pm \sqrt{144} = \pm 12$
$c = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{100 + 144} = \sqrt{244} = \sqrt{4 \cdot 61} = \pm 2 \cdot \sqrt{61}$
$c = \pm 2 \sqrt{61}$

Asymptotes

$y = \pm \frac{12}{10} x = \pm \frac{6}{5} x$

$y = \frac{6}{5} x$
$y = - \frac{6}{5} x$

Foci

$\left(0 , 2 \sqrt{61}\right)$
$\left(0 , - 2 \sqrt{61}\right)$

vertices

$\left(0 , 12\right)$
$\left(0 , - 12\right)$