# How do you find the vertices and foci of 36x^2-10y^2=360?

Nov 17, 2016

The vertices are $= \left(\sqrt{10} , 0\right)$ and $= \left(- \sqrt{10} , 0\right)$
The foci are F$= \left(\sqrt{46} , 0\right)$ and F'$= \left(- \sqrt{46} , 0\right)$
The equations of the asymptotes are

$y = \frac{6}{\sqrt{10}} x$ and $y = - \frac{6}{\sqrt{10}} x$

#### Explanation:

Let's rewrite the equation

$36 {x}^{2} - 10 {y}^{2} = 360$

Divide by 360

${x}^{2} / 10 - {y}^{2} / 36 = 1$

This is the equation of a right left hyperbola

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

The center is $= \left(0 , 0\right)$

The vertices are $\left(\pm a , 0\right)$
$= \left(\pm \sqrt{10} , 0\right)$

To calculate the foci, we need $c = \pm \sqrt{{a}^{2} + {b}^{2}}$

$c = \pm \sqrt{10 + 36} = \pm \sqrt{46}$

The foci are F$= \left(\sqrt{46} , 0\right)$ and F'$= \left(- \sqrt{46} , 0\right)$

The slope of the asymptotes are $= \pm \frac{b}{a}$

$= \pm \frac{6}{\sqrt{10}}$

The equations of the asymptotes are

$y = \frac{6}{\sqrt{10}} x$ and $y = - \frac{6}{\sqrt{10}} x$

graph{(x^2/10-y^2/36-1)(y-6/sqrt10x)(y+6/sqrt10x)=0 [-28.87, 28.87, -14.44, 14.44]}