# How do you find the vertices, asymptote, foci and graph y^2/9-x^2/100=1?

Nov 6, 2016

The vertices are $\left(0 , 3\right)$ and $\left(0 , - 3\right)$
The equations of the asymptotes are $y = \frac{3}{10} x$ and $y = - \frac{3}{10} x$
the foci are F$\left(0 , \sqrt{109}\right)$ and F'$\left(0 , - \sqrt{109}\right)$

#### Explanation:

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

This is an up down hyperbola and the center is $\left(0 , 0\right)$

The slope of the asympyotes are $\pm \frac{3}{10}$
The equations of the asymptotes are $y = \frac{3}{10} x$ and $y = - \frac{3}{10} x$
To determine the foci, we need $c = \sqrt{{a}^{2} + {b}^{2}} = \pm \sqrt{109}$
$\therefore$ the foci are F$\left(0 , \sqrt{109}\right)$ and F'$\left(0 , - \sqrt{109}\right)$
graph{(y^2/9)-(x^2/100)=1 [-18.02, 18.01, -9.01, 9.01]}