# How do you graph y^2-9x^2=9 and identify the foci and asympototes?

Nov 22, 2016

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

#### Explanation:

Let's rewrite the equation as

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

This is the equation of an up-down hyperbola.

The general equation is

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

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

The vertices are $\left(h , k \pm a\right)$, A$= \left(0 , 3\right)$ and A'$= \left(0 , - 3\right)$

The slope of the asymptotes are $\pm \frac{a}{b}$ $= \pm 3$

The equations of the asymptotes are $y = 3 x$ and $y = - 3 x$

To determine the foci, we need c, ${c}^{2} = {a}^{2} + {b}^{2}$

So, $c = \pm \sqrt{{a}^{2} + {b}^{2}} = \pm \sqrt{9 + 1} = \pm \sqrt{10}$

And the foci are F$= \left(0 , \sqrt{10}\right)$ and F'$\left(0 , - \sqrt{10}\right)$

graph{(y^2/9-x^2-1)(y-3x)(y+3x)=0 [-12.66, 12.65, -6.32, 6.34]}