# How do you find all the critical points to graph x^2 /9- y^2 /9 =1  including vertices, foci and asymptotes?

Nov 4, 2016

#### Answer:

The vertices are $\left(+ 3 , 0\right)$ and $\left(- 3 , 0\right)$
The foci are $F = \left(3 \sqrt{2} , 0\right)$ and $F ' = \left(- 3 \sqrt{2} , 0\right)$
The asymptotes are $y = x$ and $y = - x$

#### Explanation:

The equation is a left right hyperbola.
${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$
The center is $h , k$$\implies$$\left(0 , 0\right)$
The vertices are $\left(h \pm a , k\right)$$\implies$( +3,0)$\mathmr{and}$(-3,0)#

The slopes of the asymptotes are $\pm \frac{b}{a}$$= \pm \frac{3}{3} = \pm 1$
The equations of the asymptotes are $y = k \pm \frac{b}{a} \left(x - h\right)$
$\therefore y = \pm x$
To calculate the foci, we need c, ${c}^{2} = {a}^{2} + {b}^{2}$
$\therefore c = \pm \sqrt{18}$
and the foci are $F = \left(3 \sqrt{2} , 0\right)$ and $F ' = \left(- 3 \sqrt{2} , 0\right)$

graph{x^2/9-y^2/9=1 [-12.66, 12.65, -6.33, 6.33]}