# How do you find all the critical points to graph x^2/9 -y^2/16=1 including vertices, foci and asymptotes?

Jun 10, 2016

Focii are $\left(\pm a \cdot e , 0\right) = \left(\pm 5 , 0\right)$
Vertices are $\left(\pm a , 0\right) = \left(\pm 3 , 0\right) , \left(0 , \pm b\right) = \left(0 , \pm 4\right)$
Eqns. of asymptotes are $\frac{x}{a} = \pm \frac{y}{b}$, i.e.,$\frac{x}{3} = \pm \frac{y}{4}$

#### Explanation:

We compare the given equation (eqn.) with the standard eqn., ${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$ $\therefore$ ${a}^{2} = 9 , {b}^{2} = 16.$
The eccentricity e is given by : ${b}^{2} - {a}^{2} \left({e}^{2} - 1\right) .$
$\therefore 16 = 9 \left({e}^{2} - 1\right)$$\therefore e = \frac{5}{3}$
Focii are $\left(\pm a \cdot e , 0\right) = \left(\pm 5 , 0\right)$
Vertices are $\left(\pm a , 0\right) = \left(\pm 3 , 0\right) , \left(0 , \pm b\right) = \left(0 , \pm 4\right)$
Eqns. of asymptotes are $\frac{x}{a} = \pm \frac{y}{b}$, i.e.,$\frac{x}{3} = \pm \frac{y}{4}$