# How do you find all the critical points to graph 9x^2 - 4y^2 - 90x - 32y + 125 = 0 including vertices, foci and asympotes?

Nov 4, 2016

The vertices are $\left(7 , - 4\right)$ and $\left(3 , - 4\right)$
The foci are $\left(5 + \sqrt{13} , - 4\right)$ and $\left(5 - \sqrt{13} , - 4\right)$
The equations of the asymptotes are $y = - 4 + \frac{3}{2} \left(x - 5\right)$ and $y = - 4 - \frac{3}{2} \left(x - 5\right)$

#### Explanation:

This is the equation of a hyperbola
Rearranging the terms and completing the squares
$9 \left({x}^{2} - 10 x\right) - 4 \left({y}^{2} + 8 y\right) = - 125$
$9 \left({x}^{2} - 10 x + 25\right) - 4 \left({y}^{2} + 8 y + 16\right) = - 125 + 225 - 64 = 36$
$9 {\left(x - 5\right)}^{2} - 4 {\left(y + 4\right)}^{2} = 36$
dividing by $36$
${\left(x - 5\right)}^{2} / 4 - {\left(y + 4\right)}^{2} / 9 = 1$
The center of this left right hyperbola is $= \left(5 , - 4\right)$
$a = 2$ and $b = 3$
The vertices are $\left(7 , - 4\right)$ and $\left(3 , - 4\right)$
The slope of the asymptotes are $\pm \frac{3}{2}$
The equations of the asymptotes are $y = - 4 + \frac{3}{2} \left(x - 5\right)$ and $y = - 4 - \frac{3}{2} \left(x - 5\right)$
To calculate the foci, we need $c = \sqrt{4 + 9} = \sqrt{13}$4
The foci are $\left(5 + \sqrt{13} , - 4\right)$ and $\left(5 - \sqrt{13} , - 4\right)$
graph{(x-5)^2/4-(y+4)^2/9=1 [-13.86, 14.61, -9.24, 5]}