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

Jul 9, 2018

Center: $\left(0 , 0\right)$, vertices are at $\left(0 , 3\right) \mathmr{and} \left(0 , - 3\right)$, focii are at$\left(0 , 3 \sqrt{2}\right) \mathmr{and} \left(0 , - 3 \sqrt{2}\right)$, equation of asymptotes
of vertical hyperbola are
$y = \pm x$

#### Explanation:

${x}^{2} - {y}^{2} + 9 = 0 \mathmr{and} {y}^{2} - {x}^{2} = 9$ or

${y}^{2} / {3}^{2} - {x}^{2} / {3}^{2} = 1$.The standard equation of vertical

hyperbola is (y-k)^2/a^2-(x-h)^2/b^2=1; h,k being center.

Here $h = 0 , k = 0 , a = 3 , b = 3$. So center is at $\left(0 , 0\right)$

Vertices are $a = 3$ units from center . Therefore two vertices

are at $\left(0 , 3\right) \mathmr{and} \left(0 , - 3\right)$ , Focii are $c$ units from center.

${c}^{2} = {a}^{2} + {b}^{2} = 9 + 9 = 18 \therefore c = 3 \sqrt{2}$. Therefore focii are

$\left(0 , 3 \sqrt{2}\right) \mathmr{and} \left(0 , - 3 \sqrt{2}\right)$. Hyperbola has two asymptotes

that intersect at the center of the hyperbola. The asymptotes

pass through the vertices of a rectangle of dimensions

$2 a = 6 \mathmr{and} 2 b = 6$ with its center at $\left(0 , 0\right) \therefore$

slope $\pm \frac{a}{b} = \pm 1$. Equation of asymptotes are

$y - 0 = \pm 1 \left(x - 0\right) \mathmr{and} y = \pm x$

graph{x^2-y^2+9=0 [-10, 10, -5, 5]}[Ans]