# How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola (x+6)^2/36-(y+3)^2/9=1?

May 19, 2018

Vertices are at $\left(- 12 , - 3\right) \mathmr{and} \left(0 , - 3\right)$ Focii are at
$\left(\left(- 6 - 3 \sqrt{5}\right) , - 3\right) \mathmr{and} \left(\left(- 6 + 3 \sqrt{5}\right) , - 3\right)$ and
asymptotes are
$y = \frac{1}{2} x \mathmr{and} y = - \frac{1}{2} x - 6$

#### Explanation:

(x+6)^2/6^2- (y+3)^2/3^2=1; h=-6,k=-3,a=6 , b= 3

This is standard form of the equation of a hyperbola with center

(h,k) ; (x-h)^2/a^2-(y-k)^2/b^2=1 :.  Center $\left(- 6 , - 3\right)$

The vertices are $a$ units from the center, and the foci are

$c$ units from the center. Moreover ${c}^{2} = {a}^{2} + {b}^{2}$ or

${c}^{2} = {6}^{2} + {3}^{2} \therefore {c}^{2} = 45 \mathmr{and} c = \pm 3 \sqrt{5}$

Vertices are at $\left(- 12 , - 3\right) \mathmr{and} \left(0 , - 3\right)$

Focii are at $\left(\left(- 6 - 3 \sqrt{5}\right) , - 3\right) \mathmr{and} \left(\left(- 6 + 3 \sqrt{5}\right) , - 3\right)$

Each hyperbola has two asymptotes that intersect at the center of

the hyperbola. The asymptotes pass through the vertices of a

rectangle of dimensions $2 a = 12 \mathmr{and} 2 b = 6$ with its

center at $\left(- 6 , - 3\right) \therefore$ slope $= \pm \frac{b}{a} = \pm \frac{3}{6} = \pm \frac{1}{2}$.

Equation of asymptotes are $y + 3 = \pm \frac{1}{2} \left(x + 6\right)$ or

$y = - 3 \pm \frac{1}{2} \left(x + 6\right) \mathmr{and} y = \frac{1}{2} x \mathmr{and} y = - \frac{1}{2} x - 6$

Therefore, asymptotes are $y = \frac{1}{2} x \mathmr{and} y = - \frac{1}{2} x - 6$

graph{(x+6)^2/36-(y+3)^2/9=1 [-80, 80, -40, 40]}