# How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola (x+1)^2/4-(y+3)^2/9=1?

Mar 25, 2018

The vertices are $\left(- 3 , - 3\right) \mathmr{and} \left(1 , - 3\right)$ , focii are
$\left(\left(- 1 - \sqrt{13}\right) , - 3\right) \mathmr{and} \left(\left(- 1 + \sqrt{13}\right) , - 3\right)$ asymptotes
are
$y = \frac{3}{2} x - \frac{3}{2} \mathmr{and} y = - \frac{3}{2} x - \frac{9}{2}$

#### Explanation:

(x+1)^2/4- (y+3)^2/9=1; h=-1,k=-3,a=2 , 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 :.  Centre $\left(- 1 , - 3\right)$

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

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

c ((-1-2),-3) and ((-1+2,3)  or

(-3,-3) and (1,-3) ; c^2= 2^2+3^2=13

$\therefore c = \pm \sqrt{13} \therefore$ Focii are at

$\left(\left(- 1 - \sqrt{13}\right) , - 3\right) \mathmr{and} \left(\left(- 1 + \sqrt{13}\right) , - 3\right)$

Each hyperbola has two asymptotes that intersect at the center of

the hyperbola. The asymptotes pass through the vertices of a

rectangle of dimensions $2 a = 4 \mathmr{and} 2 b = 6$ with its

centre at $\left(- 1 , - 3\right) \therefore$ slope $\pm \frac{b}{a} = \pm \frac{3}{2}$ Equation of

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

or $y = \frac{3}{2} x - \frac{3}{2} \mathmr{and} y = - \frac{3}{2} x - \frac{9}{2}$ [Ans]