# How do you find the center, vertices, foci and asymptotes of (x^2/16)-(y^2/9)=1?

May 1, 2018

Center $\left(0 , 0\right)$, Vertices are -4,0) and (4,0)
Foci are $\left(- 5 , 0\right) \mathmr{and} \left(5 , 0\right)$ and equation of
asymptotes are
$y = \pm \frac{3}{4}$

#### Explanation:

 (x^2/4^2-y^2/3^2=1

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 $\left(0 , 0\right) , a = 4 , b = 3$

The vertices are $a = 4$ units from the center, and the foci are

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

$\therefore {c}^{2} = 16 + 9 = 25 \therefore c = \pm 5$ Vertices are -4,0) and (4,0)

Foci are $\left(- 5 , 0\right) \mathmr{and} \left(5 , 0\right)$ The asymptotes pass through the

vertices of a rectangle of dimensions $2 a = 8 \mathmr{and} 2 b = 6$ with its

centre at $\left(0 , 0\right) \therefore$ slope =+-b/a= +-3/4) Equation of

asymptotes are is $y - 0 = \pm \frac{3}{4} \left(x - 0\right) \mathmr{and} y = \pm \frac{3}{4}$

graph{x^2/16-y^2/9=1 [-10, 10, -5, 5]} [Ans]