How do you find the center, vertices, foci, and asymptotes of the hyperbola, and sketch its graph of x^2/9 - y^2/16 = 1?

Mar 4, 2017

The center is $\left(0 , 0\right)$
The vertices are (-3,0)$\mathmr{and}$(3,0)#

The foci are $F ' = \left(- 5 , 0\right)$ and $F = \left(5 , 0\right)$
The asymptotes are $y = \frac{4}{3} x$ and $y = - \frac{4}{3} x$

Explanation:

We compare this equation

${x}^{2} / {3}^{2} - {y}^{2} / {4}^{2} = 1$

to

${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$

The center is $C = \left(0 , 0\right)$

The vertices are $V ' = \left(- a , 0\right) = \left(- 3 , 0\right)$ and $V = \left(a , 0\right) = \left(3 , 0\right)$

To find the foci, we need the distance from the center to the foci

${c}^{2} = {a}^{2} + {b}^{2} = 9 + 16 = 25$

$c = \pm 5$

The foci are $F ' = \left(- c , 0\right) = \left(- 5 , 0\right)$ and $F = \left(c , 0\right) = \left(5 , 0\right)$

The asymptotes are

${x}^{2} / {3}^{2} - {y}^{2} / {4}^{2} = 0$

$y = \pm \frac{4}{3} x$

graph{((x^2)/9-(y^2)/16-1)(y-4/3x)(y+4/3x)=0 [-16.02, 16.02, -8.01, 8.01]}