# How do you write the equation of the hyperbola given Foci: (-8,0),(8,0) and vertices (-4sqrt3,0), (4sqrt3,0)?

Sep 14, 2017

${x}^{2} / 48 - {y}^{2} / 16 = 1.$

#### Explanation:

Recall that, for the Hyperbola $S : {x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1 ,$

Focii and Vertices are, $\left(\pm a e , 0\right) , \mathmr{and} , \left(\pm a , 0\right) ,$ resp.

Here, $e > 1 ,$ the Eccentricity, is given by, ${b}^{2} = {a}^{2} \left({e}^{2} - 1\right) .$

$\therefore a e = 8 , \mathmr{and} , a = 4 \sqrt{3.}$

Now, ${b}^{2} = {a}^{2} \left({e}^{2} - 1\right) = {a}^{2} {e}^{2} - {a}^{2} = {8}^{2} - {\left(4 \sqrt{3}\right)}^{2} = 16.$

Hence, the desired eqn. of the Hyperbola is, ${x}^{2} / 48 - {y}^{2} / 16 = 1.$