# How do you write the equation of the hyperbola given Foci: (-3,0),(3,0) and vertices (-1,0), (1,0)?

Jan 19, 2018

${x}^{2} - {y}^{2} / 8 = 1 , \mathmr{and} , 8 {x}^{2} - {y}^{2} = 8$.

#### Explanation:

The focii $\left(\pm 3 , 0\right)$ lie on the $X \text{-Axis}$, hence, the Transverse

axis of the hyperbola S is $X \text{-Axis}$.

Its eqn. is given by, $S : {x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$.

For $S$, the focii are $\left(\pm a e , 0\right)$, and the vertices are $\left(\pm a , 0\right)$.

(e eccentricity)

$\therefore a e = 3 , \mathmr{and} , a = 1 \Rightarrow e = 3$.

$\therefore {b}^{2} = {a}^{2} \left({e}^{2} - 1\right) = {1}^{2} \left({3}^{2} - 1\right) = 8$.

$\Rightarrow S : {x}^{2} / {1}^{2} - {y}^{2} / 8 = 1 , \mathmr{and} , 8 {x}^{2} - {y}^{2} = 8$.