Answer
$\displaystyle \frac{x^{2}}{2}-\frac{y^{2}}{8}=1$
Foci: $\qquad (\pm\sqrt{10},0)$
Directrices: $\quad x=\displaystyle \pm\frac{\sqrt{10}}{5}$
Work Step by Step
Eccentricity $=\displaystyle \frac{\text{distance between foci}}{\text{distance between vertices}}$
Write in standard form and read a,b.
$8x^{2}-2y^{2}=16\qquad /\div 16$
$\displaystyle \frac{x^{2}}{2}-\frac{y^{2}}{8}=1\qquad $
We have a horizontal axis with $a=\sqrt{2}, b=2\sqrt{2}$
Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{2+8}=\sqrt{10}$
Foci: $\quad (\pm c, 0)=\qquad (\pm\sqrt{10},0)$
Vertices: $\quad (\pm a, 0)=\qquad (\pm\sqrt{2}, 0)$
Asymptotes: $\quad y=\displaystyle \pm\frac{b}{a}x \Rightarrow \quad y=\pm 2x$
Eccentricity: $\displaystyle \quad e=\frac{c}{a}=\frac{\sqrt{10}}{\sqrt{2}}=\sqrt{5}$
Directrices: $\quad x=0\displaystyle \pm\frac{a}{e}=\pm\frac{\sqrt{2}}{\sqrt{5}}=\pm\frac{\sqrt{10}}{5}$
