Answer
$ \displaystyle \frac{y^{2}}{2}-\frac{x^{2}}{8}=1 $
Foci: $\qquad (0, \pm\sqrt{10})$
Directrices: $\quad y=\displaystyle \pm\frac{\sqrt{10}}{5}$
Work Step by Step
Eccentricity $=\displaystyle \frac{\text{distance between foci}}{\text{distance between vertices}}$
Write in standard form and read a,b.
$8y^{2}-2x^{2}=16\qquad /\div 16$
$ \displaystyle \frac{y^{2}}{2}-\frac{x^{2}}{8}=1 \qquad $
We have a vertical axis, $a=\sqrt{2}, b=2\sqrt{2}$
Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{2+8}=\sqrt{10}$
Foci: $\quad (0, \pm c)=\qquad (0, \pm\sqrt{10})$
Vertices: $\quad (0, \pm a)=\qquad (0, \pm\sqrt{2})$
Asymptotes: $\quad y=\displaystyle \pm\frac{a}{b}x\Rightarrow\qquad y=\displaystyle \pm\frac{x}{2}$
Eccentricity: $\displaystyle \quad e=\frac{c}{a}=\frac{2}{\sqrt{3}}$
Directrices: $\quad y=0\displaystyle \pm\frac{a}{e}=\pm\frac{\sqrt{2}}{\sqrt{5}}=\pm\frac{\sqrt{10}}{5}$
