Answer
$\displaystyle \frac{y^{2}}{8}-\frac{x^{2}}{8}=1$
Foci: $\qquad (0, \pm 4)$
Directrices: $\quad y=\pm 2$
Work Step by Step
Eccentricity $=\displaystyle \frac{\text{distance between foci}}{\text{distance between vertices}}$
Write in standard form and read a,b.
$y^{2}-x^{2}=8\qquad /\div 8$
$\displaystyle \frac{y^{2}}{8}-\frac{x^{2}}{8}=1\qquad $
We have a vertical axis, $a=2\sqrt{2}, b=2\sqrt{2}$
Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{8+8}=4$
Foci: $\quad (0, \pm c)=\qquad (0, \pm 4)$
Vertices: $\quad (0, \pm a)=\qquad (0, \pm 2\sqrt{2})$
Asymptotes: $\quad y=\displaystyle \pm\frac{a}{b}x\Rightarrow\qquad y=\pm x$
Eccentricity: $\displaystyle \quad e=\frac{c}{a}=\frac{4}{\sqrt{8}}=\sqrt{2}$
Directrices: $\quad y=0\displaystyle \pm\frac{a}{e}=\pm\frac{\sqrt{8}}{\sqrt{2}}=\pm 2$
