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