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