Answer
$F(0,\pm 1)$
Eccentricity$:\qquad e =\displaystyle \frac{1}{\sqrt{3}}$
Directrices$:\quad y=\pm 3$
Graph:
Work Step by Step
$3x^{2}+2y^{2}=6\qquad/\div 6$
$\displaystyle \frac{x^{2}}{2}+\frac{y^{2}}{3}=1,\qquad a=\sqrt{3},b=\sqrt{2}$
The major axis is vertical.
$c=\sqrt{a^{2}-b^{2}}=\sqrt{3-2}=1$
$F(0,\pm 1)$
Eccentricity$:\qquad e=\displaystyle \frac{c}{a}=\frac{1}{\sqrt{3}}$
Directrices$:\quad y=0\displaystyle \pm\frac{a}{e}$
$y=\displaystyle \pm\frac{\sqrt{3}}{(\frac{1}{\sqrt{3}})}$
$y=\pm 3$
