Answer
$\dfrac{x^2}{4}+\dfrac{y^2}{2}=1$
Work Step by Step
The eccentricity of the ellipse $\dfrac{x^2}{m^2}+\dfrac{y^2}{n^2}=1$ when $m \gt n$ is given by:
$e=\dfrac{\sqrt {m^2-n^2}}{m}$
The foci of the ellipse are: $(\pm me,0)$ and the directrices are given as: $x=\pm \dfrac{m}{e}$
Given: focus: $(-\sqrt 2,0)$ and
Directrix: $x=-2 \sqrt 2$ or, $e=\dfrac{1}{\sqrt 2}$
so, the equation of the ellipse is:
$(x+\sqrt 2)^2+(y-0)^2=\dfrac{1}{\sqrt 2} (x+2\sqrt 2)^2$
or, $\dfrac{x^2}{4}+\dfrac{y^2}{2}=1$
