Answer
$\dfrac{x^2}{4}+\dfrac{y^2}{2}=1$
Work Step by Step
The eccentricity of the ellipse $\dfrac{x^2}{p^2}+\dfrac{y^2}{q^2}=1$ when $p \gt q$ is given by:
$e=\dfrac{\sqrt {p^2-q^2}}{p}$
The foci of the ellipse are: $(\pm pe,0)$ and the directrices are given as: $x=\pm \dfrac{p}{e}$
We are given that $focus: (-\sqrt 2,0)$ and
The directrix is $x=-2 \sqrt 2 \implies e=\dfrac{1}{\sqrt 2}$
Thus, the equation of the ellipse becomes:
$(x+\sqrt 2)^2+(y-0)^2=\dfrac{1}{\sqrt 2} (x+2\sqrt 2)^2 \implies \dfrac{x^2}{4}+\dfrac{y^2}{2}=1$
