Answer
The foci are $(\pm \sqrt{2},0)$ and the vertices are $(\pm 2,0)$.
Work Step by Step
Recall: The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a\geq b$ has the foci at $(\pm c,0)$ where $c^2=a^2-b^2$ and the vertices $(\pm a,0)$.
We have
$x^2=4-2y^2$
$x^2+2y^2=4$
$\frac{x^2+2y^2}{4}=1$
$\frac{x^2}{4}+\frac{y^2}{2}=1$
$\frac{x^2}{2^2}+\frac{y^2}{\sqrt{2}^2}=1$
Then,
$a=2$ and $b=\sqrt{2}$
$c=\sqrt{a^2-b^2}=\sqrt{2^2-\sqrt{2}^2}=\sqrt{4-2}=\sqrt{2}$
So, the foci are $(\pm \sqrt{2},0)$ and the vertices are $(\pm 2,0)$.
