Answer
$\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{2}=1$
Work Step by Step
Foci on the x-axis: $\quad\Rightarrow \quad \displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \quad (a\gt b)$
Foci: $\quad(\pm c, 0) \quad\Rightarrow \quad c^{2}=2$
Vertices: $\quad (\pm a, 0)\quad\Rightarrow \quad a^{2}=4$
Center-to-focus distance: $\quad c=\sqrt{a^{2}-b^{2}} \quad\Rightarrow b^{2}=c^{2}-a^{2}$
$b^{2}= a^{2}-c^{2}$
$b^{2}=4-2=2$
So, the equation is
$\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{2}=1$
