Answer
$\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{16}=1$
Foci: $(0, -2\sqrt{3})$ and $(0, 2\sqrt{3})$.
Work Step by Step
The major axis is vertical (parallel to the y-axis). .
$\displaystyle \frac{(x-h)^{2}}{b^{2}}+ \displaystyle \frac{(y-k)^{2}}{a^{2}}=1$
From the graph,
$a=4,$
$b=2$,
center: $(0,0)$
$\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{16}=1$
Foci are $c$ units above and $c$ units below the center,
$c^{2}=a^{2}-b^{2}=16-4=12$
$c=2\sqrt{3}$
The foci are at
$(0, -2\sqrt{3})$ and $(0, 2\sqrt{3})$.
