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