Answer
$\frac{x^{2}}{12}+\frac{(y-4)^{2}}{16}=1$,
Work Step by Step
Midpoint of the vertices $(0,0)$ and $(0,8)$ is $(0,4)$ which is the center of the ellipse.
$a^2=b^2-c^2=16-4=12$
$\frac{x^{2}}{12}+\frac{(y-4)^{2}}{16}=1$,
This is an equation of an ellipse.
