Answer
$\frac{x^2}{16}+\frac{y^2}{4}=1$
Work Step by Step
Major axis is vertical.
The point $(0,4)$ is a vertice:
$a=4$
The point $(2,0)$ is one of the endpoints of the minor axis:
$b=2$
Standard form when major axis is vertical:
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
$\frac{x^2}{4^2}+\frac{y^2}{2^2}=1$
$\frac{x^2}{16}+\frac{y^2}{4}=1$
