Answer
$\frac{x^2}{4}+\frac{3y^2}{16}=1$
Work Step by Step
Major axis is vertical.
Equation of the ellipse:
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
The ellipse passes through the point $(2,0)$
$\frac{2^2}{b^2}+\frac{0^2}{a^2}=1$
$2^2=b^2$
$b=2$
The ellipse passes through the point $(-1,2)$
$\frac{(-1)^2}{2^2}+\frac{2^2}{a^2}=1$
$\frac{1}{4}+\frac{4}{a^2}=1$
$\frac{4}{a^2}=1-\frac{1}{4}=\frac{3}{4}$
$3a^2=16$
$b^2=\frac{16}{3}$
$\frac{x^2}{2^2}+\frac{y^2}{\frac{16}{3}}=1$
$\frac{x^2}{4}+\frac{3y^2}{16}=1$
