Answer
$\frac{x^2}{4}+\frac{4y^2}{9}=1$
Work Step by Step
Major axis is horizontal.
Vertices:
$(2,0)=(a,0)$
$(-2,0)=(-a,0)$
$(0,\frac{3}{2})=(0,b)$
$(0,-\frac{3}{2})=(0,-b)$
That is, $a=2$ and $b=\frac{3}{2}$
Standard form when major axis is horizontal:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\frac{x^2}{2^2}+\frac{y^2}{(\frac{3}{2})^2}=1$
$\frac{x^2}{4}+\frac{y^2}{\frac{9}{4}}=1$
$\frac{x^2}{4}+\frac{4y^2}{9}=1$
