Answer
$\frac{x^2}{4}-\frac{5y^2}{12}=1$
Work Step by Step
Transverse axis is horizontal.
$(±a,0)=(±2,0)$
$a=2$
Use the point $(3,\sqrt 3)$:
Standard form when transverse axis is horizontal:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{3^2}{2^2}-\frac{(\sqrt 3)^2}{b^2}=1$
$\frac{9}{4}-\frac{3}{b^2}=1$
$\frac{9}{4}-1=\frac{3}{b^2}$
$\frac{5}{4}=\frac{3}{b^2}$
$b^2=\frac{12}{5}$
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{x^2}{4}-\frac{y^2}{\frac{12}{5}}=1$
$\frac{x^2}{4}-\frac{5y^2}{12}=1$
