Answer
$\frac{y^2}{9}-\frac{4x^2}{9}=1$
Work Step by Step
Transverse axis is vertical.
$(0,±a)=(0,±3)$
$a=3$
Use the point $(-2,5)$:
Standard form when transverse axis is vertical:
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\frac{5^2}{3^2}-\frac{(-2)^2}{b^2}=1$
$\frac{25}{9}-\frac{4}{b^2}=1$
$\frac{25}{9}-1=\frac{4}{b^2}$
$\frac{16}{9}=\frac{4}{b^2}$
$b^2=\frac{9}{4}$
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\frac{y^2}{9}-\frac{x^2}{\frac{9}{4}}=1$
$\frac{y^2}{9}-\frac{4x^2}{9}=1$
