## Algebra and Trigonometry 10th Edition

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