## Precalculus (6th Edition) Blitzer

$\dfrac{y^2}{25}-\dfrac{x^2}{9}=1$
Center: $(0,0)$ and vertices: $(0,-5); (0,5)$ This yields: $a=2$ and Asymptote: $y=\dfrac{5}{3} x \implies b=3$ And we have a vertical traverse axis. Standard Equation for a hyperbola is : $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$ We have $a^2=25; b^2 =9$ So, the equation for the hyperbola is: $\dfrac{y^2}{25}-\dfrac{x^2}{9}=1$