Precalculus (6th Edition) Blitzer

$\dfrac{(x-2)^2}{4}-\dfrac{(y+3)^2}{9}=1$
Center: $(2,-3) \implies h=2,k=-3$ and Vertices: $2$ This yields: $a=2$ and $b=3$ Standard Equation for a hyperbola when the hyperbola opens to the left is: $\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$ We have $a^2=4; b^2 =9$ So, the equation for hyperbola is : $\dfrac{(x-2)^2}{4}-\dfrac{(y+3)^2}{9}=1$