Chapter 9 - Section 9.2 - The Hyperbola - Exercise Set - Page 982: 31

$\dfrac{(x-2)^2}{4}-\dfrac{(y+3)^2}{9}=1$

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 : $\dfrac{(x-2)^2}{4}-\dfrac{(y+3)^2}{9}=1$