Chapter 9 - Section 9.2 - The Hyperbola - Exercise Set - Page 981: 7

$\dfrac{x^2}{9}-\dfrac{y^2}{7}=1$

Foci: $(-4,0), (4,0)$ and Vertices: $(-3,0)$ and $(3,0)$ We are given that $ c=4; a=3$` Since $ c^2=a^2+b^2$ so, $ b^2=c^2-a^2=16-9=7$ Thus, the equation for the hyperbola is: $\dfrac{x^2}{9}-\dfrac{y^2}{7}=1$