## Precalculus (6th Edition) Blitzer

$\dfrac{x^2}{25}-\dfrac{y^2}{24}=1$
Foci: $(-7,0), (7,0)$ and Vertices: $(-5,0)$ and $(5,0)$ We are given that $c=7; a=5$` Since $c^2=a^2+b^2$ so, $b^2=c^2-a^2=49-25=24$ Thus, the equation for the hyperbola is: $\dfrac{x^2}{25}-\dfrac{y^2}{24}=1$