Answer
$\frac{x^2}{16}-\frac{y^2}{9}=1$
Work Step by Step
Transverse axis is horizontal.
$(±a,0)=(±4,0)$
$a=4$
$(±c,0)=(±,0)$
$c=5$
$b^2=c^2-a^2=25-16=9$
$b=3$
Standard form when transverse axis is horizontal:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$
$\frac{x^2}{16}-\frac{y^2}{9}=1$
