Calculus: Early Transcendentals 8th Edition

$\frac{(x-4)^{2}}{25}+\frac{(y+1)^{2}}{9}=1$,
$h=c=\frac{8-0}{2}=4$ $b^2=a^2-c^2=25-16=9$ $\frac{(x-4)^{2}}{25}+\frac{(y+1)^{2}}{9}=1$, This is an equation of an ellipse.