## Calculus (3rd Edition)

$$\left(\frac{x-8}{6}\right)^{2}+\left(\frac{y+8}{3}\right)^{2}=1.$$
The ellipse $$\left(\frac{x-8}{6}\right)^{2}+\left(\frac{y+4}{3}\right)^{2}=1$$ has the center $(8,-4)$. Shifting the center 4 units to down leads to the ellipse $$\left(\frac{x-8}{6}\right)^{2}+\left(\frac{y+8}{3}\right)^{2}=1.$$