## Calculus (3rd Edition)

$$\left(\frac{x+2}{6}\right)^{2}+\left(\frac{y+12}{3}\right)^{2}=1.$$
The ellipse $$\left(\frac{x-8}{6}\right)^{2}+\left(\frac{y+4}{3}\right)^{2}=1$$ has the center $(8,-4)$. Translating the center to $(-2,-12)$, leads to the new ellipse $$\left(\frac{x+2}{6}\right)^{2}+\left(\frac{y+12}{3}\right)^{2}=1.$$