## Calculus (3rd Edition)

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