Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.5 Conic Sections - Exercises - Page 636: 10

$$ \left(\frac{x-8}{6}\right)^{2}+\left(\frac{y+8}{3}\right)^{2}=1. $$

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 $$ \left(\frac{x-8}{6}\right)^{2}+\left(\frac{y+8}{3}\right)^{2}=1. $$