Answer
$$
\left(\frac{x-14}{6}\right)^{2}+\left(\frac{y+4}{3}\right)^{2}=1.
$$
Work Step by Step
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 6 units to right leads to the ellipse
$$
\left(\frac{x-14}{6}\right)^{2}+\left(\frac{y+4}{3}\right)^{2}=1.
$$
