Answer
$\frac{(x+3)^2}{36}+\frac{(y-5)^2}{4}=1$
Work Step by Step
Step 1. With the given conditions, we have $2a=12, a=6$, and $2b=4, b=2$
Step 2. The center is at $(-3,5)$, so we can write the equation as $\frac{(x+3)^2}{36}+\frac{(y-5)^2}{4}=1$
