Answer
$\frac{(x+2)^2}{16}+\frac{(y-3)^2}{64}=1$, foci: $(-2, 3\pm4\sqrt 3)$.
Work Step by Step
Step 1. Rewrite the given equation as
$4(x^2+4x+4)+(y^2-6y+9)=39+16+9=64$ or $\frac{(x+2)^2}{16}+\frac{(y-3)^2}{64}=1$,
Step 2. We have $a^2=64, b^2=16$ and $c=\sqrt {a^2-b^2}=4\sqrt 3$. The ellipse is centered at $(-2,3)$ with a vertical major axis.
Step 3. We can graph the equation as shown in the figure with foci at $(-2, 3\pm4\sqrt 3)$.
