Answer
$\frac{(x-4)^2}{9}+\frac{(y+2)^2}{4}=1$; foci at $(4\pm\sqrt 5,-2)$.
Work Step by Step
Step 1. Rewrite the given equation as
$4(x^2-8x+16)+9(y^2+4y+4)=64+36-64=36$
or
$\frac{(x-4)^2}{9}+\frac{(y+2)^2}{4}=1$,
Step 2. We have $a^2=9, b^2=4$ and $c=\sqrt {a^2-b^2}=\sqrt 5$. The ellipse is centered at $(4,-2)$ with a horizontal major axis.
Step 3. We can graph the equation as shown in the figure with foci at $(4\pm\sqrt 5,-2)$.
