Answer
$\frac{(x+5)^2}{16}+\frac{(y-1)^2}{4}=1$, see graph, foci $(-5\pm2\sqrt 3,1)$.
Work Step by Step
Step 1. Rewrite the given equation as
$(x^2+10x+25)+4(y^2-2y+1)=25+4-13=16$
or
$\frac{(x+5)^2}{16}+\frac{(y-1)^2}{4}=1$,
Step 2. We have $a^2=16, b^2=4$ and $c=\sqrt {a^2-b^2}=2\sqrt 3$. The ellipse is centered at $(-5,1)$ with a horizontal major axis.
Step 3. We can graph the equation as shown in the figure with foci at $(-5\pm2\sqrt 3,1)$.
