Answer
$\frac{(x+3)^2}{3}+\frac{(y-1)^2}{4}=1$,
See graph.
Work Step by Step
1. Given center $(-3,1)$, vertex $(-3,3)$, focus $(-3,0)$, we can identify a vertical major axis, $a=3-1=2, c=1-0=1$, thus $b=\sqrt {a^2-c^2}=\sqrt {4-1}=\sqrt {3}$,
2. Thus we have $\frac{(x+3)^2}{3}+\frac{(y-1)^2}{4}=1$,
3. See graph.
