Answer
$\dfrac{(x+3)^2}{3}+\dfrac{(y-1)^2}{4}=1$
See graph
Work Step by Step
We are given the ellipse:
Center: $(-3,1)$
Focus: $(-3,0)$
Vertex: $(-3,3)$
Because the $x$-coordinates of the vertex, center, and focus are the same, the ellipse has the equation:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Determine $h,k$ using the center:
$(h,k)=(-3,1)$
$h=-3$
$k=1$
Determine $a$ using the vertex:
$(h,k+a)=(-3,3)$
$(-3,1+a)=(-3,3)$
$1+a=3$
$a=2$
Determine $c$ using the focus:
$(h,k-c)=(-3,0)$
$(-3,1-c)=(-3,0)$
$1-c=0$
$c=1$
Determine $b$:
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b^2=2^2-1^2$
$b^2=3$
$b=\sqrt{3}$
The equation of the ellipse is:
$\dfrac{(x+3)^2}{(\sqrt 3)^2}+\dfrac{(y-1)^2}{2^2}=1$
$\dfrac{(x+3)^2}{3}+\dfrac{(y-1)^2}{4}=1$
The center is:
$(h,k)=(-3,1)$
Determine the vertices and co-vertices:
$(h,k-a)=(-3,1-2)=(-3,-1)$
$(h,k+a)=(-3,3)$
$h-b,k)=(-3-\sqrt{3},1)$
$h+b,k)=(-3+\sqrt{3},1)$
Graph the ellipse:
