Answer
$\dfrac{(x+1)^2}{9}+\dfrac{(y-2)^2}{5}=1$
See graph
Work Step by Step
We are given the ellipse:
Foci: $(1,2),(-3,2)$
Vertex: $(-4,2)$
Because the $y$-coordinates of the vertex and foci are the same, the ellipse has the equation:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Determine $h,k,c$ using the foci:
$(h-c,k)=(-3,2)\Rightarrow h-c=-3,k=2$
$(h+c,k)=(1,2)\Rightarrow h+c=1$
$k=2$
$\begin{cases}
h-c=-3\\
h+c=1
\end{cases}$
$h-c+h+c=-3+1$
$2h=-2$
$h=-1$
$h+c=1$
$-1+c=1$
$c=2$
Determine $k, a$ using the vertex:
$(h-a,k)=(-4,2)$
$(-1-a,k)=(-4,2)$
$-1-a=-4\Rightarrow a=3$
$k=2$
Determine $b$:
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b^2=3^2-2^2$
$b^2=5$
$b=\sqrt{5}$
The equation of the ellipse is:
$\dfrac{(x+1)^2}{3^2}+\dfrac{(y-2)^2}{(\sqrt 5)^2}=1$
$\dfrac{(x+1)^2}{9}+\dfrac{(y-2)^2}{5}=1$
The center is:
$(h,k)=(-1,2)$
Determine the vertices and co-vertices:
$(h-a,k)=(-4,2)$
$(h+a,k)=(-1+3,2)=(2,2)$
$h,k-b)=(-1,2-\sqrt{5})$
$h,k+b)=(-1,2+\sqrt{5})$
Graph the ellipse:
