Answer
$\dfrac{x^2}{4}+\dfrac{y^2}{13}=1$
See graph
Work Step by Step
We are given the ellipse:
Foci: $(0,-3),(0,3)$
$x$-intercepts: $-2,2$
Because the $x$-coordinates of the foci are the same, the ellipse has the equation:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Determine $h,k,b$ using the $x$-intercepts:
$(h-b,k)=(-2,0)\Rightarrow h-b=-2,k=0$
$(h+b,k)=(2,0)\Rightarrow h+b=2$
$k=0$
$\begin{cases}
h-b=-2\\
h+b=2
\end{cases}$
$h-b+h+b=-2+2$
$2h=0$
$h=0$
$h+b=2$
$0+b=2$
$b=2$
Determine $c$ using the foci:
$(h,k-c)=(0,-3)$
$(0,0-c)=(0,-3)$
$(0,-c)=(0,-3)$
$c=3$
Determine $a$:
$a^2=b^2+c^2$
$a^2=2^2+3^2$
$a^2=13$
$a=\sqrt{13}$
The equation of the ellipse is:
$\dfrac{x^2}{4}+\dfrac{y^2}{13}=1$
The center is:
$(h,k)=(0,0)$
Graph the ellipse:
