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: $\pm 2$
Because the foci have the same $x$-intercept, the parabola is vertical.
The equation is in the form:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Determine $h,k,b$ using the $x$-intercepts:
$(h,k-b)=(0,-2)\Rightarrow h=0, k-b=-2$
$(h,k+b)=(0,2)\Rightarrow k+b=2$
$h=0$
$\begin{cases}
k-b=-2\\
k+b=2
\end{cases}$
$k-b+k+b=-2+2=0$
$2k=0$
$k=0$
$k+b=2$
$0+b=2$
$b=2$
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 center is:
$(h,k)=(0,0)$
The equation of the ellipse is:
$\dfrac{x^2}{4}+\dfrac{y^2}{13}=1$
Plot the center, the vertices, the $x$-intercepts, and the foci. Then graph the ellipse:
