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