Answer
$\dfrac{(x-4)^2}{5}+\dfrac{(y-6)^2}{9}=1$
See graph
Work Step by Step
We are given the ellipse:
Focus: $(4,8)$
Vertices: $(4,3),(4,9)$
Because the $x$-coordinates of the vertex and foci are the same, the ellipse has the equation:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Determine $h,k,a$ using the vertices:
$(h,k-a)=(4,3)\Rightarrow h=4,k-a=3$
$(h,k+a)=(4,9)\Rightarrow k+a=9$
$h=4$
$\begin{cases}
k-a=3\\
k+a=9
\end{cases}$
$k-a+k+a=3+9$
$2k=12$
$k=6$
$k+a=9$
$6+a=9$
$a=3$
Determine $c$ using the focus:
$(h,k+c)=(4,8)$
$(4,6+c)=(4,8)$
$6+c=8$
$c=2$
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:
$\dfrac{(x-4)^2}{(\sqrt 5)^2}+\dfrac{(y-6)^2}{(3^2}=1$
$\dfrac{(x-4)^2}{5}+\dfrac{(y-6)^2}{9}=1$
The center is:
$(h,k)=(4,6)$
Determine the co-vertices:
$h-b,k)=(4-\sqrt{5},6)$
$h+b,k)=(4+\sqrt{5},6)$
Graph the ellipse:
