Answer
$\dfrac{x^2}{48}+\dfrac{y^2}{64}=1$
See graph
Work Step by Step
We are given the ellipse:
Focus: $(0,-4)$
Vertices: $(0,-8),(0,8)$
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 $h,k,a$ using the vertices:
$(h,k-a)=(0,-8)\Rightarrow h=0, k-a=-8$
$(h,k+a)=(0,8)\Rightarrow k+a=8$
$\begin{cases}
k-a=-8\\
k+a=8
\end{cases}$
$k-a+k+a=-8+8$
$2k=0$
$k=0$
$k+a=8$
$0+a=8$
$a=8$
Determine $c$ using the focus:
$(h,k-c)=(0,-4)$
$(0,0-c)=(0,-4)$
$(0,-c)=(0,-4)$
$c=4$
Determine $b$:
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b^2=8^2-4^2$
$b^2=48$
$b=\sqrt{48}=4\sqrt 3$
The equation of the ellipse is:
$\dfrac{x^2}{(4\sqrt 3)^2}+\dfrac{y^2}{8^2}=1$
$\dfrac{x^2}{48}+\dfrac{y^2}{64}=1$
The center is:
$(h,k)=(0,0)$
Graph the ellipse:
