Answer
$\dfrac{x^2}{12}+\dfrac{y^2}{16}=1$
See graph
Work Step by Step
We are given the ellipse:
Foci: $(0,-2),(0,2)$
Length of the major axis: $8$
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 $a$ using the length of the major axis:
$2a=8$
$a=4$
Determine $h,k,c$ from the foci:
$(h,k-c)=(0,-2)\Rightarrow h=0,k-c=-2$
$(h,k+c)=(0,2)\Rightarrow k+c=2$
$\begin{cases}
k-c=-2\\
k+c=2
\end{cases}$
$k-c+k+c=-2+2$
$2k=0$
$k=0$
$k+c=2$
$0+c=2$
$c=2$
Determine $b$:
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b^2=4^2-2^2$
$b^2=12$
$b=\sqrt{12}=2\sqrt 3$
The equation of the ellipse is:
$\dfrac{x^2}{(2\sqrt 3)^2}+\dfrac{y^2}{4^2}=1$
$\dfrac{x^2}{12}+\dfrac{y^2}{16}=1$
The center is:
$(h,k)=(0,0)$
Graph the ellipse:
