Answer
$\dfrac{x^2}{5}+\dfrac{y^2}{9}=1$
Work Step by Step
The standard form of an ellipse that has foci $(0, \pm c)$ is
$$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$
where $\quad \quad a \gt b, \quad c^2=a^2-b^2, \quad\text{and}\quad \text{Center at }(0, 0)$
The given ellipse has its foci at $(0, -2)$ and $(0, 2)$ so its standard equation is in the same form as above with center at $(0, 0)$ and $c=2$..
The major axis is $6$ units long, which means that $2a=6$.
Hence,
$$2a=6
\\a=3\\
\implies a^2=9$$
Solve for $b^2$:
$$c^2=a^2-b^2 \\ 2^2 =(3)^2-b^2 \\
b^2=9-4\\
b^2=5$$
So, the standard form of an ellipse becomes:
$$\dfrac{x^2}{5}+\dfrac{y^2}{9}=1$$
