Answer
$\frac{x^2}{9} + \frac{y^2}{13} = 1$
Work Step by Step
RECALL:
The standard equation of an ellipse with whose center is at (0, 0) is:
(i) $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ (horizontal major axis)
(ii) $\frac{x^2}{b^2} + \frac{y^2}{a^2}=1$ (vertical major axis)
where
$a \gt b$
2a = length of major axis
2b = length of minor axis
c = distance of each focus from the center
$a^2=b^2+c^2$
The foci are (0, -2) and (0, 2), which are equidistant from the origin (0, 0). This means that the ellipse has its center at (0, 0).
The foci, when connect together, form a vertical segment. This means that the major axis is also vertical.
The length of minor axis is 6 $\longrightarrow 2b=6 \longrightarrow b=3$
Each focus is 2 units away from the center so $c=2$.
Solve for $b$ using the formula $a^2=b^2=c^2$ to have:
$\\a^2 = b^2+c^2
\\a^2=3^2+2^2
\\a^2=9+4
\\a^2=13
\\a=\sqrt{13}$
Therefore, the equation of the ellipse is:
$\\\frac{x^2}{b^2} + \frac{y^2}{a^2}=1
\\\frac{x^2}{3^2} + \frac{y^2}{(\sqrt{13})^2}=1
\\\frac{x^2}{9} + \frac{y^2}{13} = 1$
