Answer
$\frac{x^2}{32}+\frac{y^2}{36}=1$
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).
Each focus is 2 units away from the center therefore $c=2$.
RECALL:
The Eccentricity (e) of an ellipse is given by the formula:
$\\e=\frac{c}{a}$
Substitute the known values of $e$ and $c$ to have:
$\\e= \frac{c}{a}
\\\frac{1}{3} = \frac{2}{a}
\\a(1) = 3(2)
\\a=6$
Solve for $b$ using the formula $a^2=b^2+c^2$ to have:
$a^2=b^2+c^2
\\6^2=b^2+2^2
\\36=b^2+4
\\36-4=b^2
\\32=b^2
\\\sqrt{32}=b
\\\sqrt{16(2)}=b^2
\\4\sqrt2=b$
When connected together using a line, the foci form a vertical line. Thus, the major axis is also vertical.
Therefore, the equation of the ellipse is:
$\\\frac{x^2}{b^2} + \frac{y^2}{a^2}=1
\\\frac{x^2}{(4\sqrt2)^2}+\frac{y^2}{6^2}=1
\\\frac{x^2}{32}+\frac{y^2}{36}=1$