Answer
$\frac{x^2}{25} + \frac{y^2}{9} = 1$
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
$a \gt b$
a = distance from the center to an endpoint of the major axis
b = distance from the center to an endpoint of the minor axis
2c = distance between the foci
$a^2=b^2+c^2$
The endpoints of the minor axis are (0, -3) and (0, 3). These points are equidistant from (0, 0) so the center of the ellipse is at (0, 0).
Note that:
1. Each endpoint of the minor axis is 3 units way from the center so $b=3$.
2. The distance between foci is 8 units so $2c=8 \longrightarrow c=4$.
3. The minor axis is vertical therefore the major axis is horizontal.
Solve for $a$ using the formula $a^2=b^2=c^2$ to have:
$\\a^2 = b^2+c^2
\\a^2=3^2+4^2
\\a^2=9+16
\\a^2=25
\\a=\sqrt{25}
\\a=5$
Therefore, the equation of the ellipse is:
$\\\frac{x^2}{a^2} + \frac{y^2}{b^2}=1
\\\frac{x^2}{5^2} + \frac{y^2}{3^2}=1
\\\frac{x^2}{25} + \frac{y^2}{9} = 1$
