Answer
$\frac{x^2}{100} + \frac{y^2}{91} = 1$
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
2b = 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 major axis are (-10, 0) and (10, 0), which are equidistant from the origin (0, 0). This means that the ellipse has its center at (0, 0).
Note that:
1. Each endpoint of the major axis is 10 units away from the center so $a=10$.
2. The foci are 6 units apart $\longrightarrow 2c=6 \longrightarrow c=3$
3. The major axis is horizontal.
Solve for $b$ using the formula $a^2=b^2=c^2$ to have:
$\\a^2 = b^2+c^2
\\10^2=b^2+3^2
\\100=b^2+9
\\91=b^2
\\\sqrt{91}=b$
Therefore, the equation of the ellipse is:
$\\\frac{x^2}{a^2} + \frac{y^2}{b^2}=1
\\\frac{x^2}{10^2} + \frac{y^2}{(\sqrt{91})^2}=1
\\\frac{x^2}{100} + \frac{y^2}{91} = 1$