Find an equation in standard form for the ellipse that satisfies the given conditions.
-Vertices at $( \pm 12,0 )$ and foci at $( \pm 6,0 )$
A) $\frac { x ^ { 2 } } { 11,664 } + \frac { y ^ { 2 } } { 12 } = 1$
B) $\frac { x ^ { 2 } } { 144 } + \frac { y } { 108 } = 1$
C) $\frac { x ^ { 2 } } { 144 } + \frac { y ^ { 2 } } { 108 } = 1$
D) $\frac { x } { 12 } + \frac { y ^ { 2 } } { 108 } = 2$

Question

Quizplus · Accepted Answer

The standard form of an ellipse with a horizontal major axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $2a$ is the length of the major axis and $2b$ is the length of the minor axis. The distance between the center and a focus is $c$, and $a^2 = b^2 + c^2$. Given vertices at $(\pm 12, 0)$, we have $a = 12$, so $a^2 = 144$. The foci are at $(\pm 6, 0)$, so $c = 6$, and $c^2 = 36$. Solving for $b^2$, we get $b^2 = a^2 - c^2 = 144 - 36 = 108$. Therefore, the correct equation is $\frac{x^2}{144} + \frac{y^2}{108} = 1$.

Find an Equation in Standard Form for the Ellipse That (±12,0)( \pm 12,0 )(±12,0)

Find an Equation in Standard Form for the Ellipse That $( \pm 12,0 )$