Find the standard-form equation of the ellipse centered at the origin and satisfying the given conditions.
-An ellipse with vertices $( \pm 6,0 )$ and foci at $( \pm 5,0 )$
A) $\frac { x ^ { 2 } } { 11 } + \frac { y ^ { 2 } } { 36 } = 1$
B) $\frac { x ^ { 2 } } { 11 } + \frac { y ^ { 2 } } { 61 } = 1$
C) $\frac { x ^ { 2 } } { 11 } + \frac { y ^ { 2 } } { 11 } = 1$
D) $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 11 } = 1$

Question

Quizplus · Accepted Answer

Since the ellipse is centered at the origin, the equation will be in the form:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
where $a$ is the distance from the center to the vertices, and $b$ is the distance from the center to the co-vertices.

We're given that the vertices are $(\pm 6, 0)$, so $a=6$. We're also given that the foci are at $(\pm 5,0)$, so $c=5$. Recall that the relationship between $a$, $b$, and $c$ for an ellipse is given by $c^2=a^2-b^2$.

Substituting the values we know, we have:
$$5^2=6^2-b^2 \Rightarrow b^2=6^2-5^2=11$$
Thus, the equation of the ellipse is:
$$\frac{x^2}{6^2}+\frac{y^2}{\sqrt{11}^2}=1 \Rightarrow \frac{x^2}{36}+\frac{y^2}{11}=1$$

The correct answer choice is D.

Find the Standard-Form Equation of the Ellipse Centered at the Origin