Find the standard-form equation of the ellipse centered at the origin and satisfying the given conditions.
-An ellipse with intercepts $( \pm 4,0 )$ and $( 0 , \pm 2 )$
A) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 2 } = 1$
B) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$
C) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1$
D) $\frac { x ^ { 2 } } { 2 } + \frac { y ^ { 2 } } { 4 } = 1$

Question

Quizplus · Accepted Answer

Since the ellipse is centered at the origin, the standard-form equation of the ellipse is:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
where $a$ and $b$ are the lengths of the semi-axes.
The intercepts of the ellipse are $(\pm 4, 0)$ and $(0, \pm 2)$. This means that the semi-axes are $a=4$ and $b=2$. 
Substituting these values into the equation gives:
$$\frac{x^2}{4^2}+\frac{y^2}{2^2}=1$$
Simplifying:
$$\frac{x^2}{16}+\frac{y^2}{4}=1$$
Therefore, the standard-form equation of the ellipse is $\boxed{	extbf{(C)}\ \frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1}$.

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