Find the standard form of the equation of the ellipse with the given characteristics and center at the origin.
A) $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1$
B) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 4 } = 1$
C) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 16 } = 1$
D) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$
E) $\frac { x } { 4 } + \frac { y } { 16 } = 1$

Question

Find the standard form of the equation of the ellipse with the given characteristics and center at the origin. 
A) $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1$ 
B) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 4 } = 1$ 
C) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 16 } = 1$ 
D) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 16 } = 1$ 
E) $\frac { x } { 4 } + \frac { y } { 16 } = 1$

Quizplus · Accepted Answer

Since the center is at the origin, we have $h=k=0$. Also, since the major axis is horizontal, we have $a>\!b$. Thus, $a^{2}=16$ and $b^{2}=4$. Therefore, the standard form of the equation of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, which simplifies to $\boxed{\frac{x^{2}}{4}+\frac{y^{2}}{16}=1}$. Hence, the answer is $\boxed{\mathrm{(D)}}$.

Find the Standard Form of the Equation of the Ellipse x24−y216=1\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 14x2​−16y2​=1

Find the Standard Form of the Equation of the Ellipse $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1$