Find the standard form of the equation of the ellipse with the given characteristics and center at the origin.
Vertices: $( 0 , \pm 8 )$ ; Foci: $( 0 , \pm 4 )$
A) $\frac { x ^ { 2 } } { 48 } + \frac { y ^ { 2 } } { 64 } = 1$
B) $\frac { x ^ { 2 } } { 8 } + \frac { y ^ { 2 } } { 4 } = 1$
C) $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 16 } = 1$
D) $\frac { x ^ { 2 } } { 64 } - \frac { y ^ { 2 } } { 48 } = 1$
E) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 48 } = 1$

Question

Find the standard form of the equation of the ellipse with the given characteristics and center at the origin. 
Vertices: $( 0 , \pm 8 )$ ; Foci: $( 0 , \pm 4 )$ 
A) $\frac { x ^ { 2 } } { 48 } + \frac { y ^ { 2 } } { 64 } = 1$
B) $\frac { x ^ { 2 } } { 8 } + \frac { y ^ { 2 } } { 4 } = 1$
C) $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 16 } = 1$
D) $\frac { x ^ { 2 } } { 64 } - \frac { y ^ { 2 } } { 48 } = 1$
E) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 48 } = 1$

Quizplus · Accepted Answer

The major axis is vertical, and the distance between the center and a vertex is 8, so $a = 8$. The distance between the center and a focus is 4, so $c = 4$. Therefore, $b^2 = a^2 - c^2 = 8^2 - 4^2 = 48$, so $b = \sqrt{48} = 4\sqrt{3}$. The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, so the equation of this ellipse is $\frac{x^2}{64} + \frac{y^2}{48} = 1$.

Find the Standard Form of the Equation of the Ellipse (0,±8)( 0 , \pm 8 )(0,±8)

Find the Standard Form of the Equation of the Ellipse $( 0 , \pm 8 )$