Find an equation for the ellipse described.
-Focus at (0, -6); vertices at (0, -7)and (0, 7)
A) $\frac { x ^ { 2 } } { 13 } + \frac { y ^ { 2 } } { 49 } = 1$
B) $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 13 } = 1$
C) $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 49 } = 1$
D) $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 13 } = 1$

Question

Find an equation for the ellipse described.
-Focus at (0, -6); vertices at (0, -7)and (0, 7) 
A) $\frac { x ^ { 2 } } { 13 } + \frac { y ^ { 2 } } { 49 } = 1$
B) $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 13 } = 1$
C) $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 49 } = 1$
D) $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 13 } = 1$

Quizplus · Accepted Answer

The vertices at (0, -7) and (0, 7) indicate that the major axis is vertical and the length of the major axis is 14, so $a = 7$ and $a^2 = 49$. The focus at (0, -6) indicates that $c = 6$, and since $c^2 + b^2 = a^2$, we find $b^2 = a^2 - c^2 = 49 - 36 = 13$. Therefore, the equation of the ellipse is $\frac{x^2}{13} + \frac{y^2}{49} = 1$.

Find an Equation for the Ellipse Described x213+y249=1\frac { x ^ { 2 } } { 13 } + \frac { y ^ { 2 } } { 49 } = 113x2​+49y2​=1

Find an Equation for the Ellipse Described $\frac { x ^ { 2 } } { 13 } + \frac { y ^ { 2 } } { 49 } = 1$