Find an equation of the ellipse that satisfies the given conditions. Foci: (0, ± 1), vertices (0, ± 6)
A) $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 35 } = 1$
B) $\frac { x ^ { 2 } } { 36 } - \frac { y ^ { 2 } } { 35 } = 1$
C) $\frac { x ^ { 2 } } { 35 } + \frac { y ^ { 2 } } { 36 } = 1$
D) $\frac { x ^ { 2 } } { 35 } - \frac { y ^ { 2 } } { 36 } = 1$

Question

Quizplus · Accepted Answer

Since the foci are along the y-axis, the major axis of the ellipse is vertical. The distance between the foci is $\sqrt{(0-0)^2+(1-(-1))^2}=2$, which is also the length of the major axis. Thus $2a=12$ and $a=6$. The coordinates of the vertices indicate that $b=5$. Therefore, the equation of the ellipse is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, which simplifies to $\frac{x^2}{35}+\frac{y^2}{36}=1$.

Find an Equation of the Ellipse That Satisfies the Given x236+y235=1\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 35 } = 136x2​+35y2​=1

Find an Equation of the Ellipse That Satisfies the Given $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 35 } = 1$