Write an equation for the ellipse.
-$e = \frac { 9 } { 41 } ;$ foci at $( 9,0 ) , ( - 9,0 )$
A) $\frac { x ^ { 2 } } { 1600 } + \frac { y ^ { 2 } } { 1681 } = 1$

B) $\frac { x ^ { 2 } } { 1681 } + \frac { y ^ { 2 } } { 1600 } = 1$

C) $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 1600 } = 1$

D) $\frac { x ^ { 2 } } { 1600 } + \frac { y ^ { 2 } } { 81 } = 1$

Question

Write an equation for the ellipse.
-$e = \frac { 9 } { 41 } ;$ foci at $( 9,0 ) , ( - 9,0 )$
A) $\frac { x ^ { 2 } } { 1600 } + \frac { y ^ { 2 } } { 1681 } = 1$

B) $\frac { x ^ { 2 } } { 1681 } + \frac { y ^ { 2 } } { 1600 } = 1$

C) $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 1600 } = 1$

D) $\frac { x ^ { 2 } } { 1600 } + \frac { y ^ { 2 } } { 81 } = 1$

Quizplus · Accepted Answer

The eccentricity $e$ of an ellipse is given by $e = \frac{c}{a}$, where $c$ is the distance from the center to a focus, and $a$ is the distance from the center to a vertex. Given $e = \frac{9}{41}$ and the foci at $(9,0)$ and $(-9,0)$, we have $c = 9$. Thus, $a = \frac{c}{e} = \frac{9}{\frac{9}{41}} = 41$, so $a^2 = 1681$. The major axis is along the x-axis because the foci are on the x-axis, making $a^2 = 1681$ the denominator under $x^2$. The minor axis length $b$ can be found using $b^2 = a^2 - c^2 = 1681 - 81 = 1600$, making $b^2 = 1600$ the denominator under $y^2$. Therefore, the correct equation is $\frac{x^2}{1681} + \frac{y^2}{1600} = 1$.

Write an Equation for the Ellipse e=941;e = \frac { 9 } { 41 } ;e=419​;

Write an Equation for the Ellipse $e = \frac { 9 } { 41 } ;$