Write an equation for the ellipse.
-eccentricity $\frac { 5 } { 13 }$; foci at $( 5,0 ) , ( - 5,0 )$
A) $\frac { x ^ { 2 } } { 169 } + \frac { y ^ { 2 } } { 144 } = 1$
B) $\frac { x ^ { 2 } } { 144 } + \frac { y ^ { 2 } } { 25 } = 1$
C) $\frac { x ^ { 2 } } { 144 } + \frac { y ^ { 2 } } { 169 } = 1$
D) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 144 } = 1$

Question

Quizplus · Accepted Answer

The eccentricity 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 semi-major axis length. Given $e = \frac{5}{13}$ and the foci at $(5,0)$ and $(-5,0)$, we have $c = 5$. Thus, $a = \frac{c}{e} = \frac{5}{\frac{5}{13}} = 13$, so $a^2 = 169$. The distance between the foci is $2c = 10$, and since $c^2 = a^2 - b^2$, we find $b^2 = a^2 - c^2 = 169 - 25 = 144$. Therefore, the equation of the ellipse is $\frac{x^2}{169} + \frac{y^2}{144} = 1$.

Write an Equation for the Ellipse 513\frac { 5 } { 13 }135​

Write an Equation for the Ellipse $\frac { 5 } { 13 }$