Write the standard form of an equation of the ellipse subject to the given conditions.
-Foci: $( - 13,4 )$ and $( 19,4 ) ;$ Eccentricity: $\frac { 4 } { 5 }$
A) $\frac { ( x - 3 ) ^ { 2 } } { 400 } + \frac { ( y - 4 ) ^ { 2 } } { 144 } = 1$
B) $\frac { ( x + 3 ) ^ { 2 } } { 144 } + \frac { ( y + 4 ) ^ { 2 } } { 400 } = 1$
C) $\frac { ( x + 3 ) ^ { 2 } } { 400 } + \frac { ( y + 4 ) ^ { 2 } } { 144 } = 1$
D) $\frac { ( x - 3 ) ^ { 2 } } { 144 } + \frac { ( y - 4 ) ^ { 2 } } { 400 } = 1$

Question

Quizplus · Accepted Answer

The center of the ellipse is the midpoint of the foci, which is $\left(\frac{-13 + 19}{2}, \frac{4 + 4}{2}\right) = (3, 4)$. The distance between the foci is $32$, so $2c = 32$, giving $c = 16$. The eccentricity $e = \frac{4}{5}$ gives us $c = ae$, so $a = \frac{c}{e} = \frac{16}{\frac{4}{5}} = 20$. Thus, $a^2 = 400$. Since $c^2 = a^2 - b^2$, we find $b^2 = a^2 - c^2 = 400 - 256 = 144$. The major axis is horizontal, so the equation is $\frac{(x - 3)^2}{400} + \frac{(y - 4)^2}{144} = 1$.

Write the Standard Form of an Equation of the Ellipse (−13,4)( - 13,4 )(−13,4)

Write the Standard Form of an Equation of the Ellipse $( - 13,4 )$