Find the standard-form equation for an ellipse which satisfies the given conditions.
-An ellipse centered at the origin having focus $( \sqrt { 77 } , 0 )$ and directrix $x = \frac { 81 } { \sqrt { 77 } }$
A) $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 4 } = 1$
B) $\frac { x ^ { 2 } } { 81 } + y ^ { 2 } = 1$
C) $\frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 81 } = 1$
D) $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 2 } = 1$

Question

Quizplus · Accepted Answer

The standard form of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $2a$ is the length of the major axis and $2b$ is the length of the minor axis. The distance between the center and a focus is $c$, where $c^2 = a^2 - b^2$. Given the focus $(\sqrt{77}, 0)$, we have $c = \sqrt{77}$. The directrix is given as $x = \frac{81}{\sqrt{77}}$, and the eccentricity $e = \frac{c}{a}$. Since the distance from the center to the directrix is the reciprocal of the eccentricity times the semi-major axis, $a/e = \frac{81}{\sqrt{77}}$, we can solve for $a$ using $e = c/a$. Substituting $c = \sqrt{77}$ gives $a = 9$, so $a^2 = 81$. Using $c^2 = a^2 - b^2$, we find $b^2 = a^2 - c^2 = 81 - 77 = 4$. Therefore, the equation of the ellipse is $\frac{x^2}{81} + \frac{y^2}{4} = 1$.

Find the Standard-Form Equation for an Ellipse Which Satisfies the Given