Find the directrices of the ellipse.
-$16 x ^ { 2 } + 81 y ^ { 2 } = 1296$
A) $x = \pm \frac { 81 } { \sqrt { 97 } }$
B) $y = \pm \frac { 9 } { \sqrt { 65 } }$
C) $y = \pm \frac { 81 } { \sqrt { 65 } }$
D) $x = \pm \frac { 81 } { \sqrt { 65 } }$

Question

Quizplus · Accepted Answer

The standard form of the equation of an ellipse is $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$, where $(h,k)$ is the center and $2a$ and $2b$ are the lengths of the major and minor axes, respectively. The directrices are given by $x=\pm\frac{a^2}{c}$, where $c$ is the distance from the center to each focus. 
Rewrite the given equation as $\frac{x^2}{(\frac{1296}{16})}+\frac{y^2}{(\frac{1296}{81})}=1$, so the center is at $(0,0)$ and $2a=2\sqrt{\frac{1296}{16}}=72$ and $2b=2\sqrt{\frac{1296}{81}}=18$. The distance from the center to each focus is $c=\sqrt{a^2-b^2}=54$, so the equation of the directrices is $x=\pm\frac{a^2}{c}=\pm\frac{72^2}{54}=\pm\frac{81}{\sqrt{65}}$. Therefore, the answer is choice D.

Find the Directrices of the Ellipse 16x2+81y2=129616 x ^ { 2 } + 81 y ^ { 2 } = 129616x2+81y2=1296

Find the Directrices of the Ellipse $16 x ^ { 2 } + 81 y ^ { 2 } = 1296$