The distance between the directrices of the ellipse $\frac{x^2}{36}+\frac{y^2}{20}=1$ is
The distance between the directrices of the ellipse $\frac{x^2}{36}+\frac{y^2}{20}=1$ is
$9$
$6 \sqrt{5}$
$8$
$3 \sqrt{5}$
Solution
for $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \Rightarrow \frac{x^2}{36}+\frac{y^2}{20}=1$
eccentricity $(e)=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{20}{36}}$
$=\frac{2}{3}$
Equation of directrix is $x= \pm a e$
$\begin{aligned} & x=6 \frac{2}{3} \text { and } x=-6 \frac{2}{3} \\ & x=4 \text { and } x=-4\end{aligned}$
distance between $x=4$ and $x=-4=8$