Question

If the distance between the directrices of an ellipse be three times the distance between its foci, then the eccentricity is ___________.

Answer 1

for an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where a > b
Distance between directrix is $\frac{2a}{e}$ where e is eccentricity and Distance between foci is 2ae
According to given condition,
distance between directrix = 3(distance between foci)
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{2a}{e}=3\times 2ae \\ \mathrm{i}.\mathrm{e}{e}^2=\frac{1}{3} \\ \mathrm{i}.\mathrm{e}e=\frac{1}{\sqrt{3}} \end{gathered}$$