Find the distance between the directrix of the ellipse x^2 +36 + y^2 +20=1 

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$$\begin{gathered} Considering that we are given with, \\ \frac{x^2}{36}+\frac{y^2}{20}=1 \\ Now, Comparing it with the s\tan dard equation of ellipse, \\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \\ we get, \\ {a}^2=36 \\ a=\pm 6 \\ and, \\ {b}^2=20 \\ b=\pm 2\sqrt{5} \\ now, \\ Eccetricity \text{'}e\text{'}=\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{36-20}}{6}=\frac{\sqrt{16}}{6} \\ e=\frac{4}{6}=\frac{2}{3} \\ Now, \\ the dis\tan ce between the directrices is given by= \frac{a}{e} \\ =\frac{6}{{\displaystyle \frac{2}{3}}}=\frac{6\times 3}{2}=9 \end{gathered}$$

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