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# Find the distance between the directrices of the ellipse $\large\frac{x^2}{36}+\frac{y^2}{20}$$=1. \begin{array}{1 1}-9\\-\large\frac{1}{9}\\9\\\large\frac{1}{9}\end{array} ## 1 Answer Comment A) Toolbox: • Distance between the directrices is \large\frac{a}{e} • Eccentricity =\large\frac{\sqrt{a^2-b^2}}{a} Answer : 9 Given equation of the ellipse is \large\frac{x^2}{36}+\frac{y^2}{20}$$=1$
$\therefore a^2=36$
$\Rightarrow a=\pm 6$
$b^2=20$
$\Rightarrow b=\pm 2\sqrt 5$
$\therefore e=\large\frac{\sqrt{a^2-b^2}}{a}$
$e=\large\frac{\sqrt{36-20}}{6}$
$\;\;\;=\large\frac{4}{6}$
$\;\;\;=\large\frac{2}{3}$
Hence the distance between the directrices is $\large\frac{a}{e}$
$\Rightarrow \large\frac{6}{\Large\frac{2}{3}}$$=9$