The distance between the directrices of the e | vedclass

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(C) Given the equation of the ellipse: $\frac{x^2}{36} + \frac{y^2}{20} = 1$.Comparing this with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we get $a^2 =

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$18$

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(C) Given the equation of the ellipse: $\frac{x^2}{36} + \frac{y^2}{20} = 1$.Comparing this with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we get $a^2 = 36$ and $b^2 = 20$,so $a = 6$ and $b = 2\sqrt{5}$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{20}{36}} = \sqrt{\frac{16}{36}} = \frac{4}{6} = \frac{2}{3}$.The equations of the directrices for an ellipse with $a > b$ are $x = \pm \frac{a}{e}$.The distance between the directrices is $2 	imes \frac{a}{e} = 2 	imes \frac{6}{2/3} = 2 	imes 9 = 18$.

The distance between the directrices of the ellipse $\frac{x^2}{36} + \frac{y^2}{20} = 1$ is

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