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 the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we

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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 the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we have $a^2=36$ and $b^2=20$,which gives $a=6$.The eccentricity $e$ is calculated as $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{20}{36}} = \sqrt{\frac{16}{36}} = \frac{4}{6} = \frac{2}{3}$.The equations of the directrices for an ellipse with $a > b$ are $x = \pm \frac{a}{e}$.Substituting the values,we get $x = \pm \frac{6}{2/3} = \pm 9$.The distance between the directrices is the difference between these two values: $|9 - (-9)| = 18$.

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

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