If the distance between the foci of an ellips | vedclass

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(C) Given the distance between the foci is $2ae = 6$,so $ae = 3$ $(1)$.The distance between the directrices is $\frac{2a}{e} = 12$,so $a = 6e$ $(2)$.S

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$3\sqrt{2}$

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(C) Given the distance between the foci is $2ae = 6$,so $ae = 3$ $(1)$.The distance between the directrices is $\frac{2a}{e} = 12$,so $a = 6e$ $(2)$.Substituting $(2)$ into $(1)$,we get $6e^2 = 3$,which implies $e^2 = \frac{1}{2}$,so $e = \frac{1}{\sqrt{2}}$.Then $a = 6 	imes \frac{1}{\sqrt{2}} = 3\sqrt{2}$.Using the relation $b^2 = a^2(1 - e^2)$,we have $b^2 = (3\sqrt{2})^2(1 - \frac{1}{2}) = 18 	imes \frac{1}{2} = 9$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{2(9)}{3\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.

If the distance between the foci of an ellipse is $6$ and the distance between its directrices is $12$,then the length of its latus rectum is

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