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(B) The distance between the foci of an ellipse is $2ae$.The length of the latus rectum of an ellipse is $\frac{2b^2}{a}$.According to the problem,the

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$\sqrt{2}-1$

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(B) The distance between the foci of an ellipse is $2ae$.The length of the latus rectum of an ellipse is $\frac{2b^2}{a}$.According to the problem,the distance between the foci is half the length of the latus rectum:$2ae = \frac{1}{2} 	imes \frac{2b^2}{a}$$2ae = \frac{b^2}{a}$Using the relation $b^2 = a^2(1-e^2)$,we get:$2ae = \frac{a^2(1-e^2)}{a}$$2ae = a(1-e^2)$$2e = 1-e^2$$e^2 + 2e - 1 = 0$Using the quadratic formula $e = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:$e = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$Since the eccentricity $e$ must be positive,$e = \sqrt{2}-1$.

If the distance between the foci of an ellipse is half the length of its latus rectum,then the eccentricity of the ellipse is

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