The eccentricity of an ellipse whose length o | vedclass

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(A) The length of the latus rectum of an ellipse is $\frac{2b^2}{a}$ and the distance between its foci is $2ae$.Given that $\frac{2b^2}{a} = 2ae$,we h

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$2 \sin 18^{\circ}$

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(A) The length of the latus rectum of an ellipse is $\frac{2b^2}{a}$ and the distance between its foci is $2ae$.Given that $\frac{2b^2}{a} = 2ae$,we have $b^2 = a^2e$.Using the relation $b^2 = a^2(1 - e^2)$,we get $a^2(1 - e^2) = a^2e$.Dividing by $a^2$,we obtain $1 - e^2 = e$,which simplifies to $e^2 + e - 1 = 0$.Solving this quadratic equation for $e$ (where $e > 0$),we get $e = \frac{-1 + \sqrt{5}}{2} = \frac{\sqrt{5} - 1}{2}$.Since $\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$,it follows that $e = 2 \sin 18^{\circ}$.

The eccentricity of an ellipse whose length of latus rectum is equal to the distance between its foci is

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