If the end points of the latus rectum of an e | vedclass

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(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The coordinates of the end points of the latus rectum are $(ae, \frac{b^2

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$\frac{\sqrt{5} - 1}{2}$

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(D) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The coordinates of the end points of the latus rectum are $(ae, \frac{b^2}{a})$ and $(ae, -\frac{b^2}{a})$.Since these points are vertices of a square,the distance between them must be equal to the distance between the two end points of the latus rectum on the same side of the major axis,which is $2ae$.Thus,$2 	imes \frac{b^2}{a} = 2ae$.$\frac{b^2}{a} = ae$.Using $b^2 = a^2(1 - e^2)$,we get $\frac{a^2(1 - e^2)}{a} = ae$.$a(1 - e^2) = ae \Rightarrow 1 - e^2 = e$.$e^2 + e - 1 = 0$.Solving for $e$ using the quadratic formula,$e = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.Since $e > 0$,we have $e = \frac{\sqrt{5} - 1}{2}$.

If the end points of the latus rectum of an ellipse are the vertices of a square,then the eccentricity of the ellipse will be -

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