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(B) For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the coordinates of the vertices of an inscribed rectangle are $(\pm a \cos 	heta, \pm b \sin \

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$\frac{\sqrt{3}}{2}$

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(B) For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the coordinates of the vertices of an inscribed rectangle are $(\pm a \cos 	heta, \pm b \sin 	heta)$.Area of the rectangle $A = (2a \cos 	heta)(2b \sin 	heta) = 2ab \sin 2	heta$.For the area to be maximum,$\sin 2	heta = 1$,which implies $	heta = \frac{\pi}{4}$.Given length $L = 2a \cos 	heta = 2a \cos(\frac{\pi}{4}) = 2a(\frac{1}{\sqrt{2}}) = a\sqrt{2} = 8\sqrt{2}$,so $a = 8$.Given breadth $B = 2b \sin 	heta = 2b \sin(\frac{\pi}{4}) = 2b(\frac{1}{\sqrt{2}}) = b\sqrt{2} = 4\sqrt{2}$,so $b = 4$.Thus,$\frac{b}{a} = \frac{4}{8} = \frac{1}{2}$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

If the length and breadth of a rectangle of maximum area that can be inscribed in an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are $8 \sqrt{2}$ and $4 \sqrt{2}$ respectively,then the eccentricity of that ellipse is

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