If l and b are respectively the length and br | vedclass

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(C) The equation of the ellipse is $\frac{x^2}{64} + \frac{y^2}{16} = 1$. Let a vertex of the rectangle in the first quadrant be $(x, y) = (8 \cos 	h

Vedclass · Accepted Answer

$(8 \sqrt{2}, 4 \sqrt{2})$

Vedclass · Answer

(C) The equation of the ellipse is $\frac{x^2}{64} + \frac{y^2}{16} = 1$. Let a vertex of the rectangle in the first quadrant be $(x, y) = (8 \cos 	heta, 4 \sin 	heta)$.The length of the rectangle is $l = 2x = 16 \cos 	heta$ and the breadth is $b = 2y = 8 \sin 	heta$.The area $A = l 	imes b = (16 \cos 	heta)(8 \sin 	heta) = 128 \sin 	heta \cos 	heta = 64 \sin 2 	heta$.For the area to be maximum,$\sin 2 	heta$ must be $1$,so $2 	heta = \frac{\pi}{2}$,which gives $	heta = \frac{\pi}{4}$.Substituting $	heta = \frac{\pi}{4}$:$l = 16 \cos \frac{\pi}{4} = 16 	imes \frac{1}{\sqrt{2}} = 8 \sqrt{2}$.$b = 8 \sin \frac{\pi}{4} = 8 	imes \frac{1}{\sqrt{2}} = 4 \sqrt{2}$.Thus,$(l, b) = (8 \sqrt{2}, 4 \sqrt{2})$.

If $l$ and $b$ are respectively the length and breadth of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$,then $(l, b) =$

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