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(B) The equation of the given ellipse is $x^2+4y^2=64$,which can be written as $\frac{x^2}{64}+\frac{y^2}{16}=1$. Let the vertex $P$ of the rectangle

Vedclass · Accepted Answer

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

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(B) The equation of the given ellipse is $x^2+4y^2=64$,which can be written as $\frac{x^2}{64}+\frac{y^2}{16}=1$. Let the vertex $P$ of the rectangle in the first quadrant be $(8\cos	heta, 4\sin	heta)$. The coordinates of the four vertices of the rectangle are $(8\cos	heta, 4\sin	heta)$,$(-8\cos	heta, 4\sin	heta)$,$(-8\cos	heta, -4\sin	heta)$,and $(8\cos	heta, -4\sin	heta)$. The length of the sides of the rectangle are $L = 2(8\cos	heta) = 16\cos	heta$ and $W = 2(4\sin	heta) = 8\sin	heta$. The area $A$ of the rectangle is $A = L 	imes W = (16\cos	heta)(8\sin	heta) = 128\sin	heta\cos	heta = 64\sin(2	heta)$. For the area to be maximum,$\sin(2	heta)$ must be $1$,which implies $2	heta = \frac{\pi}{2}$ or $	heta = \frac{\pi}{4}$. Substituting $	heta = \frac{\pi}{4}$ into the expressions for the sides: $L = 16\cos(\frac{\pi}{4}) = 16(\frac{1}{\sqrt{2}}) = 8\sqrt{2}$. $W = 8\sin(\frac{\pi}{4}) = 8(\frac{1}{\sqrt{2}}) = 4\sqrt{2}$. Thus,the lengths of the sides are $8\sqrt{2}$ and $4\sqrt{2}$.

The lengths of the sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are

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