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

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Aicharya · Accepted Answer

Equation of given ellipse is $\begin{array}{rlrl} & x^2+4 y^2 =64 \ \Rightarrow & \frac{x^2}{64}+\frac{y^2}{16} =1 \end{array}$ Let the vertex $P(8 \cos \theta, 4 \sin \theta)$ of the rectangle $P Q R S$ having greatest area. $\begin{aligned} \therefore \text { Area } & =A=4(8 \cos \theta)(4 \sin \theta) \ & =64 \sin 2 \theta \end{aligned}$ For greatest $\operatorname{area} \sin 2 \theta=1$, so $A=64$ sq. units and $\theta=\frac{\pi}{4}$. Therefore point $P(4 \sqrt{2}, 2 \sqrt{2})$ So, length 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+4…

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