A rectangle of maximum area is inscribed in an ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, then its dimensions are

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Aicharya · Accepted Answer

Length of rectangle $=10 \cos 	heta$ and breadth of rectangle $=8 \sin 	heta$ $	herefore$ Area of rectangle $=(10 \cos 	heta)(8 \sin 	heta)=40(\sin 	heta)$ Maximum area will occur when $\sin 2 	heta=1$ $\begin{aligned} & 	herefore \sin 2 	heta=\sin \frac{\pi}{2} \quad \Rightarrow 	heta=\frac{\pi}{4} \ & 	herefore P=\left(\frac{5}{\sqrt{2}}, \frac{4}{\sqrt{2}}ight) \Rightarrow 	ext { Dimensions of rectangle are } 5 \sqrt{2}, 4 \sqrt{2} \end{aligned}$

A rectangle of maximum area is inscribed in an ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, then its…

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