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Consider an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ .
What is the area of the greatest rectangle that can be inscribed in the ellipse?
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What is the area of the greatest rectangle that can be inscribed in the ellipse?
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The correct answer is:
$2 a b$
Given equation of ellipse, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Let $\mathrm{A}(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)$ be any point on ellipse(1st quadrant)
Coordinate of $\mathrm{B}=[\mathrm{a} \cos (\pi-\theta), \mathrm{b} \sin (\pi-\theta)]$
$=(-a \cos \theta, b \sin \theta)$(2nd quadrant)
Coordinate of $C=[a \cos (\pi+\theta), b \sin (\pi+\theta)]$(3rd quadrant)
Coordinate of $\mathrm{D}=[\mathrm{a} \cos (2 \pi-\theta), \mathrm{b} \sin (2 \pi-\theta)]$
$=(a \cos \theta,-b \sin \theta) \quad$ (4th quadrant)
Area of the rectangle $\mathrm{ABCD}$ $=(a \cos \theta+a \cos \theta)(b \sin \theta+b \sin \theta)$
$=2 a \cos \theta \times 2 b \sin \theta=2 a b \sin 2 \theta$
$=2 a b \times 1=2 a b$
Let $\mathrm{A}(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)$ be any point on ellipse(1st quadrant)
Coordinate of $\mathrm{B}=[\mathrm{a} \cos (\pi-\theta), \mathrm{b} \sin (\pi-\theta)]$
$=(-a \cos \theta, b \sin \theta)$(2nd quadrant)
Coordinate of $C=[a \cos (\pi+\theta), b \sin (\pi+\theta)]$(3rd quadrant)
Coordinate of $\mathrm{D}=[\mathrm{a} \cos (2 \pi-\theta), \mathrm{b} \sin (2 \pi-\theta)]$
$=(a \cos \theta,-b \sin \theta) \quad$ (4th quadrant)
Area of the rectangle $\mathrm{ABCD}$ $=(a \cos \theta+a \cos \theta)(b \sin \theta+b \sin \theta)$
$=2 a \cos \theta \times 2 b \sin \theta=2 a b \sin 2 \theta$
$=2 a b \times 1=2 a b$
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