Let P be any point on the ellipse frac{x^2}{a | vedclass

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(D) The coordinates of the foci $S_1$ and $S_2$ are $(-ae, 0)$ and $(ae, 0)$ respectively.The distance between the foci is $S_1S_2 = 2ae$.Let $P$ be a

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$abe$

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(D) The coordinates of the foci $S_1$ and $S_2$ are $(-ae, 0)$ and $(ae, 0)$ respectively.The distance between the foci is $S_1S_2 = 2ae$.Let $P$ be a point on the ellipse given by $(a \cos 	heta, b \sin 	heta)$.The area of $\Delta PS_1S_2$ is given by $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Here,the base $S_1S_2 = 2ae$ and the height is the absolute value of the $y$-coordinate of $P$,which is $|b \sin 	heta|$.$	ext{Area} = \frac{1}{2} 	imes (2ae) 	imes |b \sin 	heta| = ae |b \sin 	heta| = abe |\sin 	heta|$.Since the maximum value of $|\sin 	heta|$ is $1$,the maximum area is $abe$ square units.

Let $P$ be any point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. If $S_1$ and $S_2$ are its foci,then the maximum area of $\Delta PS_1S_2$ is (in square units):

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