Let S and S' be the foci of the ellipse frac{ | vedclass

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(A) The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{16} = 1$,where $a^2 = 25$ and $b^2 = 16$.The eccentricity $e$ is given by $e = \sqrt{1

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$12$

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(A) The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{16} = 1$,where $a^2 = 25$ and $b^2 = 16$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.The foci $S$ and $S'$ are $(\pm ae, 0) = (\pm 5 	imes \frac{3}{5}, 0) = (\pm 3, 0)$.Let $P = (5 \cos 	heta, 4 \sin 	heta)$ be a point on the ellipse.The area of $	riangle PSS'$ with vertices $(5 \cos 	heta, 4 \sin 	heta)$,$(-3, 0)$,and $(3, 0)$ is:$	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$= \frac{1}{2} |5 \cos 	heta (0 - 0) + (-3)(0 - 4 \sin 	heta) + 3(4 \sin 	heta - 0)|$$= \frac{1}{2} |0 + 12 \sin 	heta + 12 \sin 	heta| = \frac{1}{2} |24 \sin 	heta| = 12 |\sin 	heta|$.The maximum value of $|\sin 	heta|$ is $1$.Therefore,the maximum area is $12 	imes 1 = 12$ square units.

Let $S$ and $S'$ be the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ and $P$ be a variable point on the ellipse. The maximum area of the triangle $PSS'$ is ............. square units.

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