If P is any point on the ellipse frac{x^2}{25 | vedclass

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(B) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 9$,so $a = 5$ and $b = 3$.The foci $S$ and $S^{\prime}$ are

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$12$

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(B) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 9$,so $a = 5$ and $b = 3$.The foci $S$ and $S^{\prime}$ are at $(\pm ae, 0)$,where $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.Thus,the foci are $S(4, 0)$ and $S^{\prime}(-4, 0)$.The distance between the foci is $SS^{\prime} = 2ae = 2(5)(\frac{4}{5}) = 8$.The area of $\Delta SPS^{\prime} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes SS^{\prime} 	imes |y_P|$,where $y_P$ is the $y$-coordinate of point $P$.The area is maximum when $|y_P|$ is maximum.For the ellipse,the maximum value of $|y_P|$ is $b = 3$.Therefore,the maximum area $= \frac{1}{2} 	imes 8 	imes 3 = 12$ sq. units.

If $P$ is any point on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and $S, S^{\prime}$ are its foci,then the maximum area (in sq. units) of $\Delta SPS^{\prime} =$

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