If P is any point on the ellipse 16x^2 + 25y^ | vedclass

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(C) The equation of the ellipse is $16x^2 + 25y^2 = 400$. Dividing by $400$, we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$.Here, $a^2 = 25$ and $b^2 =

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$5\sqrt{7}/4$

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(C) The equation of the ellipse is $16x^2 + 25y^2 = 400$. Dividing by $400$, we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$.Here, $a^2 = 25$ and $b^2 = 16$, so $a = 5$ and $b = 4$.The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.The foci are $S(ae, 0) = (3, 0)$ and $S'(-ae, 0) = (-3, 0)$.Let $P = (x_0, y_0)$. The area of $\Delta PSS'$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (SS') 	imes |y_0| = 9$.The distance $SS' = 2ae = 2(3) = 6$.So, $\frac{1}{2} 	imes 6 	imes |y_0| = 9 \implies 3|y_0| = 9 \implies |y_0| = 3$.Since $P$ lies on the ellipse, $16x_0^2 + 25(3)^2 = 400$.$16x_0^2 + 225 = 400 \implies 16x_0^2 = 175 \implies x_0^2 = \frac{175}{16}$.$x_0 = \pm \frac{\sqrt{25 	imes 7}}{4} = \pm \frac{5\sqrt{7}}{4}$.Thus, the abscissa is $\frac{5\sqrt{7}}{4}$.

If $P$ is any point on the ellipse $16x^2 + 25y^2 = 400$ with foci $S$ and $S'$, and the area of $\Delta PSS'$ is $9 \text{ square units}$, then the abscissa of point $P$ is:

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