Let A, B, and C be three points on the ellips | vedclass

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(C) The equation of the ellipse is $\frac{x^2}{5^2} + \frac{y^2}{4^2} = 1$. Let the coordinates of $A$ and $C$ be $(-5\cos 	heta, -4\sin 	heta)$ and

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$15\sqrt{3}$

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(C) The equation of the ellipse is $\frac{x^2}{5^2} + \frac{y^2}{4^2} = 1$. Let the coordinates of $A$ and $C$ be $(-5\cos 	heta, -4\sin 	heta)$ and $(5\cos 	heta, -4\sin 	heta)$ respectively,where $	heta \in (0, \pi/2)$. The point $B$ is the endpoint of the minor axis with a positive ordinate,so $B = (0, 4)$. The base of $\Delta ABC$ is the length of segment $AC$,which is $5\cos 	heta - (-5\cos 	heta) = 10\cos 	heta$. The height of $\Delta ABC$ is the vertical distance from $B(0, 4)$ to the line $y = -4\sin 	heta$,which is $4 - (-4\sin 	heta) = 4(1 + \sin 	heta)$. The area $S$ of $\Delta ABC$ is given by $S = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (10\cos 	heta) 	imes (4(1 + \sin 	heta)) = 20\cos 	heta(1 + \sin 	heta)$. To maximize $S$,we differentiate with respect to $	heta$: $\frac{dS}{d	heta} = 20[-\sin 	heta(1 + \sin 	heta) + \cos 	heta(\cos 	heta)] = 20[-\sin 	heta - \sin^2 	heta + \cos^2 	heta] = 20[-\sin 	heta - \sin^2 	heta + (1 - \sin^2 	heta)] = 20[1 - \sin 	heta - 2\sin^2 	heta]$. Setting $\frac{dS}{d	heta} = 0$,we get $2\sin^2 	heta + \sin 	heta - 1 = 0$,which factors as $(2\sin 	heta - 1)(\sin 	heta + 1) = 0$. Since $	heta \in (0, \pi/2)$,$\sin 	heta = 1/2$,so $	heta = \pi/6$. Substituting $\sin 	heta = 1/2$ and $\cos 	heta = \sqrt{3}/2$ into the area formula: $S_{\max} = 20 	imes \frac{\sqrt{3}}{2} 	imes (1 + \frac{1}{2}) = 20 	imes \frac{\sqrt{3}}{2} 	imes \frac{3}{2} = 15\sqrt{3}$.

Let $A, B,$ and $C$ be three points on the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$. The line joining $A$ and $C$ is parallel to the $x$-axis,and $B$ is the endpoint of the minor axis whose ordinate is positive. Find the maximum area of $\Delta ABC$.

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