Let P be a variable point on the ellipse frac | vedclass

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(B) The coordinates of the foci are $F_1(-ae, 0)$ and $F_2(ae, 0)$.Let the point $P$ be $(x, y)$. The base of the triangle $PF_1F_2$ is the distance b

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$abe$

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(B) The coordinates of the foci are $F_1(-ae, 0)$ and $F_2(ae, 0)$.Let the point $P$ be $(x, y)$. The base of the triangle $PF_1F_2$ is the distance between the foci,which is $F_1F_2 = 2ae$.The height of the triangle is the perpendicular distance from $P$ to the $x$-axis,which is $|y|$.The area $A$ of the triangle $PF_1F_2$ is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2ae) 	imes |y| = ae|y|$.For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the maximum value of $|y|$ is $b$ (at $x = 0$).Therefore,the maximum area $A = ae 	imes b = abe$.

Let $P$ be a variable point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with foci $F_1$ and $F_2$. If $A$ is the area of the triangle $PF_1F_2$,then the maximum value of $A$ is

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