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(B) The foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$) are $F_1(-ae, 0)$ and $F_2(ae, 0)$.Let $P(x, y)$ be a point on the

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$abe$

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(B) The foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$) are $F_1(-ae, 0)$ and $F_2(ae, 0)$.Let $P(x, y)$ be a point on the ellipse. 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 $	riangle PF_1F_2$ is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2ae) 	imes |y| = ae|y|$.Since $P(x, y)$ lies on the ellipse,$y^2 = b^2(1 - \frac{x^2}{a^2})$,so $|y| = \frac{b}{a}\sqrt{a^2 - x^2}$.Substituting this into the area formula,$A = ae 	imes \frac{b}{a}\sqrt{a^2 - x^2} = be\sqrt{a^2 - x^2}$.The area $A$ is maximized when $\sqrt{a^2 - x^2}$ is maximum,which occurs at $x = 0$.Thus,the maximum value of $A$ is $be 	imes \sqrt{a^2 - 0^2} = be 	imes a = abe$.

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 triangle $PF_1F_2$,then the maximum value of $A$ is:

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