P is a variable point on the ellipse frac{x^2 | vedclass

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(C) Let the coordinates of point $P$ be $(a \cos 	heta, b \sin 	heta)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The foci of the ellip

Vedclass · Accepted Answer

$aeb$

Vedclass · Answer

(C) Let the coordinates of point $P$ be $(a \cos 	heta, b \sin 	heta)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The foci of the ellipse are $F_1(ae, 0)$ and $F_2(-ae, 0)$. The base of the triangle $P F_1 F_2$ is the distance between the foci,which is $F_1 F_2 = 2ae$. The height of the triangle is the absolute value of the $y$-coordinate of point $P$,which is $h = |b \sin 	heta|$. The area $A$ of $\Delta P F_1 F_2$ is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2ae) 	imes |b \sin 	heta| = aeb |\sin 	heta|$. For the maximum value of $A$,we need the maximum value of $|\sin 	heta|$,which is $1$ (at $	heta = \frac{\pi}{2}$ or $\frac{3\pi}{2}$). Therefore,$A_{	ext{max}} = aeb$. Thus,option $C$ is correct.

$P$ is 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 $P F_1 F_2$,then the maximum value of $A$ is

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