Question

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 $\Delta P{F}_1{F}_2$, then the maximum value of A is

• A. $b\sqrt{a^2-b^2}$
• B. $b\sqrt{b^2-a^2}$
• C. $a\sqrt{a^2-b^2}$
• D. $a\sqrt{b^2-a^2}$

Answer 1

The correct option is A

$b\sqrt{a^2-b^2}$

Given, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Foci $F_1$ and $F_2$ are (-ae, 0) and (ae, 0), respectively. Let P(x, y) be any variable point on the ellipse.
The area A of the triangle $P{F}_1{F}_2$ is given by


$A=\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ -ae& 0& 1\\ ae& 0& 1\end{array}\right|$
$=\frac{1}{2}(-y)(-ae\times 1-ae\times 1)$
$=-\frac{1}{2}y(-2ae)=aey=ae.b\sqrt{1-\frac{x^2}{a^2}}$
So, A is maximum when x = 0
$\therefore$ Maximum of A $=abe=ab\sqrt{1-\frac{b^2}{a^2}}=ab\sqrt{\frac{a^2-b^2}{a^2}}$
$=b\sqrt{a^2-b^2}$