Question

$P$ is a variable point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$ whose foci are $F$ and $F^{\prime }$. The maximum area (in unit${}^2$) of the $\Delta PF{F}^{{}^{\prime }}$ is

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

Answer 1

The correct option is A $2b\sqrt{a^2-b^2}$

Given ellipse may be written as , $\frac{x^2}{(\sqrt{2a}{)}^2}+\frac{y^2}{(\sqrt{2b}{)}^2}=1.$
$\Rightarrow e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow F_1\equiv (-\sqrt{2}ae,0),F_2\equiv (\sqrt{2}ae,0)$
Let any point on the ellipse is, $P(\sqrt{2}a\cos \theta ,\sqrt{2}b\sin \theta )$
Thus area of triangle is $\Delta P{F}_1{F}_2=\frac{1}{2}|\left|\begin{array}{ccc}\sqrt{2}a\cos \theta & \sqrt{2}b\sin \theta & 1\\ -\sqrt{2}ae& 0& 1\\ \sqrt{2}ae& 0& 1\end{array}\right||=2abe\sin \theta$
We know maximum value of $\sin \theta$ is $1.$
Hence maximum area is, $=2abe=2b\sqrt{a^2-b^2}$