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

• A. $2abe$
• B. $abe$
• C. $\frac{1}{2}abe$
• D. None of these

Answer 1

The correct option is B $abe$

Let $P\equiv (a\cos \theta ,b\sin \theta )$

Then, $A=$ area of $△P{F}_1{F}_2$

$=\frac{1}{2}\left|\begin{array}{ccc}a\cos \theta & b\sin \theta & 1\\ ae& 0& 1\\ -ae& 0& 1\end{array}\right|=|\frac{1}{2}.2ae.b\sin \theta |=abe\left|\sin \theta \right|$

Clearly, $A$ is maximum when $\left|\sin \theta \right|=1$

$\therefore$ Maximum value of $A=abe$.