Question

Let $P$ be a variable point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with foci $S_1$ and $S_2$. If $A$ be the area of $\Delta P{S}_1{S}_2$, then the maximum value of $A$, is

• A. $ab$
• B. $abe$
• C. $\frac{1}{2}ab$
• D. $\frac{1}{2}abe$

Answer 1

The correct option is B $abe$

Let $P(a\cos \theta ,b\sin \theta )$ be the variable point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Then,
$A=Area\left(\Delta P{S}_1{S}_2\right)$
$\Rightarrow A=\frac{1}{2}\left|\begin{array}{ccc}a\cos \theta & b\sin \theta & 1\\ ae& 0& 1\\ -ae& 0& 1\end{array}\right|=\frac{1}{2}b\sin \theta \times 2ae=abe\sin \theta$
Clearly, it is maximum when $\theta =\frac{\pi }{2}$ and the maximum value of $A$ is $abe.$