Question

Let ` $P$' be a variable point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with foci $S(ae,0)$ and $S^{\prime }(-ae,0)$ . lf $A$ is the area of the triangle $PS{S}^1$, then the maximum value of $A$ (where $e$ is eccentricity and $b^2=a^2(1-e^2))$ is

• A. $\frac{ab}{2}$
• B. 2 abe
• C. abe
• D. 4abe

Answer 1

The correct option is C abe

Let $P(acos\theta ,bsin\theta )$ be a point on the ellipse.
Area $A=\frac{1}{2}\left|\begin{array}{ccc}acos\theta & bsin\theta & 1\\ ae& 0& 1\\ -ae& 0& 1\end{array}\right|$
$A=abesin\theta$
Maximum value of $sin\theta$ is 1.
So, maximum value of area is abe.