Question

$P$ is 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 $PF_1F_2$. then the maximum value of $A$ is

• A. $\frac{e}{ab}$
• B. $\frac{ab}{e}$
• C. $aeb$
• D. $\frac{ab}{a}$

Answer 1

The Correct Option is C

Let point $P(a \cos \theta, b \sin \theta)$ on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $\therefore$ Area of $\Delta P F_{1} F_{2}=\frac{1}{2}(2 a e) b|\sin \theta|=a e b|\sin \theta|=A$ For maximum value of $A, \theta=\frac{\pi}{2}$ or $\frac{3 \pi}{2}$ so $A_{\max }=a e b .$