\(P\) is a variable point on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with foci \(F_1\) and \(F_2\)…
\(P\) 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 \(P F_1 F_2\), then the maximum value of \(A\) is
\(\frac{e}{a b}\)
\(\frac{a e}{b}\)
aeb
\(\frac{a b}{e}\)
Solution
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 \(A_{\max }=a e b\).
Hence, option (c) correct.