Q96 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Mathematics

\(P\) 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

  • A
    \(\frac{e}{a b}\)
  • B
    \(\frac{a e}{b}\)
  • Correct answer
    C
    aeb
  • D
    \(\frac{a b}{e}\)

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.

Hence, the correct answer is option: 'C'.

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