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Question

Mathematics Question on Ellipse

PP is a variable point on the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 with foci F1F_1 and F2F_2 . If AA is the area of the triangle PF1F2PF_1F_2. then the maximum value of AA is

A

eab\frac{e}{ab}

B

abe\frac{ab}{e}

C

aebaeb

D

aba\frac{ab}{a}

Answer

aebaeb

Explanation

Solution

Let point P(acosθ,bsinθ)P(a \cos \theta, b \sin \theta) on the ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \therefore Area of ΔPF1F2=12(2ae)bsinθ=aebsinθ=A\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,θ=π2A, \theta=\frac{\pi}{2} or 3π2\frac{3 \pi}{2} so Amax=aeb.A_{\max }=a e b .