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Question

P is a variable point on the ellipse x2a2+y2b2=2 whose foci are F and F. The maximum area (in unit2) of the ΔPFF is

A
2ba2b2
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B
2ba2b2
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C
ba2b2
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D
2aa2b2
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Solution

The correct option is A 2ba2b2
Given ellipse may be written as , x2(2a)2+y2(2b)2=1.
e=1b2a2
F1(2ae,0),F2(2ae,0)
Let any point on the ellipse is, P(2acosθ,2bsinθ)
Thus area of triangle is ΔPF1F2=12|∣ ∣ ∣2acosθ2bsinθ12ae012ae01∣ ∣ ∣|=2abesinθ
We know maximum value of sinθ is 1.
Hence maximum area is, =2abe=2ba2b2

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