Let the point P(α,β) on the ellipse 4x2+3y2=12, in the first quadrant such that the area enclosed by the lines y=x,y=β,x=α and the x− axis is maximum, then
A
P≡(32,1)
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B
P≡(√32,√3)
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C
eccentric angle of P is π3
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D
eccentric angle of P is π6
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Solution
The correct option is D eccentric angle of P is π6 Any point P(α,β) on the ellipse x23+y24=1 is (√3cosθ,2sinθ)
As P lies in first quadrant, thus θ∈(0,π2)
Line PQ:y=2sinθ
Line PR:x=√3cosθ
Line OQ:y=x and Q≡(2sinθ,2sinθ)
△= Area of trapezium PQORP =12(OR+PQ)PR=12(√3cosθ+(√3cosθ−2sinθ))2sinθ=2√3cosθsinθ−2sin2θ=√3sin2θ+cos2θ−1=2cos(2θ−π3)−1
Which is maximum when cos(2θ−π3) is maximum ⇒(2θ−π3)=0 ⇒θ=π6
Hence, point P is (32,1).