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Question

Let θ(0,π2) and x=X cosθ+y sinθ,y=X sinθY cosθ such that x2+2xy+y2=aX2+bY2 Where a and b are constants then


A

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B

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C

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D

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Solution

The correct option is A


x2+y2=x2+y2
xy=(x2y2) sinθ cosθXY(cos2θsin2θ)
x2+4xy+y2=X2+Y2+2(X2Y2)sin2θ2XYcos2theta
=1+2sinθ)X2+(1+2sinθ)Y22cos2θ XY
From the question
a=1+2sin 2θ,b=12sin 2θ,cos 2θ=0
cos 2θ=0θ=π4, Then a=1+2sinπ2
b=12sinπ2
a=3,b=1


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