If x cos α+y sinα=x cos β+y sinβ=2a , then cosα cosβ is
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a
4ax2+y2
b
4a2-y2x2+y2
c
4ayx2+y2
d
4a2-x2x2+y2
answer is B.
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Detailed Solution
we have, x cos α+y sinα=x cos β+y sinβ=2a⇒α,β are roots of x cos θ+y sinθ=2anow,x cos θ+y sinθ=2ax cos θ-2a2=y2sin2θx2 cos2 θ-4ax cosθ+4a2=y21-cos2θ⇒x2+y2 cos2 θ-4ax cosθ+4a2-y2=0 Clearly, cosα ,cosβ are the roots of this equation cosα cosβ=4a2-y2x2+y2