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Question

α and β are the eccentric angles of the extremities of a focal chord of the ellipse x2a2+y2b2=1, then cos2(αβ2)cos2(α+β2)=

A
a2+b2a2
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B
a2a2+b2
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C
a2a2b2
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D
a2b2a2
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Solution

The correct option is B a2b2a2
(α,β) are eccentric angles of extremities of focal chord.
Equation of chord with (α,β) as extremities is xacos(α+β2)+ybsin(α+β2)=cos(αβ2)
Since, it passes through (ae,0)
ecos(α+β2)=cos(αβ2)

cos2(αβ2)cos2(α+β2)=e2=a2b2a2

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