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Question

If α and β are the eccentric angles of the extremities of a focal chord of an ellipse of eccentricity e then cos(αβ2)=

A
ecos(αβ2)
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B
ecos(α+β2)
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C
ecos(αβ3)
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D
ecos(2α+2β2)
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Solution

The correct option is B ecos(α+β2)
Equation of the chord joining the points α,β on the ellipse is ybsinα=bcos(α+β2)asin(α+β2)(xacosα)
This line passes through the focus (ae, 0) then we have cos(αβ2)=ecos(α+β2) after simplification

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