If $α$ and $β$ are the eccentric angles of the extremities of a focal chord of an ellipse of eccentricity e then $cos (\frac{α - β}{2}) =$

e cos (\frac{α - β}{2})

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e cos (\frac{α + β}{2})

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e cos (\frac{α - β}{3})

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e cos (\frac{2 α + 2 β}{2})

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Solution

The correct option is B $e cos (\frac{α + β}{2})$
Equation of the chord joining the points $α, β$ on the ellipse is $y - b s i n α = - \frac{b c o s (\frac{α + β}{2})}{a s i n (\frac{α + β}{2})} (x - a c o s α)$
This line passes through the focus (ae, 0) then we have $cos (\frac{α - β}{2}) = e cos (\frac{α + β}{2})$ after simplification

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