If $α, β$ , are the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $x^{2} + 4 y^{2} = 4$ then $\sqrt{3} \cos (\frac{α + β}{2}) =$

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$2 \cos (\frac{α - β}{2})$

$2 \sin (\frac{α - β}{2})$

$2 \sec (\frac{α + β}{2})$

$2 \sin (\frac{α + β}{2}) =$

answer is A.

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Detailed Solution

Given ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$

$I f α, β a r e e c e n t r i c a n g l e s o f a f o c a l c h o r d t h e n$

$e = \frac{\cos (\frac{α - β}{2})}{\cos (\frac{α + β}{2})} \sqrt{\frac{4 - 1}{4}} = \frac{\cos (\frac{α - β}{2})}{\cos (\frac{α + β}{2})} \sqrt{3} \cos (\frac{α + β}{2}) = 2 \cos (\frac{α - β}{2})$

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