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Question: If a, b are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentr...

If a, b are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is –

A

cos⁡α+cos⁡βcos⁡(α+β)\frac{\cos\alpha + \cos\beta}{\cos(\alpha + \beta)}

B

sin⁡α–sin⁡βsin⁡(α–β)\frac{\sin\alpha –\sin\beta}{\sin(\alpha –\beta)}

C

cos⁡α–cos⁡βcos⁡(α–β)\frac{\cos\alpha –\cos\beta}{\cos(\alpha –\beta)}

D

sin⁡α+sin⁡βsin⁡(α+β)\frac{\sin\alpha + \sin\beta}{\sin(\alpha + \beta)}

Answer

sin⁡α+sin⁡βsin⁡(α+β)\frac{\sin\alpha + \sin\beta}{\sin(\alpha + \beta)}

Explanation

Solution

(a cos a, b sin a), (a cos b, b sin b), (ae, 0) are collinear.

b(sin⁡β–sin⁡α)a(cos⁡β–cos⁡α)=bsin⁡α–0acos⁡α–ae\frac{b(\sin\beta –\sin\alpha)}{a(\cos\beta –\cos\alpha)} = \frac{b\sin\alpha –0}{a\cos\alpha –ae}

\ (cosa – e) (sinb – sina) = sin a(cosb – cosa)

\ e =cos⁡α(sin⁡β–sin⁡α)–sin⁡α(cos⁡β–cos⁡α)sin⁡β–sin⁡α\frac{\cos\alpha(\sin\beta –\sin\alpha)–\sin\alpha(\cos\beta –\cos\alpha)}{\sin\beta –\sin\alpha}

= sin⁡(α–β)sin⁡α–sin⁡β\frac{\sin(\alpha –\beta)}{\sin\alpha –\sin\beta}= sin⁡α+sin⁡βsin⁡(α+β)\frac{\sin\alpha + \sin\beta}{\sin(\alpha + \beta)}