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Question: If \(\alpha\)and \(\beta\)are the eccentric angles of the extremities of a focal chord of an ellipse...

If α\alphaand β\betaare 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

The equation of a chord joining points having eccentric angles

α\alpha and β\beta is given by

xacos⁡(α+β2)+ybsin⁡(α+β2)=cos⁡(α−β2)\frac{\mathbf{x}}{\mathbf{a}}\mathbf{\cos}\left( \frac{\mathbf{\alpha + \beta}}{\mathbf{2}} \right)\mathbf{+}\frac{\mathbf{y}}{\mathbf{b}}\mathbf{\sin}\left( \frac{\mathbf{\alpha + \beta}}{\mathbf{2}} \right)\mathbf{=}\mathbf{\cos}\left( \frac{\mathbf{\alpha}\mathbf{-}\mathbf{\beta}}{\mathbf{2}} \right)

If it passes through (ae,0)(ae,0) then ecos⁡(α+β2)=cos⁡(α−β2)e\cos\left( \frac{\alpha + \beta}{2} \right) = \cos\left( \frac{\alpha - \beta}{2} \right)

⇒ e=cos⁡(α−β2)cos⁡(α+β2)e = \frac{\cos\left( \frac{\alpha - \beta}{2} \right)}{\cos\left( \frac{\alpha + \beta}{2} \right)}⇒e=2sin⁡(α+β2)cos⁡(α−β2)2sin⁡(α+β2)cos⁡(α+β2)e = \frac{2\sin\left( \frac{\alpha + \beta}{2} \right)\cos\left( \frac{\alpha - \beta}{2} \right)}{2\sin\left( \frac{\alpha + \beta}{2} \right)\cos\left( \frac{\alpha + \beta}{2} \right)} ⇒

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