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

If a and b are the eccentric angles of 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

Equation of chord be

xa\frac{x}{a}cos α+β2\frac{\alpha + \beta}{2} + yb\frac{y}{b}sinα+β2\frac{\alpha + \beta}{2} = cosα–β2\frac{\alpha –\beta}{2}

since it passes through (ae, 0)

so e cos α+β2\frac{\alpha + \beta}{2} = cosα–β2\frac{\alpha –\beta}{2}

e = cos⁡α–β2cos⁡α+β2\frac{\cos\frac{\alpha –\beta}{2}}{\cos\frac{\alpha + \beta}{2}} = 2cos⁡α–β2sin⁡α+β2sin⁡(α+β)\frac{2\cos\frac{\alpha –\beta}{2}\sin\frac{\alpha + \beta}{2}}{\sin(\alpha + \beta)}

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