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(D) The equation of a chord joining points having eccentric angles $\alpha$ and $\beta$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is give

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$\frac{\sin \alpha + \sin \beta}{\sin (\alpha + \beta)}$

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(D) The equation of a chord joining points having eccentric angles $\alpha$ and $\beta$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\frac{x}{a} \cos \left(\frac{\alpha+\beta}{2}ight) + \frac{y}{b} \sin \left(\frac{\alpha+\beta}{2}ight) = \cos \left(\frac{\alpha-\beta}{2}ight)$.Since this is a focal chord,it passes through the focus $(ae, 0)$.Substituting $(ae, 0)$ into the equation,we get $e \cos \left(\frac{\alpha+\beta}{2}ight) = \cos \left(\frac{\alpha-\beta}{2}ight)$.Thus,$e = \frac{\cos ((\alpha-\beta)/2)}{\cos ((\alpha+\beta)/2)}$.Using the product-to-sum identities or simplifying,we note that for a focal chord,$	an(\alpha/2) 	an(\beta/2) = \frac{e-1}{e+1}$.Alternatively,using the identity $\frac{\sin \alpha + \sin \beta}{\sin (\alpha + \beta)} = \frac{2 \sin ((\alpha+\beta)/2) \cos ((\alpha-\beta)/2)}{2 \sin ((\alpha+\beta)/2) \cos ((\alpha+\beta)/2)} = \frac{\cos ((\alpha-\beta)/2)}{\cos ((\alpha+\beta)/2)} = e$.

If $\alpha$ and $\beta$ are the eccentric angles of the extremities of a focal chord of an ellipse,then the eccentricity of the ellipse is

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