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Question

Question: If α and β are the eccentric angles of the extremities of a focal chord of an standard ellipse, then...

If α and β are the eccentric angles of the extremities of a focal chord of an standard ellipse, then the eccentricity of the ellipse is

A

cosα+cosβcos(α+β)\frac{\cos\alpha + \cos\beta}{\cos(\alpha + \beta)}

B

sinαsinβsin(αβ)\frac{\sin\alpha - \sin\beta}{\sin(\alpha - \beta)}

C

cosαcosβcos(αβ)\frac{\cos\alpha - \cos\beta}{\cos(\alpha - \beta)}

D

sinα+sinβsin(α+β)\frac{\sin\alpha + \sin\beta}{\sin(\alpha + \beta)}

Answer

sinα+sinβsin(α+β)\frac{\sin\alpha + \sin\beta}{\sin(\alpha + \beta)}

Explanation

Solution

xacosα+β2+ybsinα+β2=cosαβ2\frac{\mathbf{x}}{\mathbf{a}}\mathbf{\cos}\frac{\mathbf{\alpha + \beta}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{b}}\mathbf{\sin}\frac{\mathbf{\alpha + \beta}}{\mathbf{2}}\mathbf{=}\mathbf{\cos}\frac{\mathbf{\alpha - \beta}}{\mathbf{2}}; aeacosα+β2=cosαβ2\frac{\mathbf{ae}}{\mathbf{a}}\mathbf{\cos}\frac{\mathbf{\alpha + \beta}}{\mathbf{2}}\mathbf{=}\mathbf{\cos}\frac{\mathbf{\alpha - \beta}}{\mathbf{2}}e=cosαβ2cosα+β2.2sinα+β22sinα+β2=sinα+sinβsin(α+β)\mathbf{e =}\frac{\mathbf{\cos}\frac{\mathbf{\alpha}\mathbf{-}\mathbf{\beta}}{\mathbf{2}}}{\mathbf{\cos}\frac{\mathbf{\alpha + \beta}}{\mathbf{2}}}\mathbf{.}\frac{\mathbf{2}\mathbf{\sin}\frac{\mathbf{\alpha + \beta}}{\mathbf{2}}}{\mathbf{2}\mathbf{\sin}\frac{\mathbf{\alpha + \beta}}{\mathbf{2}}}\mathbf{=}\frac{\mathbf{\sin}\mathbf{\alpha}\mathbf{+}\mathbf{\sin}\mathbf{\beta}}{\mathbf{\sin}\mathbf{(}\mathbf{\alpha + \beta)}}.