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

 

Option: 1

\mathrm{\frac{\cos \alpha+\cos \beta}{\cos (\alpha-\beta)}}


Option: 2

\mathrm{\frac{\sin \alpha-\sin \beta}{\sin (\alpha-\beta)}}


Option: 3

\mathrm{\frac{\cos \alpha-\cos \beta}{\cos (\alpha-\beta)}}


Option: 4

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


Answers (1)

best_answer

The equation of a chord joining points having eccentric angles  \mathrm{\alpha \text { and } \beta}and is given by  
\mathrm{\frac{x}{a} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{y}{b} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)}
If it passes through\mathrm{(a e, 0)} then \mathrm{e \cos \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)}
\mathrm{e=\frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)} 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)} \Rightarrow e=\frac{\sin \alpha+\sin \beta}{\sin (\alpha+\beta)}}

 

Posted by

SANGALDEEP SINGH

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