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If \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

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


Option: 2

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


Option: 3

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

 


Option: 4

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


Answers (1)

best_answer

(d) The equation of a chord joining points having eccentric angles \alpha \: \text{and}\: \beta is given by

\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 thought (a e, 0), then 

\begin{aligned} \cos \left(\frac{\alpha+\beta}{2}\right) & =\cos \left(\frac{\alpha-\beta}{2}\right) \\ \\\Rightarrow \quad & \quad=\frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)} \end{aligned}

\begin{array}{ll} \Rightarrow \: \: \: \: \: 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 \quad e=\frac{\sin \alpha+\sin \beta}{\sin (\alpha+\beta)} \end{array}

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vishal kumar

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