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If the chord joining two points whose eccentric angles are \alpha and \beta, cut the major axis of an ellipse at a distance c from the centre, then the value of \tan \frac{\alpha}{2} \tan \frac{\beta}{2} is

Option: 1

\mathrm{\frac{c-a}{c+a}}


Option: 2

\mathrm{\frac{c+a}{c-a}}


Option: 3

\mathrm{\frac{c-2 a}{c+2 a}}


Option: 4

\mathrm{\frac{2 c+a}{2 c-a}}


Answers (1)

best_answer

:The equation of the chord joining points whose

\mathrm{\text { eccentric angles are } \alpha \text { and } \beta \text { on the ellipse } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text {, is }}

\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)}

This will cut the major axis at the point (c, 0), if

\mathrm{\frac{c}{a} \cos \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)}

\mathrm{\begin{aligned} \Rightarrow & \frac{\cos \left(\frac{\alpha+\beta}{2}\right)}{\cos \left(\frac{\alpha-\beta}{2}\right)}=\frac{a}{c} \\ \Rightarrow & \frac{\cos \left(\frac{\alpha+\beta}{2}\right)+\cos \left(\frac{\alpha-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)-\cos \left(\frac{\alpha-\beta}{2}\right)}=\frac{a+c}{a-c} \\ \Rightarrow & \frac{2 \cos \alpha / 2 \cos \beta / 2}{-2 \sin \alpha / 2 \sin \beta / 2}=\frac{a+c}{a-c} \Rightarrow \tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{c-a}{c+a} \end{aligned}}

Posted by

Shailly goel

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