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If \mathrm{(a \sec \theta, b \tan \theta)} and\mathrm{(\operatorname{asec} \phi, \operatorname{btan} \phi)} be the coordinate of the ends of a focal chord of \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}, then \mathrm{\tan \frac{\theta}{2} \tan \frac{\phi}{2}} equals to

Option: 1

\mathrm{\frac{\mathrm{e}-1}{\mathrm{e}+1}}


Option: 2

\mathrm{\frac{1-\mathrm{e}}{1+\mathrm{e}}}


Option: 3

\mathrm{\frac{1+\mathrm{e}}{1-\mathrm{c}}}


Option: 4

\mathrm{\frac{e+1}{e-1}}


Answers (1)

best_answer

Equation of chord connecting the points\mathrm{(\operatorname{asec} \theta, b \tan \theta) \text { and }(\operatorname{asec} \phi, b \tan \phi)} is 
\mathrm{\frac{x}{a} \cos \left(\frac{\theta+\phi}{2}\right)-\frac{y}{b} \sin \left(\frac{\theta+\phi}{2}\right)=\cos \left(\frac{\theta-\phi}{2}\right)}
If it passes through\mathrm{(ae, 0);} we have, ecos\mathrm{\left(\frac{\theta-\phi}{2}\right)=\cos \left(\frac{\theta+\phi}{2}\right)}
\mathrm{\mathrm{e}=\frac{\cos \left(\frac{\theta+\phi}{2}\right)}{\cos \left(\frac{\theta-\phi}{2}\right)}=\frac{1-\tan \frac{\theta}{2} \cdot \tan \frac{\phi}{2}}{1+\tan \frac{\theta}{2} \tan \frac{\phi}{2}} \Rightarrow \tan \frac{\theta}{2} \cdot \tan \frac{\phi}{2}=\frac{1-\mathrm{e}}{1+\mathrm{e}}}
Hence (b) is the correct answer.

Posted by

sudhir.kumar

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