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  • If the chord joining points <img alt="\mathrm{A\left(a \sec \theta_1, b \tan \theta_1\right) }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BA%5Cleft%28a%20%5Csec%20%5Ctheta_1%2

If the chord joining points \mathrm{A\left(a \sec \theta_1, b \tan \theta_1\right) } and \mathrm{B\left(a \sec \theta_2, b \tan \theta_2\right)} on the hyperbola passes through the focus of the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}, then \mathrm{\tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2}} is equal to

Option: 1

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


Option: 2

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


Option: 3

\mathrm{\frac{2(1-e)}{1+e}}


Option: 4

\mathrm{\frac{1-e}{2(1+e)}}


Answers (1)

best_answer

The equation of the chord AB is

\mathrm{ \frac{x}{a} \cos \left(\frac{\theta_1-\theta_2}{2}\right)-\frac{y}{b} \sin \frac{\theta_1+\theta_2}{2}=\cos \left(\frac{\theta_1+\theta_2}{2}\right) }

It passes through the focus \mathrm{ S(a e, 0).} The condition is

\mathrm{ e \cos \left(\frac{\theta_1-\theta_2}{2}\right)=\cos \frac{\theta_1+\theta_2}{2} }

Or \mathrm{ \frac{e}{1}=\frac{\cos \frac{\theta_1+\theta_2}{2}}{\cos \frac{\theta_1-\theta_2}{2}}}

\mathrm{ \begin{aligned} \Rightarrow \frac{1-e}{1+e} & =\frac{\cos \frac{\theta_1-\theta_2}{2}-\cos \frac{\theta_1+\theta_2}{2}}{\cos \frac{\theta_1-\theta_2}{2}+\cos \frac{\theta_1+\theta_2}{2}} \\\\ & =\frac{2 \sin \frac{\theta_1}{2} \sin \frac{\theta_2}{2}}{2 \cos \frac{\theta_1}{2} \cos \frac{\theta_2}{2}}=\tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2} \end{aligned} }

The answer is (b) 

Posted by

sudhir.kumar

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