A) \[\frac{e-1}{e+1}\]
B) \[\frac{1-e}{1+e}\]
C) \[\frac{1+e}{1-e}\]
D) \[\frac{e+1}{e-1}\]
Correct Answer: A
Solution :
The equation of the chord joining \[(a\text{ }sec\theta ,\text{ }b\text{ }tan\theta )\]and\[(a\text{ }sec\phi ,\text{ }b\text{ }tan\phi )\]is \[y-b\tan \phi =\frac{b\tan \theta -b\tan \phi }{a\sec \theta -a\sec \phi }(x-a\sec \theta )\] which reduces to \[\frac{x}{a}\cos \left( \frac{\theta -\phi }{2} \right)-\frac{y}{b}\sin \left( \frac{\theta +\phi }{2} \right)=\cos \left( \frac{\theta +\phi }{2} \right)\] this passes through\[(ae,0)\] \[\Rightarrow \] \[\frac{e-1}{e+1}=\frac{\cos \left( \frac{\theta +\phi }{2} \right)-\cos \left( \frac{\theta -\phi }{2} \right)}{\cos \left( \frac{\theta +\phi }{2} \right)+\cos \left( \frac{\theta -\phi }{2} \right)}\] \[\Rightarrow \] \[\frac{e-1}{e+1}=\tan \frac{\theta }{2}.\tan \frac{\phi }{2}\]
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