A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: C
Solution :
[c] Equation of tangent at a point \[(a\,\sec \theta ,\,\,b\,\,tan\theta )\] to the hyperbola is: \[\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1\] Solving this tangent with circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}},\]we get \[{{y}^{2}}\left[ 1+\frac{{{a}^{2}}{{\cos }^{2}}\theta {{\tan }^{2}}\theta }{{{b}^{2}}} \right]+\frac{2{{a}^{2}}{{\cos }^{2}}\theta \tan \theta }{b}y-{{a}^{2}}{{\sin }^{2}}\theta =0\]This equation has roots \[{{y}_{1}}\] and \[{{y}_{2}}\]. \[\Rightarrow \,\,\,\frac{{{y}_{1}}+{{y}_{2}}}{{{y}_{1}}{{y}_{2}}}=\frac{{{a}^{2}}b{{\sin }^{2}}\theta }{2{{a}^{2}}{{\cos }^{2}}\theta \tan \theta }=\frac{b\tan \theta }{2}=\frac{k}{2}\]
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