A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{6}\]
C) \[\frac{\pi }{8}\]
D) \[\frac{\pi }{4}\]
Correct Answer: B
Solution :
Equation of tangent is drawn at a\[(3\sqrt{3}\cos \theta ,\,\,sin\theta )\]to the curve \[\frac{{{x}^{2}}}{27}+\frac{{{y}^{2}}}{1}=1\] is\[\frac{x\cos \theta }{3\sqrt{3}}+\frac{y\sin \theta }{1}=1\] Thus, sum of intercepts \[=(3\sqrt{3}\sec \theta +\operatorname{cosec}\theta )=f(\theta )\](say) \[\Rightarrow \] \[f(\theta )=3\sqrt{3}\sec \theta \tan \theta -\cos ec\theta \cot \theta \] \[=\frac{3\sqrt{3}\,{{\sin }^{3}}\,\theta -{{\cos }^{3}}\theta }{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\] Put \[f(\theta )=0\] \[\Rightarrow \] \[{{\sin }^{3}}\theta =\frac{1}{{{3}^{3/2}}}{{\cos }^{3}}\theta \] \[\Rightarrow \] \[\tan \theta =\frac{1}{\sqrt{3}}\] \[\Rightarrow \] \[\theta =\frac{\pi }{6}\]You need to login to perform this action.
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