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Manipal Engineering Manipal Engineering Solved Paper-2012

  • question_answer
    Tangent is drawn to ellipse\[\frac{{{x}^{2}}}{27}+\frac{{{y}^{2}}}{1}=1\]at\[(3\sqrt{3}\cos \theta ,\,\,\sin \theta )\], where\[\left( \theta \in \left( 0,\,\,\frac{\pi }{2} \right) \right)\]. Then, the value of\[\theta \]such that the sum of intercept on axes made by this tangent is minimum, is

    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         \[f(\theta )=0\] \[\Rightarrow \]               \[{{\sin }^{3}}\theta =\frac{1}{{{3}^{3/2}}}{{\cos }^{3}}\theta \] \[\Rightarrow \]               \[\tan \theta =\frac{1}{\sqrt{3}}\] \[\Rightarrow \]                     \[\theta =\frac{\pi }{6}\]


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