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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2010

  • question_answer
    The eccentric angle of the point \[(2,\sqrt{3})\]  lying on \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{4}=1\] is

    A) \[\frac{\pi }{4}\]                                              

    B) \[\frac{\pi }{2}\]

    C) \[\frac{\pi }{3}\]                                              

    D) \[\frac{\pi }{6}\]

    Correct Answer: C

    Solution :

    Given equation of an ellipse is \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{4}=1\] and point \[P(2,\sqrt{3})\] Let \[\theta \] be the eccentric angle. The parametric coordinate of an ellipse is                 \[\left. \begin{matrix}    x=4\,\cos \theta   \\    y=2\sin \theta   \\ \end{matrix} \right\}\]  ……(i) Given that, eccentric angle at P is,                 \[2=4\,\cos \theta \Rightarrow \cos \theta =\frac{1}{2}\]                 \[\sqrt{2}=2{{\sin }^{\theta }}\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}\] Hence,                  \[\theta =\frac{\pi }{3}\]


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