JEE Main & Advanced Sample Paper JEE Main Sample Paper-17

  • question_answer
    The eccentricity of the ellipse \[3{{x}^{2}}+4y=12\] is  changing at the rate of \[0.1/\sec .\]The time at which it will touch the auxiliary circle is

    A)  2 seconds                          

    B)  3 seconds

    C)  5 seconds                          

    D)  6 seconds

    Correct Answer: C

    Solution :

     Eccentricity of ellipse\[=\frac{1}{2}\] \[\Rightarrow \]               \[\frac{de}{dt}=-0.1\] Eccentricity of auxiliary circle\[~=0\] \[\Rightarrow \]               \[\int\limits_{1/2}^{0}{de=-0.1\int\limits_{0}^{T}{dt}}\] T is time at which it will touch the auxiliary circle. \[\Rightarrow \]               \[-\frac{1}{2}=-0.1(T-0)\] \[\Rightarrow \]\[T=5\,\text{seconds}\text{.}\]


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