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JEE Main & Advanced Sample Paper JEE Main Sample Paper-16

  • question_answer
    Any ordinate MP of ellipse \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{16}=1\]meets the auxiliary circle at Q, then locus of point of intersection of normal?s at P and Q 10 curves is

    A) \[{{x}^{2}}+{{y}^{2}}=8\]                             

    B) \[{{x}^{2}}+{{y}^{2}}=34\]

    C) \[{{x}^{2}}+{{y}^{2}}=64\]                           

    D) \[{{x}^{2}}+{{y}^{2}}=81\]

    Correct Answer: D

    Solution :

    Normal to ellipse at P is \[\frac{5x}{\cos \theta }-\frac{4y}{\sin \theta }=9\]..(i) equation of normal to circle at Q is \[y=x\tan \theta \]                                                          ..(ii) eliminating\['\theta '\]from (i) and (ii) we get \[{{x}^{2}}+{{y}^{2}}=81\]


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