JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    The locus of the foot of perpendicular drawn from the centre of the ellipse \[{{x}^{2}}+3{{y}^{2}}=6\] on any tangent to it is :   JEE Main  Solved  Paper-2014

    A) \[{{({{x}^{2}}-{{y}^{2}})}^{2}}=6{{x}^{2}}+2{{y}^{2}}\]

    B) \[{{({{x}^{2}}-{{y}^{2}})}^{2}}=6{{x}^{2}}-2{{y}^{2}}\]

    C) \[{{({{x}^{2}}+{{y}^{2}})}^{2}}=6{{x}^{2}}+2{{y}^{2}}\]

    D) \[{{({{x}^{2}}+{{y}^{2}})}^{2}}=6{{x}^{2}}-2{{y}^{2}}\]

    Correct Answer: C

    Solution :

    Given ellipse is \[\frac{{{x}^{2}}}{6}+\frac{{{y}^{2}}}{2}=1\]                                                         ?.(1) The equation of any tangent to it is \[y=2x\pm \sqrt{6{{m}^{2}}+2}\]                                              ?(2) Also perpendicular to (2) through the center of ellipse is\[y=-\frac{1}{m}x\]                                        ?(3) eliminating ?m? from (2) and (3) Gives the required locus as \[{{({{x}^{2}}+{{y}^{2}})}^{2}}=6{{x}^{2}}+2{{y}^{2}}\]

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