JEE Main & Advanced Mathematics Conic Sections Question Bank Ellipse

  • question_answer
    The locus of the point of intersection of the perpendicular tangents to the ellipse \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{4}=1\] is  [Karnataka CET 2003]

    A)            \[{{x}^{2}}+{{y}^{2}}=9\]           

    B)            \[{{x}^{2}}+{{y}^{2}}=4\]

    C)            \[{{x}^{2}}+{{y}^{2}}=13\]         

    D)            \[{{x}^{2}}+{{y}^{2}}=5\]

    Correct Answer: C

    Solution :

               The locus of point of intersection of two perpendicular tangents drawn on the ellipse is \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}},\] which is called ?director- circle?.            Given ellipse is \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{4}=1\], \Locus is \[{{x}^{2}}+{{y}^{2}}=13.\]


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