CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

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

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

    B)  \[{{x}^{2}}+{{y}^{2}}=9\]

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

    D)  \[{{x}^{2}}+{{y}^{2}}=13\]

    Correct Answer: D

    Solution :

    The locus of the point of intersection of the perpendicular tangents to ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is a director circle and whose equation is given by \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\] \[\therefore \] Here the equation of director circle                 \[{{x}^{2}}+{{y}^{2}}=9+4\]                 \[{{x}^{2}}+{{y}^{2}}=13\]


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