JEE Main & Advanced Mathematics Rectangular Cartesian Coordinates Question Bank Transformation of axes and Locus

  • question_answer
    The coordinates of the point A and B are \[(ak,0)\] and\[\left( \frac{a}{k},0 \right),\,\,(k=\pm 1)\]. If a point P moves so that \[PA=kPB,\] then the equation to the locus of P is

    A) \[{{k}^{2}}({{x}^{2}}+{{y}^{2}})-{{a}^{2}}=0\]

    B) \[{{x}^{2}}+{{y}^{2}}-{{k}^{2}}{{a}^{2}}=0\]

    C) \[{{x}^{2}}+{{y}^{2}}+{{a}^{2}}=0\]

    D) \[{{x}^{2}}+{{y}^{2}}-{{a}^{2}}=0\]

    Correct Answer: D

    Solution :

    \[{{(x-ak)}^{2}}+{{y}^{2}}={{k}^{2}}\,\left[ {{\left( x-\frac{a}{k} \right)}^{2}}+{{y}^{2}} \right]\] \[\Rightarrow \,\,(1-{{k}^{2}})\,({{x}^{2}}+{{y}^{2}})-2akx+2akx+{{a}^{2}}{{k}^{2}}-{{a}^{2}}=0\] \[\Rightarrow \,\,{{x}^{2}}+{{y}^{2}}-{{a}^{2}}=0\].


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