Q82 / 180 · COMEDK 2013 · Mathematics

What is the equation of the locus of a point which moves such that $4$ times its distance from the $X$-axis is the square of its distance from the origin?

  • A
    $x^{2}+y^{2}-4 y=0$
  • Correct answer
    B
    $x^{2}+y^{2}-4|y|=0$
  • C
    $x^{2}+y^{2}-4 x=0$
  • D
    $x^{2}+y^{2}-4|x|=0$

Solution

Let the required point be $P(x, y)$. Then, $$ \begin{aligned} 4|y| &=\left(\sqrt{\left.(x-0)^{2}+(y-0)^{2}\right)^{2}}\right.\\ \Rightarrow \quad 4|y| &=x^{2}+y^{2} \\ \Rightarrow x^{2}+y^{2}-4|y| &=0 \end{aligned} $$

Hence, the correct answer is option: 'B'.

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