COMEDK · Maths · 18. Heights and Distances
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 \(x^{2}+y^{2}-4 y=0\)
- B \(x^{2}+y^{2}-4|y|=0\)
- C \(x^{2}+y^{2}-4 x=0\)
- D \(x^{2}+y^{2}-4|x|=0\)
Answer & Solution
Correct Answer
(B) \(x^{2}+y^{2}-4|y|=0\)
Step-by-step Solution
Detailed explanation
Let the required point be \(P(x, y)\). Then, \[ \begin{aligned} 4|y| &=\left(\sqrt{\left.(x-0)^{2}+(y-0)^{2}\right)^{2}}\right.\\ \Rightarrow \quad 4|y| &=x^{2}+y^{2} \\ \Rightarrow x^{2}+y^{2}-4|y| &=0 \end{aligned} \]
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