| 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) \[{{x}^{2}}+{{y}^{2}}-4y=0\]
B) \[{{x}^{2}}+{{y}^{2}}-4\left| y \right|=0\]
C) \[{{x}^{2}}+{{y}^{2}}-4x=0\]
D) \[{{x}^{2}}+{{y}^{2}}-4\left| x \right|=0\]
Correct Answer: B
Solution :
| Let \[(x,y)\]be the point. |
| According to the question, |
| \[4\sqrt{{{(x-x)}^{2}}+{{y}^{2}}}={{x}^{2}}+{{y}^{2}}\] |
| \[\Rightarrow \] \[4\left| y \right|={{x}^{2}}+{{y}^{2}}\] |
| \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-4\left| y \right|=0\] |
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