A) \[{{x}^{2}}-12y=36\]
B) \[{{y}^{2}}+12x=36\]
C) \[{{y}^{2}}-12x=36\]
D) \[{{x}^{2}}+12y=36\]
Correct Answer: B
Solution :
Let point be \[P\,(x,\,\,y)\]. So, distance from the origin \[OP=\sqrt{{{x}^{2}}+{{y}^{2}}}\] and distance from the line \[=(x-2)\] \[\therefore \,\,\,\sqrt{{{x}^{2}}+{{y}^{2}}}+(x-2)=4\,\,\,\Rightarrow \,\,\sqrt{{{x}^{2}}+{{y}^{2}}}=(-x+6)\] \[\Rightarrow \,\,{{x}^{2}}+{{y}^{2}}={{x}^{2}}+36-12x\,\,\Rightarrow \,\,{{y}^{2}}+12x=36\].
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