A) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=6\]
B) \[{{x}^{2}}-4x+2z+6=0\]
C) \[{{y}^{2}}-2y-4x+2z+6=0\]
D) \[{{x}^{2}}+{{y}^{2}}-{{z}^{2}}=0\]
Correct Answer: C
Solution :
[c] The variable point is \[P(x,y,z)\]. Its distance from the y-axis \[=\sqrt{{{x}^{2}}+{{z}^{2}}}\] Its distance from (2, 1, -1) \[=\sqrt{{{(x-2)}^{2}}+{{(y-1)}^{2}}+{{(z+1)}^{2}}}\] Given: \[\sqrt{{{x}^{2}}+{{z}^{2}}}=\sqrt{{{(x-2)}^{2}}+{{(y-1)}^{2}}+{{(z+1)}^{2}}}\] \[\Rightarrow \,\,\,\,{{y}^{2}}-2y-4x+2z+6=0\]You need to login to perform this action.
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