The equation of locus of a point where distance from the y-axis is equal to its distance from the point A(2,1,-1) is-

$x^{2} + y^{2} + z^{2} = 6$

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$x^{2} - 4 x + 2 z + 6 = 0$

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$y^{2} - 2 y - 4 x + 2 z + 6 = 0$

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$x^{2} + y^{2} - z^{2} = 0$

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Solution

The correct option is C
$y^{2} - 2 y - 4 x + 2 z + 6 = 0$

Let the variable point be $P (x, y, z) .$
Its distance from $Y - a x i s = \sqrt{x^{2} + z^{2}}$
Its distance from $A = \sqrt{(x - 2)^{2} + (y - 1)^{2} + (z + 1)^{2}}$
Given $\sqrt{x + z^{2}} = \sqrt{(x - 2)^{2} + (y - 1)^{2} + (z + 1)^{2}} \Rightarrow y^{2} - 2 y - 4 x + 2 z + 6 = 0$

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