Question
The equation of locus of a point whose distance from the $$y$$-axis is equal to its distance from the point $$\left( {2,\,1,\, - 1} \right)$$ is :
A.
$${x^2} + {y^2} + {z^2} = 6$$
B.
$${x^2} - 4x + 2z + 6 = 0$$
C.
$${y^2} - 2y - 4x + 2z + 6 = 0$$
D.
$${x^2} + {y^2} - {z^2} = 0$$
Answer :
$${y^2} - 2y - 4x + 2z + 6 = 0$$
Solution :
The variable point is $$P\left( {x,\,y,\,z} \right)$$
Its distance from the $$y$$-axis $$ = \sqrt {{x^2} + {z^2}} $$
Its distance from $$\left( {2,\,1,\, - 1} \right)$$
$$\eqalign{
& = \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 1} \right)}^2} + {{\left( {z + 1} \right)}^2}} \cr
& {\text{Given : }}\sqrt {{x^2} + {z^2}} \cr
& = \sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 1} \right)}^2} + {{\left( {z + 1} \right)}^2}} \cr
& = {y^2} - 2y - 4x + 2z + 6 = 0 \cr} $$