Question

If the distance of $P$ from $(1,1,1)$ is equal to double the distance of $P$ from the $y$-axis then the locus of $P$ is

• A. $3x^2-y^2+3z^2+2x+2y+2z-3=0$
• B. $3x^2+y^2+3m{z}^2+2x+2y+2z-3=0$
• C. $3x^2+3y^2+3z^2+2x+2y+2z-3=0$
• D. $3x^2-y^2+3z^2+2x+2y+2z+3=0$

Answer 1

The correct option is A $3x^2-y^2+3z^2+2x+2y+2z-3=0$

Let $P(x,y,z)$ be a point.
Its distance from the point $(1,1,1)$ is ${\left(\right(x-1{)}^2+(y-1{)}^2+(z-1{)}^2)}^{1/2}$
This distance is equal to twice the distance from the y-axis i.e. ${(x^2+z^2)}^{1/2}$.
Thus, ${\left(\right(x-1{)}^2+(y-1{)}^2+(z-1{)}^2)}^{1/2}$ $=$ 2*${(x^2+z^2)}^{1/2}$.
Hence, the locus of the point is
$3x^2-y^2+3z^2+2x+2y+2z-3=0$