Coordinate Geometry
3D Geometry
MMTS_Full_Test_09
Grade 12
Question:
Find the locus of a point whose distance from $x$-axis is twice the distance from the point $(1,-1,2)$
$y^2+2x-2y-4z+6=0$
$x^2+2x-2y-4z+6=0$
$x^2-2x+2y-4z+6=0$
$z^2-2x+2y-4z+6=0$
Step-by-Step Solution
Key Concept: Distance from x-axis = $\sqrt{y^2+z^2}$; set equal to $2\cdot$ distance from $(1,-1,2)$
$y^2+z^2=4[(x-1)^2+(y+1)^2+(z-2)^2]$. Expand: $y^2+2x-2y-4z+6=0$.
Correct Answer: 1