Three Dimensional Geometry Question 319
Question: The equation of locus of a point whose distance from the y-axis is equal to its distance from the point (2, 1, -1) is
Options:
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 $
Show Answer
Answer:
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 $
BETA
