Three Dimensional Geometry Question 235
Question: If the sum of the squares of the distance of the point (x, y, z) from the points (a, 0, 0) and (-a, 0, 0) is $ 2c^{2} $ , then which one of the following is correct?
Options:
A) $ x^{2}+a^{2}=2c^{2}-y^{2}-z^{2} $
B) $ x^{2}+a^{2}=c^{2}-y^{2}-z^{2} $
C) $ x^{2}-a^{2}=c^{2}-y^{2}-z^{2} $
D) $ x^{2}+a^{2}=c^{2}+y^{2}+z^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let the point be P(x, y, z) and two points, (a, 0, 0) and (-a, 0, 0) be A and B As given in the problem, $ PA^{2}+PB^{2}=2c^{2} $ So, $ {{(x+a)}^{2}}+{{(y-0)}^{2}}+{{(z-0)}^{2}} $ $ +{{(x-a)}^{2}}+{{(y-0)}^{2}}+{{(z-0)}^{2}}=2c^{2} $ or, $ {{(x+a)}^{2}}+y^{2}+z^{2}+{{(x-a)}^{2}}+y^{2}+z^{2}=2c^{2} $ $ x^{2}+2a+a^{2}+y^{2}+z^{2}+x^{2}-2a+a^{2}+y^{2}+z^{2}=2c^{2} $
$ \Rightarrow 2(x^{2}+y^{2}+z^{2}+a^{2})=2c^{2} $
$ \Rightarrow x^{2}+y^{2}+z^{2}+a^{2}=c^{2} $
$ \Rightarrow x^{2}+a^{2}=c^{2}-y^{2}-z^{2} $
BETA
