Three Dimensional Geometry Question 21
Question: A point moves so that the sum of its distances from the points (4, 0, 0) and (?4, 0, 0) remains 10. The locus of the point is
[MP PET 1988]
Options:
A) $ 9x^{2}-25y^{2}+25z^{2}=225 $
B) $ 9x^{2}+25y^{2}-25z^{2}=225 $
C) $ 9x^{2}+25y^{2}+25z^{2}=225 $
D) $ 9x^{2}+25y^{2}+25z^{2}+225=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \sqrt{{{(x-4)}^{2}}+y^{2}+z^{2}}+\sqrt{{{(x+4)}^{2}}+y^{2}+z^{2}}=10 $
$ \Rightarrow 2,(x^{2}+y^{2}+z^{2})+2\sqrt{[{{(x-4)}^{2}}+y^{2}+z^{2}],[{{(x+4)}^{2}}+y^{2}+z^{2}]} $ $ =100-32=68 $
$ \Rightarrow {{(x^{2}+y^{2}+z^{2}-34)}^{2}} $ $ =[{{(x-4)}^{2}}+y^{2}+z^{2}],[{{(x+4)}^{2}}+y^{2}+z^{2}] $
$ \Rightarrow {{(x^{2}+y^{2}+z^{2})}^{2}}-68,(x^{2}+y^{2}+z^{2})+{{(34)}^{2}} $ $ =[(x^{2}+y^{2}+z^{2}+16)-8x][(x^{2}+y^{2}+z^{2}+16)+8x] $ $ ={{(x^{2}+y^{2}+z^{2}+16)}^{2}}-64x^{2} $ $ =,(x^{2}+y^{2}+z^{2})+32,(x^{2}+y^{2}+z^{2})-64x^{2}+{{(16)}^{2}} $
$ \Rightarrow 9x^{2}+25y^{2}+25z^{2}-225=0 $ .
BETA
