Coordinate Geometry Question 125
Question: Locus of centroid of the triangle whose vertices are $ (a\cos t,a\sin t),\ (b\sin t,-b\cos t) $ and (1, 0), where t is a parameter; is
[AIEEE 2003]
Options:
A) $ {{(3x-1)}^{2}}+{{(3y)}^{2}}=a^{2}-b^{2} $
B) $ {{(3x-1)}^{2}}+{{(3y)}^{2}}=a^{2}+b^{2} $
C) $ {{(3x+1)}^{2}}+{{(3y)}^{2}}=a^{2}+b^{2} $
D) $ {{(3x+1)}^{2}}+{{(3y)}^{2}}=a^{2}-b^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ 3h=a\cos t+b\sin t+1,3k=a\sin t-b\cos t $
$ a^{2}+b^{2}={{(3h-1)}^{2}}+{{(3k)}^{2}} $
$ {{(3x-1)}^{2}}+{{(3y)}^{2}}=a^{2}+b^{2} $ .
BETA
