Question
Locus of centroid of the triangle whose vertices are $$\left( {a\,\cos \,t,\,a\,\sin \,t} \right),\,\left( {b\,\sin \,t,\, - b\,\cos \,t} \right)$$ and (1, 0), where $$t$$ is a parameter, is-
A.
$${\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}$$
B.
$${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}$$
C.
$${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$
D.
$${\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$
Answer :
$${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$
Solution :
$$\eqalign{
& x = \frac{{a\,\cos \,t + b\,\sin \,t + 1}}{3} \Rightarrow a\,\cos \,t + b\,\sin \,t = 3x - 1 \cr
& y = \frac{{a\,\sin \,t - b\,\cos \,t}}{3} \Rightarrow a\,\sin \,t - b\,\cos \,t = 3y \cr} $$
Squaring & adding, $${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$