# Locus of centriod 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

$\begin{array}{1 1}(A)\;(3x+1)^2+(3y)^2=a^2-b^2 \\(B)\;(3x-1)^2+(3y)^2=a^2-b^2 \\(C)\; (3x-1)^2+(3y^2)=a^2+b^2 \\(D)\;(3x+1)^2+(3y)^2=a^2+b^2 \end{array}$

$x=\large\frac{a \cos t+ b \sin t+1}{3}$
$\qquad= a \cos t + b \sin t =3x-1$
$y= \large\frac{a \sin t- b \cos t}{3}$
$\quad= a \sin t - b \cos t=3y$
squaring and adding we get,
$3(x-1)^2+(3y)^2=a^2+b^2$
Hence C is the correct answer.
answered Apr 8, 2014 by