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Q.
Equation of the locus of the centroid of the triangle whose vertices are $(a \cos k, a \sin k)$ $(b \sin k,-b \cos k)$ and $(1,0)$, where $k$ is a
parameter, is
We have,
$ A =(a \cos k, a \sin k) $
$ B =(b \sin k,-b \cos k) $
$C =(1,0) $
Let $(x, y)$ be the centroid of $\Delta A B C$.
$\therefore x=\frac{a \cos k+b \sin k+1}{3}$
and $y=\frac{a \sin k-b \cos k+0}{3}$
$\Rightarrow 3 x-1=a \cos k+b \sin k\,\,\,...(i)$
and $3 y=a \sin k-b \cos k\,\,\,...(ii)$
On squaring and then adding Eqs. (i) and (ii) we get
$(3 x-1)^{2}+(3 y)^{2}=a^{2}(\left.\sin ^{2} k+\cos ^{2} k\right) +b^{2}\left(\sin ^{2} k+\cos ^{2} k\right)$
$\therefore (3 x-1)^{2}+9 y^{2}=a^{2}+b^{2}$
$\therefore (1-3 x)^{2}+9 y^{2}=a^{2}+b^{2} $