Equation of the locus of the centroid of the triangle whose vertices are $(a \cos k, a \sin k),(b \sin k,-b…
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
$(1-3 x)^2+9 y^2=a^2+b^2$
$(3 x-1)^2+9 y^2=2 a^2+2 b^2$
$(3 x+1)^2+(3y)^2=2 a^2+2 b^2$
$(3 x+1)^2+(3y)^2=3 a^2+3 b^2$
Solution
Let A, B ad C the vertices of triangle
$\begin{aligned}
& \mathrm{A}=(a \cos k, a \sin k) \\
& \mathrm{B}=(b \sin k-b \cos k) \\
& \mathrm{C}=(1,0)
\end{aligned}$
Let $\mathrm{G}(x, y)$ be the centroid,
$\therefore \quad x=\frac{a \cos k+b \sin k+1}{3}$
$\Rightarrow \quad 3 x-1=a \cos k+b \sin k$ ... (i)
and $y=\frac{a \sin k-b \cos k+0}{3}$
and $3 y=a \sin k-b \cos k$... (ii)
On squaring and then adding Eqs. (i) and (ii) we get
$\begin{aligned}
& (3 x-1)^2+3 y^2=a^2\left(\sin ^2 k+\cos ^2 k\right)+b^2\left(\sin ^2 k+\cos ^2 k\right) \\
& \therefore \quad(3 x-1)^2+9 y^2=a^2+b^2 \\
& \therefore \quad(1-3 x)^2+9 y^2=a^2+b^2
\end{aligned}$