Question

Find the locus of the centroid of a triangle whose verices are $(a\cos t,a\sin t),(a\sin t,-a\cos t)$ and $(1,0)$, where '$t$' is the parameter.

Answer 1

The centroid of a triangle whose verices are $(a\cos t,a\sin t),(a\sin t,-a\cos t)$ and $(1,0)$ is $(\frac{a(\cos t+\sin t)+1}{3},\frac{a(\sin t-\cos t)+0}{3})$.

Let the cenroid be $(x,y)$.

Then,

$x=\frac{a(\cos t+\sin t)+1}{3}$ and $y=\frac{a(-\cos t+\sin t)+0}{3}$

or, $\sin t+\cos t=\frac{3x-1}{a}$.....(1) and $\sin t-\cos t=\frac{3y}{a}$.....(2).

Now squaring (1) and (2) and then adding we get,

$2({\sin }^{2}t+{\cos }^{2}t)=\frac{(3x-1{)}^2}{a^2}+\frac{y^2}{a^2}$

or, $(3x-1{)}^2+y^2=2a^2$.

This is the required locus of the centroid of the triangle.