Animated Solution for Mathematics - Straight Lines: Locus of centroid 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

Imagine you are standing on a coordinate plane, watching a triangle evolve. Its vertices are not static; they are alive, shifting with the flow of a parameter $t$.

We have vertex $A=(a\cos t,a\sin t)$, vertex $B=(b\sin t,-b\cos t)$, and a steadfast anchor at $C=(1,0)$. As $t$ changes, the triangle stretches, rotates, and moves. Our mission is to find the path, or the locus, of its centroid $G(x,y)$.

The centroid is the geometric center, the average of the vertices. It is the point where the medians intersect.

Mathematically, for any triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, the centroid is defined by:

Substituting our specific vertices, we obtain the position of our centroid at any moment $t$:

Now, we face the challenge: $t$ is everywhere. To find the locus, we need an equation that links $x$ and $y$ directly, without the interference of $t$.

Let's isolate the trigonometric terms. Multiplying by 3, we get:

This is where the magic happens. We have two equations, and we need to combine them. The standard JEE strategy here is to square both equations and add them to force the identity ${\sin }^{2}t+{\cos }^{2}t=1$ to emerge.

Let's perform the squaring:

Now, add them together. Look at the cross-terms: $+2ab\sin t\cos t$ and $-2ab\sin t\cos t$. They vanish, canceling each other out perfectly.

We are left with:

Since ${\sin }^{2}t+{\cos }^{2}t=1$, the entire right side collapses into $a^2+b^2$.

We arrive at our destination:

This is the equation of the locus. It describes a circle.

As $t$ varies, the centroid traces this circular path. You have successfully navigated the complexity of the moving vertices to find the elegant, underlying structure.