© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the coordinates of the centroid
The vertices of the triangle are given as
$ (a\cos t,\;a\sin t),\;(b\sin t,\;-b\cos t),\;(1,\;0) $.
The centroid $G(x, y)$ of a triangle is the average of the coordinates of its vertices.
Hence,
$$
x = \frac{a\cos t + b\sin t + 1}{3},
\quad
y = \frac{a\sin t - b\cos t}{3}.
$$
Step 2: Rearrange the centroid coordinates
From
$ x = \frac{a\cos t + b\sin t + 1}{3} $,
multiply both sides by 3:
$$
3x = a\cos t + b\sin t + 1.
$$
Rearrange to isolate $a\cos t + b\sin t$:
$$
a\cos t + b\sin t = 3x - 1.
$$
Similarly, from
$ y = \frac{a\sin t - b\cos t}{3} $,
multiply both sides by 3:
$$
3y = a\sin t - b\cos t.
$$
Step 3: Use trigonometric identities to find the locus
We have
$ a\cos t + b\sin t = 3x - 1 \quad (1)$
and
$ a\sin t - b\cos t = 3y \quad (2). $
Let us try to combine these two expressions in a way to form a recognizable equation.
3.1 Square and add both expressions
Square equation (1):
$$
(a\cos t + b\sin t)^2 = (3x - 1)^2
$$
Square equation (2):
$$
(a\sin t - b\cos t)^2 = (3y)^2
$$
Now, add these two results:
$$
(a\cos t + b\sin t)^2 + (a\sin t - b\cos t)^2
= (3x - 1)^2 + (3y)^2.
$$
3.2 Simplify the left-hand side
Expand each term:
$
(a\cos t + b\sin t)^2
= a^2\cos^2 t + 2ab\cos t\,\sin t + b^2\sin^2 t,
$
$
(a\sin t - b\cos t)^2
= a^2\sin^2 t - 2ab\sin t\,\cos t + b^2\cos^2 t.
$
Add them:
$$
\begin{aligned}
(a\cos t + b\sin t)^2 + (a\sin t - b\cos t)^2
& = a^2(\cos^2 t + \sin^2 t) + b^2(\sin^2 t + \cos^2 t) \\
& \quad + 2ab(\cos t\,\sin t - \sin t\,\cos t).
\end{aligned}
$$
Notice that $ \cos^2 t + \sin^2 t = 1 $ and the terms $ 2ab(\cos t\,\sin t - \sin t\,\cos t) $ cancel out to 0. Thus,
$$
(a\cos t + b\sin t)^2
+ (a\sin t - b\cos t)^2
= a^2 + b^2.
$$
Step 4: Final locus equation
Substituting back, we get
$$
a^2 + b^2 = (3x - 1)^2 + (3y)^2.
$$
Hence, the locus of the centroid is
$$
(3x - 1)^2 + (3y)^2 = a^2 + b^2.
$$
Answer
The correct locus is
$
(3x - 1)^2 + (3y)^2 = a^2 + b^2.
$
