Coordinate Geometry – Locus of Centroid – JEE Main 28 Jan 2025 Shift 1

Solution:

First, we find $n$ and $r$ using the ratio of binomial coefficients.
$$ \frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{56}{28} = 2 $$
$$ \frac{n – r + 1}{r} = 2 \Rightarrow n – r + 1 = 2r \Rightarrow n + 1 = 3r \quad \dots(1) $$

$$ \frac{{}^{n}C_{r+1}}{{}^{n}C_r} = \frac{70}{56} = \frac{5}{4} $$
$$ \frac{n – r}{r + 1} = \frac{5}{4} \Rightarrow 4n – 4r = 5r + 5 \Rightarrow 4n – 5 = 9r \quad \dots(2) $$
Substitute $3r = n+1$ from (1) into (2):
$4n – 5 = 3(3r) = 3(n+1)$
$4n – 5 = 3n + 3 \Rightarrow n = 8$.
From (1): $3r = 9 \Rightarrow r = 3$.

Now determine coordinates of vertex C:
$C(3r-n, r^2-n-1) = C(3(3)-8, 3^2-8-1) = C(1, 0)$.
Vertices are $A(4\cos t, 4\sin t)$, $B(2\sin t, -2\cos t)$, $C(1, 0)$.

Let centroid be $G(x, y)$.
$$ x = \frac{4\cos t + 2\sin t + 1}{3} \Rightarrow 3x – 1 = 4\cos t + 2\sin t $$
$$ y = \frac{4\sin t – 2\cos t + 0}{3} \Rightarrow 3y = 4\sin t – 2\cos t $$
Squaring and adding both equations:
$$ (3x-1)^2 + (3y)^2 = (4\cos t + 2\sin t)^2 + (4\sin t – 2\cos t)^2 $$
$$ = [16\cos^2 t + 4\sin^2 t + 16\sin t \cos t] + [16\sin^2 t + 4\cos^2 t – 16\sin t \cos t] $$
$$ = 16(\cos^2 t + \sin^2 t) + 4(\sin^2 t + \cos^2 t) $$
$$ = 16 + 4 = 20 $$
Comparing with $(3x-1)^2 + (3y)^2 = \alpha$, we get $\alpha = 20$.

Ans. (1)