JEE Main 2025 (Online) 28th January Morning ShiftQuestion 1
Let ${}^nC_{r-1}=28$, ${}^nC_r=56$ and ${}^nC_{r+1}=70$. Let $A(4\cos t, 4\sin t)$, $B(2\sin t, -2\cos t)$ and $C(3r-n, r^2-n-1)$ be the vertices of a triangle $ABC$, where $t$ is a parameter. If $(3x-1)^2+(3y)^2=\alpha$, is the locus of the centroid of triangle $ABC$, then $\alpha$ equals
Circle
Solution
Concept. Ratios of consecutive binomial coefficients give $n$ and $r$. Then the centroid coordinates are sines and cosines of $t$; squaring and adding removes $t$. Formulas used. (1) $\dfrac{{}^nC_r}{{}^nC_{r-1}}=\dfrac{n-r+1}{r}$(2) Centroid $=\big(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\big)$Step 1 - Use (1). $\frac{56}{28}=\frac{n-r+1}{r}\Rightarrow 2r=n-r+1\Rightarrow n=3r-1$. $\frac{70}{56}=\frac{n-r}{r+1}\Rightarrow 5r+5=4n-4r\Rightarrow 9r+5=4(3r-1)\Rightarrow r=3$, $n=8$. Step 2 - Point $C$. $C=(9-8,\ 9-8-1)=(1,0)$. Step 3 - Centroid, (2). $$3x=4\cos t+2\sin t+1,\qquad 3y=4\sin t-2\cos t.$$ So $3x-1=4\cos t+2\sin t$ and $3y=4\sin t-2\cos t$. Step 4 - Square and add. The cross terms $\pm16\sin t\cos t$ cancel: $$(3x-1)^2+(3y)^2=16(\cos^2t+\sin^2t)+4(\sin^2t+\cos^2t)=20.$$ Answer: A