Straight Lines
Straight Lines
nta_pyq_2025_apr
Grade 11
Question:
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
Step-by-Step Solution
Key Concept: First find $n$ and $r$ from the binomial coefficient ratios; compute the fixed vertex $C$; then express the centroid $(h,k)$ parametrically and eliminate $t$ using $\sin^2 t+\cos^2 t=1$.
${}^nC_{r-1}/{}^nC_r = r/(n-r+1) = 28/56 = 1/2 \Rightarrow 2r = n-r+1$.
${}^nC_r/{}^nC_{r+1} = (r+1)/(n-r) = 56/70 = 4/5 \Rightarrow 5r+5=4n-4r$.
Solving: $n=8$, $r=3$. So $C(1,0)$.
Centroid: $h = \tfrac{4\cos t+2\sin t+1}{3}$, $k=\tfrac{4\sin t-2\cos t}{3}$.
$3h-1 = 4\cos t+2\sin t$, $3k = 4\sin t-2\cos t$.
$(3h-1)^2+(3k)^2 = 16+4 = 20 \Rightarrow \alpha=20$.
Correct Answer: 4