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Mathematics · 2025

JEE Main · 28 January 2025, Shift 1 · Q11

Let ^n C_r-1=28,^n C_r=56 and ^n C_r+1=70. Let A ( 4 cos t, 4 sin t ), B ( 2 sin t,-2 cos t ) and C(3 r-n, r^2-n-1) be the vertices of a triangle A B…

Let $\displaystyle { }^n C_{r-1}=28,{ }^n C_r=56$ and $\displaystyle { }^n C_{r+1}=70$. Let $\displaystyle A$ ( $\displaystyle 4 \cos t, 4 \sin t$ ), $\displaystyle B$ ( $\displaystyle 2 \sin t,-2 \cos t$ ) and $\displaystyle C\left(3 r-n, r^2-n-1\right)$ be the vertices of a triangle $\displaystyle A B C$, where $\displaystyle t$ is a parameter. If $\displaystyle (3 x-1)^2+(3 y)^2$ $\displaystyle =\alpha$, is the locus of the centroid of triangle ABC, then $\displaystyle \alpha$ equals
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.