Jee Main 2025 — Medium JEE math MCQ
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
- A. 6
- B. 18
- C. 8
- D. 20
Solution
The correct option is **D**. (D. 20)
