Question
Let y^ 2 =12 x be the parabola with its vertex at O. Let P be a point on the parabola and A be a point on the x-axis such that OPA =90^ . Then the locus of the centroid of such triangles OPA is :
JEE Main
Mathematics
Parabola
2026
JEE Main 2026 (21 January Shift 2)
JEE Main Mathematics Question (2026) — Solution
Options
- A. y^ 2 -2 x+8=0
- B. y^ 2 -9 x+6=0
- C. y^ 2 -4 x+8=0
- D. y^ 2 -6 x+4=0
Answer
A. y^ 2 -2 x+8=0
Step-by-step solution
Parabola y^2 = 12x, a = 3. Let P = (3t^2, 6t), O = (0,0), A = (h, 0). OPA = 90^ PO PA = 0: (-3t^2)(h - 3t^2) + (-6t)(-6t) = 0 h = 3t^2 + 12. Centroid G = ( 0 + 3t^2 + h 3 , 0 + 6t + 0 3 ) = (2t^2 + 4, \ 2t). Let G = (X, Y): t = Y/2, X = Y^2/2 + 4. Locus: y^2 - 2x + 8 = 0.
Related: Mathematics — Parabola · All PYQ Banks