Let y2 = 12x 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:
- y2 - 4x + 8 = 0
- y2 - 2x + 8 = 0
- y2 - 9x + 6 = 0
- y2 - 6x + 4 = 0
Answer
(2) y2 - 2x + 8 = 0
Solution
For y2 = 12x, a = 3; let P(3t2, 6t) and A(h, 0).
Slope of OP = 6t3t2 = 2t, and since OP ⊥ PA, the slope of PA is - t2:
0 - 6th - 3t2 = - t2
h - 3t2 = 12 h = 3t2 + 12
For the centroid (x, y): x = 0 + 3t2 + 3t2 + 123 = 2t2 + 4, y = 0 + 6t + 03 = 2t
Eliminating t = y2 gives x = y22 + 4
y2 - 2x + 8 = 0
Answer: (2)