Let y²=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 :

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 .