Let $y^2 = 12x$ be the parabola with its vertex at $O$. Let $P$ be a point on the parabola and $A$ be a point on the y-axis such that $OPA = 90^\circ$. Then the locus of the centroid of triangle $OPA$ is:

Let $P(3t^2, 6t)$ $m_{AP} = \frac{t}{2}$ Equation of $AP$ is $y - 6t = \frac{t}{2}(x - 3t^2)$ Put $y = 0 \Rightarrow x = 12 + 3t^2$ $\Rightarrow A(12 + 3t^2, 0)$ Let centroid of $\triangle OPA$ be $(h,k)$ $\Rightarrow 3h = 0 + 3t^2 + 12 + 3t^2$ $\Rightarrow 3k = 0 + 6t + 0$ $\Rightarrow t = \frac{k}{2},; h = 2t^2 + 4$ $\Rightarrow h = \frac{k^2}{2} + 4$ $\Rightarrow$ locus of $(h,k)$ is $y^2 = 2x - 8$