Let y^{2}=12x be the parabola with its vertex | vedclass

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(C) The parabola is $y^{2}=12x$, so $4a=12 \Rightarrow a=3$. Any point $P$ on the parabola can be represented as $(3t^{2}, 6t)$.The slope of $OP$ is $

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$y^{2}-2x+8=0$

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(C) The parabola is $y^{2}=12x$, so $4a=12 \Rightarrow a=3$. Any point $P$ on the parabola can be represented as $(3t^{2}, 6t)$.The slope of $OP$ is $m_{OP} = \frac{6t-0}{3t^{2}-0} = \frac{2}{t}$.Since $\angle OPA=90^{\circ}$, $OP \perp PA$, so the slope of $PA$ is $m_{PA} = -\frac{t}{2}$.The equation of line $PA$ is $y-6t = -\frac{t}{2}(x-3t^{2})$.For point $A$ on the $x$-axis, set $y=0$: $-6t = -\frac{t}{2}(x-3t^{2})$ $\Rightarrow 12 = x-3t^{2}$ $\Rightarrow x = 3t^{2}+12$.Thus, $A = (3t^{2}+12, 0)$.Let the centroid of $	riangle OPA$ be $G(h, k)$.$h = \frac{0+3t^{2}+3t^{2}+12}{3} = \frac{6t^{2}+12}{3} = 2t^{2}+4$.$k = \frac{0+6t+0}{3} = 2t$.From $k=2t$, we have $t = \frac{k}{2}$.Substituting into $h$: $h = 2(\frac{k}{2})^{2}+4 = \frac{k^{2}}{2}+4$.$2h = k^{2}+8 \Rightarrow k^{2} = 2h-8$.The locus is $y^{2} = 2x-8$, which can be written as $y^{2}-2x+8=0$.

Let $y^{2}=12x$ be the parabola with its vertex at $O(0,0)$. Let $P$ be a point on the parabola and $A$ be a point on the $x$-axis such that $\angle OPA=90^{\circ}$. Then the locus of the centroid of such triangles $OPA$ is:

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