Question

{"given":"The parabola is $y^2 = 12x$, with vertex at $O(0,0)$. A point P on the parabola is given by $P(3t^2, 6t)$. Point A is on the x-axis, so $A(x_A, 0)$. The condition is $\\angle OPA = 90^{\\circ}$.","key_observation":"The condition $\\angle OPA = 90^{\\circ}$ implies that the product of the slopes of OP and PA is $-1$. This allows us to find the coordinates of A in terms of $t$, and then the centroid's coordinates $(h,k)$ can be expressed in terms of $t$. Eliminating $t$ will give the locus.","option_analysis":[{"label":"(A)","text":"$y^{2} - 6x + 4 = 0$","verdict":"incorrect","explanation":"This option does not match the derived locus equation $y^2 - 2x + 8 = 0$."},{"label":"(B)","text":"$y^{2} - 9x + 6 = 0$","verdict":"incorrect","explanation":"This option does not match the derived locus equation $y^2 - 2x + 8 = 0$."},{"label":"(C)","text":"$y^{2} - 2x + 8 = 0$","verdict":"correct","explanation":"The derived locus of the centroid is $y^2 - 2x + 8 = 0$, which matches this option."},{"label":"(D)","text":"$y^{2} - 4x + 8 = 0$","verdict":"incorrect","explanation":"This option does not match the derived locus equation $y^2 - 2x + 8 = 0$."}],"answer":"C","formula_steps":["The parabola is $y^2 = 12x$. Comparing with $y^2 = 4ax$, we get $a=3$.","A point P on the parabola is $(at^2, 2at)$, so $P(3t^2, 6t)$. The vertex is $O(0,0)$.","The slope of OP is $m_{OP} = \\frac{6t - 0}{3t^2 - 0} = \\frac{2}{t}$.","Since $\\angle OPA = 90^{\\circ}$, $OP \\perp PA$. Thus, $m_{OP} \\cdot m_{PA} = -1$.","So, $m_{PA} = -\\frac{1}{m_{OP}} = -\\frac{t}{2}$.","The equation of line PA passing through $P(3t^2, 6t)$ with slope $m_{PA} = -\\frac{t}{2}$ is $y - 6t = -\\frac{t}{2}(x - 3t^2)$.","Point A lies on the x-axis, so its y-coordinate is 0. Substitute $y=0$ into the equation of PA:","$-6t = -\\frac{t}{2}(x_A - 3t^2)$. Assuming $t \\neq 0$, divide by $-t$:","$6 = \\frac{1}{2}(x_A - 3t^2) \\Rightarrow 12 = x_A - 3t^2 \\Rightarrow x_A = 12 + 3t^2$.","So, $A(12 + 3t^2, 0)$.","Let $G(h,k)$ be the centroid of $\\triangle OPA$. The vertices are $O(0,0)$, $P(3t^2, 6t)$, $A(12 + 3t^2, 0)$.","The coordinates of the centroid are $h = \\frac{0 + 3t^2 + (12 + 3t^2)}{3}$ and $k = \\frac{0 + 6t + 0}{3}$.","This simplifies to $h = \\frac{6t^2 + 12}{3} = 2t^2 + 4$ and $k = \\frac{6t}{3} = 2t$.","From $k = 2t$, we get $t = \\frac{k}{2}$.","Substitute $t$ into the equation for $h$: $h = 2\\left(\\frac{k}{2}\\right)^2 + 4$.","$h = 2\\left(\\frac{k^2}{4}\\right) + 4 \\Rightarrow h = \\frac{k^2}{2} + 4$.","To find the locus, replace $(h,k)$ with $(x,y)$: $x = \\frac{y^2}{2} + 4$.","Multiply by 2: $2x = y^2 + 8$.","Rearrange the equation: $y^2 - 2x + 8 = 0$."]}