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

### Related Formula textCentroid G(x,y) = left( fracx_1 + x_2 + x_33, fracy_1 + y_2 + y_33 right)$$\text{Centroid } G(x,y) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$$ textProduct of slopes for perpendicular lines m_1 m_2 = -1$$\text{Product of slopes for perpendicular lines } m_1 m_2 = -1$$ ### Core Logic

Define coordinates: O(0,0)$O(0,0)$, P(3t^2, 6t)$P(3t^2, 6t)$ (since y^2 = 12x implies 4a = 12 implies a=3$y^2 = 12x \implies 4a = 12 \implies a=3$). Determine the slope of OP$OP$ and use perpendicularity to find the equation of PA$PA$ and locate point A$A$ on the x-axis. ### Step 1: Locate A via Perpendicularity Slope of OP$OP$ is m_OP = frac6t - 03t^2 - 0 = frac2t$m_{OP} = \frac{6t - 0}{3t^2 - 0} = \frac{2}{t}$. Since angle OPA = 90^circ$\angle OPA = 90^\circ$, slope of AP$AP$ is m_AP = -fract2$m_{AP} = -\frac{t}{2}$. Equation of AP$AP$: y - 6t = -fract2(x - 3t^2)$$y - 6t = -\frac{t}{2}(x - 3t^2)$$ To find A$A$ on the x-axis, put y = 0$y = 0$: -6t = -fract2(x - 3t^2) implies 12 = x - 3t^2 implies x = 12 + 3t^2$$-6t = -\frac{t}{2}(x - 3t^2) \implies 12 = x - 3t^2 \implies x = 12 + 3t^2$$ So, A$A$ is (12 + 3t^2, 0)$(12 + 3t^2, 0)$. ### Step 2: Locus of the Centroid Let the centroid of triangle OPA$\triangle OPA$ be G(h, k)$G(h, k)$. h = frac0 + 3t^2 + (12 + 3t^2)3 = frac6t^2 + 123 = 2t^2 + 4$$h = \frac{0 + 3t^2 + (12 + 3t^2)}{3} = \frac{6t^2 + 12}{3} = 2t^2 + 4$$ k = frac0 + 6t + 03 = 2t$$k = \frac{0 + 6t + 0}{3} = 2t$$ From k = 2t implies t = frack2$k = 2t \implies t = \frac{k}{2}$. Substitute t$t$ into the equation for h$h$: h = 2left(frack2right)^2 + 4 = frack^22 + 4$$h = 2\left(\frac{k}{2}\right)^2 + 4 = \frac{k^2}{2} + 4$$ 2h = k^2 + 8 implies k^2 = 2h - 8$$2h = k^2 + 8 \implies k^2 = 2h - 8$$ Replacing (h, k)$(h, k)$ with (x, y)$(x, y)$, the locus is y^2 = 2x - 8 implies y^2 - 2x + 8 = 0$y^2 = 2x - 8 \implies y^2 - 2x + 8 = 0$. ### Pattern Recognition For any right-angled configuration involving the origin and axis on a parabola, parametric geometry simplifies equations significantly. Find coordinates O, P, A$O, P, A$, set up the centroid algebraic relations, and eliminate parameter t$t$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines