Parabola Part 2

Concept:Parametric form of parabola $y^2 = 12x$ and condition for perpendicular lines via dot product.Explanation:For the parabola $y^2 = 12x$, we have $4a = 12$, so $a = 3$.A point $P$ on the parabola is taken as $P(3t^2,\ 6t)$ in parameter $t$.Let $A$ lie on the $x$-axis, so $A(x,0)$.Condition: $\angle OPA = 90^\circ$ means $OP \perp PA$.Compute vectors: $\overrightarrow{OP} = (3t^2,\ 6t)$ and $\overrightarrow{PA} = (x-3t^2,\ -6t)$.Dot product zero: $(3t^2)(x-3t^2) + (6t)(-6t) = 0$.Simplify: $3t^2(x-3t^2) - 36t^2 = 0$.Divide by $3t^2$ (for $t \neq 0$): $x - 3t^2 - 12 = 0$.Thus $x = 12 + 3t^2$.Centroid $G$ of $\triangle OPA$: $h = \frac{0+3t^2+x}{3}$, $k = \frac{0+6t+0}{3}$.Substitute $x$: $h = \frac{3t^2 + (12+3t^2)}{3} = \frac{12+6t^2}{3} = 4 + 2t^2$.And $k = 2t$.Eliminate $t$: from $k = 2t$, $t = k/2$.Then $h = 4 + 2\left(\frac{k}{2}\right)^2 = 4 + \frac{k^2}{2}$.Multiply by $2$: $2h = 8 + k^2$ ⟹ $k^2 - 2h + 8 = 0$.Replace $(h,k)$ by $(x,y)$: $y^2 - 2x + 8 = 0$.Answer:Option B: $y^2 - 2x + 8 = 0$.