JEE Main 2026 — 21 January, Evening Shift — Question 5

$\mathrm{m}_{\text {AP }}=\frac{-\mathrm{t}}{2}$

Equation of AP is $y-6 t=\frac{-t}{2}\left(x-3 t^{2}\right)$

Put $\mathrm{y}=0$

$\Rightarrow \mathrm{x}=12+3 \mathrm{t}^{2}$

$\Rightarrow \mathrm{A}\left(12+3 \mathrm{t}^{2}, 0\right)$

Let centroid of $\triangle$ OPA be $\mathrm{G}(\mathrm{h}, \mathrm{k})$

$\Rightarrow 3 \mathrm{~h}=0+3 \mathrm{t}^{2}+12+3 \mathrm{t}^{2}$

$3 \mathrm{k}=0+6 \mathrm{t}+0$

$\Rightarrow \mathrm{t}=\frac{\mathrm{k}}{2}, \mathrm{~h}=2 \mathrm{t}^{2}+4$

$\Rightarrow \mathrm{h}=2 \frac{\mathrm{k}^{2}}{4}+4$

⇒ Locus of $(\mathrm{h}, \mathrm{k})$ is $\mathrm{y}^{2}=2 \mathrm{x}-8$

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