Q26 / 75 · JEE Main 2026 (21 January Shift 2) · Mathematics
Let $y^{2}=12 x$ 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 $\angle \mathrm{OPA}=90^{\circ}$. Then the locus of the centroid of such triangles OPA is :
- Correct answerA$y^{2}-2 x+8=0$
- B$y^{2}-9 x+6=0$
- C$y^{2}-4 x+8=0$
- D$y^{2}-6 x+4=0$
Solution
Parabola $y^2 = 12x$, $a = 3$. Let $P = (3t^2, 6t)$, $O = (0,0)$, $A = (h, 0)$.
$\angle OPA = 90^\circ \Rightarrow \vec{PO} \cdot \vec{PA} = 0$:
$(-3t^2)(h - 3t^2) + (-6t)(-6t) = 0 \Rightarrow h = 3t^2 + 12$.
Centroid $G = \left(\frac{0 + 3t^2 + h}{3}, \frac{0 + 6t + 0}{3}\right) = (2t^2 + 4, \ 2t)$.
Let $G = (X, Y)$: $t = Y/2$, $X = Y^2/2 + 4$.
Locus: $y^2 - 2x + 8 = 0$.
$\angle OPA = 90^\circ \Rightarrow \vec{PO} \cdot \vec{PA} = 0$:
$(-3t^2)(h - 3t^2) + (-6t)(-6t) = 0 \Rightarrow h = 3t^2 + 12$.
Centroid $G = \left(\frac{0 + 3t^2 + h}{3}, \frac{0 + 6t + 0}{3}\right) = (2t^2 + 4, \ 2t)$.
Let $G = (X, Y)$: $t = Y/2$, $X = Y^2/2 + 4$.
Locus: $y^2 - 2x + 8 = 0$.
Hence, the correct answer is option: 'A'.