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JEE Main 202621 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Parabola

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Parabola

Let y^ 2 =12 x be the parabola with its vertex at O. Let P be a...

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 :

Options

A $y^{2}-2 x+8=0$
B $y^{2}-9 x+6=0$
C $y^{2}-4 x+8=0$
D $y^{2}-6 x+4=0$

Correct answer

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

Step-by-step 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

.

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