JEE Mathematics 2026 Question with Solution

Given: The parabola is $y^2 = 12x$ with vertex at $O(0,0)$. A point $P$ lies on the parabola and $A$ lies on the $x$-axis such that $\angle OPA = 90^\circ$.

Find: The locus of the centroid of triangle $OPA$.

For $y^2 = 4ax$, we have $4a = 12$, so $a = 3$.

Hence a general point on the parabola is

$$P = (3t^2, 6t)$$

and let

$$A = (x_0, 0).$$

Since $\angle OPA = 90^\circ$, the lines $OP$ and $PA$ are perpendicular. Therefore,

$$(\text{slope of } OP)(\text{slope of } PA) = -1.$$

Now,

$$\text{slope of } OP = \frac{6t}{3t^2} = \frac{2}{t}$$

and

$$\text{slope of } PA = \frac{0-6t}{x_0-3t^2} = \frac{-6t}{x_0-3t^2}.$$

So,

$$\frac{2}{t} \cdot \frac{-6t}{x_0-3t^2} = -1$$

which gives

$$\frac{-12}{x_0-3t^2} = -1.$$

Hence,

$$x_0 - 3t^2 = 12$$

so

$$x_0 = 3t^2 + 12.$$

Let the centroid of triangle $OPA$ be $G(h,k)$. Then

$$h = \frac{0 + 3t^2 + (3t^2+12)}{3} = 2t^2 + 4$$

and

$$k = \frac{0 + 6t + 0}{3} = 2t.$$

Thus,

$$t = \frac{k}{2}.$$

Substitute into the expression for $h$:

$$h = 2\left(\frac{k}{2}\right)^2 + 4 = \frac{k^2}{2} + 4.$$

Therefore,

$$2h = k^2 + 8$$

or

$$k^2 - 2h + 8 = 0.$$

Replacing $h, k$ by $x, y$, the locus is

$$y^2 - 2x + 8 = 0.$$

Therefore, the correct option is D.

Given: $P$ is taken on the parabola $y^2 = 12x$ and $A$ is on the $x$-axis with $\angle OPA = 90^\circ$.

Find: The equation of the locus of the centroid.

Write the point on the parabola in parametric form as

$$P(3t^2, 6t).$$

Since $A$ is on the $x$-axis, let

$$A(x_0, 0).$$

The right angle is at $P$, so vectors along $PO$ and $PA$ are perpendicular. This is equivalent to the slope condition used above, giving

$$x_0 = 3t^2 + 12.$$

Now the centroid of triangle with vertices $O(0,0)$, $P(3t^2,6t)$, and $A(3t^2+12,0)$ is

$$G\left(\frac{0+3t^2+(3t^2+12)}{3}, \frac{0+6t+0}{3}\right).$$

So,

$$G(2t^2+4, 2t).$$

If $G = (x,y)$, then

$$x = 2t^2 + 4, \qquad y = 2t.$$

Eliminate $t$ using

$$t = \frac{y}{2}.$$

Then

$$x = 2\left(\frac{y}{2}\right)^2 + 4 = \frac{y^2}{2} + 4.$$

Multiply by $2$:

$$2x = y^2 + 8.$$

Hence,

$$y^2 - 2x + 8 = 0.$$

Therefore, the locus of the centroid is $y^2 - 2x + 8 = 0$.