JEE Mathematics 2026 Question with Solution

Given: The parabola is $y^2 = 4x$. A chord through the origin is considered, and $P(h,k)$ is its mid-point. Find: The locus of the point dividing $OP$ internally in the ratio $3:1$.

A chord of the parabola through the origin can be written as

$$y = mx$$

Substituting in $y^2 = 4x$,

$$m^2x^2 = 4x$$

so the second point of intersection is

$$x = \frac{4}{m^2}, \quad y = \frac{4}{m}$$

Therefore the second point is $\left(\frac{4}{m^2}, \frac{4}{m}\right)$.

The mid-point $P(h,k)$ of the chord joining $O(0,0)$ and $\left(\frac{4}{m^2}, \frac{4}{m}\right)$ is

$$h = \frac{2}{m^2}, \quad k = \frac{2}{m}$$

Eliminating $m$,

$$k^2 = 2h$$

Hence the locus $S$ is

$$y^2 = 2x$$

Now let $Q(x,y)$ be the point dividing $OP$ internally in the ratio $3:1$. By section formula,

$$x = \frac{3h}{4}, \quad y = \frac{3k}{4}$$

So,

$$h = \frac{4x}{3}, \quad k = \frac{4y}{3}$$

Substituting in $k^2 = 2h$,

$$\left(\frac{4y}{3}\right)^2 = 2\left(\frac{4x}{3}\right)$$ $$\frac{16y^2}{9} = \frac{8x}{3}$$ $$2y^2 = 3x$$

Therefore, the required locus is $2y^2 = 3x$, so the correct option is C.

Given: A chord of $y^2 = 4x$ passes through the origin. Find: The locus after first taking the mid-point and then dividing the segment from the origin in the ratio $3:1$.

1. Take the chord through the origin as

$$y = mx$$

2. Intersect it with the parabola:

$$y^2 = 4x$$ $$(mx)^2 = 4x$$ $$m^2x^2 = 4x$$

One intersection is $x=0$, that is the origin. The other is

$$x = \frac{4}{m^2}$$

and then

$$y = m\left(\frac{4}{m^2}\right) = \frac{4}{m}$$

3. Mid-point of the chord from $O(0,0)$ to $\left(\frac{4}{m^2}, \frac{4}{m}\right)$ is

$$\left(\frac{2}{m^2}, \frac{2}{m}\right)$$

Write this as $P(h,k)$. Then

$$h = \frac{2}{m^2}, \quad k = \frac{2}{m}$$

From $k = \frac{2}{m}$, we get

$$k^2 = \frac{4}{m^2}$$

But

$$2h = 2\cdot \frac{2}{m^2} = \frac{4}{m^2}$$

So,

$$k^2 = 2h$$

Thus $S$ is

$$y^2 = 2x$$

4. Let the required point be $Q(x,y)$ dividing $OP$ in the ratio $3:1$ internally. Since $O=(0,0)$ and $P=(h,k)$,

$$Q = \left(\frac{3h}{4}, \frac{3k}{4}\right)$$

Hence,

$$h = \frac{4x}{3}, \quad k = \frac{4y}{3}$$

5. Replace in the locus of $P$:

$$k^2 = 2h$$ $$\left(\frac{4y}{3}\right)^2 = 2\left(\frac{4x}{3}\right)$$ $$\frac{16y^2}{9} = \frac{8x}{3}$$

Multiplying by $9$,

$$16y^2 = 24x$$ $$2y^2 = 3x$$

Therefore, the required locus is $2y^2 = 3x$.