JEE Main 2026 — Parabola Question with Solution

Let the vertex of the parabola $y^2=4x$ be $O(0,0)$.

Let the coordinates of the points $P$ and $Q$ on the parabola be $(t_1^2,2t_1)$ and $(t_2^2,2t_2)$ respectively.

The slope of the chord $OP$ is $m_1=\frac{2t_1-0}{t_1^2-0}=\frac{2}{t_1}$.

The slope of the chord $OQ$ is $m_2=\frac{2t_2-0}{t_2^2-0}=\frac{2}{t_2}$.

Since $OP$ and $OQ$ are perpendicular to each other, $m_1{m}_2=-1$.

$\Rightarrow \left(\frac{2}{t_1}\right)\left(\frac{2}{t_2}\right)=-1$

$\Rightarrow t_1{t}_2=-4$

Let $M(h,k)$ be the mid-point of the line segment $PQ$. Then,

$h=\frac{t_1^2+t_2^2}{2}\Rightarrow t_1^2+t_2^2=2h$

$k=\frac{2t_1+2t_2}{2}\Rightarrow t_1+t_2=k$

Squaring the equation for $k$, we get:

$k^2=(t_1+t_2{)}^2=t_1^2+t_2^2+2t_1{t}_2$

Substituting the values of $t_1^2+t_2^2$ and $t_1{t}_2$, we obtain:

$k^2=2h+2(-4)$

$k^2=2h-8$

$k^2=2(h-4)$

Replacing $h$ with $x$ and $k$ with $y$, the locus of the mid-point is:

$y^2=2(x-4)$

This represents a parabola of the form $Y^2=4AX$, where the length of the latus rectum is $4A=2$.

Answer: $2$