JEE Main 2026 — Mathematics Coordinate Geometry

Let the vertex of the parabola $y^2 = 4x$y2=4x be $O(0, 0)$O(0,0).

Let the coordinates of the points $P$P and $Q$Q on the parabola be $(t_1^2, 2t_1)$(t12​,2t1​) and $(t_2^2, 2t_2)$(t22​,2t2​) respectively.

The slope of the chord $OP$OP is $m_1 = \dfrac{2t_1 - 0}{t_1^2 - 0} = \dfrac{2}{t_1}$m1​=t12​−02t1​−0​=t1​2​.

The slope of the chord $OQ$OQ is $m_2 = \dfrac{2t_2 - 0}{t_2^2 - 0} = \dfrac{2}{t_2}$m2​=t22​−02t2​−0​=t2​2​.

Since $OP$OP and $OQ$OQ are perpendicular to each other, $m_1 m_2 = -1$m1​m2​=−1.

$$\Rightarrow \left(\dfrac{2}{t_1}\right) \left(\dfrac{2}{t_2}\right) = -1$$⇒(t1​2​)(t2​2​)=−1

$$\Rightarrow t_1 t_2 = -4$$⇒t1​t2​=−4

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

$$h = \dfrac{t_1^2 + t_2^2}{2} \Rightarrow t_1^2 + t_2^2 = 2h$$h=2t12​+t22​​⇒t12​+t22​=2h

$$k = \dfrac{2t_1 + 2t_2}{2} \Rightarrow t_1 + t_2 = k$$k=22t1​+2t2​​⇒t1​+t2​=k

Squaring the equation for $k$k, we get:

$$k^2 = (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1 t_2$$k2=(t1​+t2​)2=t12​+t22​+2t1​t2​

Substituting the values of $t_1^2 + t_2^2$t12​+t22​ and $t_1 t_2$t1​t2​, we obtain:

$$k^2 = 2h + 2(-4)$$k2=2h+2(−4)

$$k^2 = 2h - 8$$k2=2h−8

$$k^2 = 2(h - 4)$$k2=2(h−4)

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

$$y^2 = 2(x - 4)$$y2=2(x−4)

This represents a parabola of the form $Y^2 = 4AX$Y2=4AX, where the length of the latus rectum is $4A = 2$4A=2.

Answer: $2$2