JEE Main Mathematics Question (2026) — Solution

Step-by-step 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 = 2t_1 - 0 t_1^2 - 0 = 2 t_1 . The slope of the chord OQ is m_2 = 2t_2 - 0 t_2^2 - 0 = 2 t_2 . Since OP and OQ are perpendicular to each other, m_1 m_2 = -1. ( 2 t_1 ) ( 2 t_2 ) = -1 t_1 t_2 = -4 Let M(h, k) be the mid-point of the line segment PQ. Then, h = t_1^2 + t_2^2 2 t_1^2 + t_2^2 = 2h k = 2t_1 + 2t_2 2 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