JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let the coordinates of points P and Q on the parabola y^2 = 12x be (x_1, y_1) and (x_2, y_2). Given that the ordinates are in the ratio 1:2, we have y_2 = 2y_1. Since P and Q lie on the parabola, their abscissae are x_1 = y_1^2 12 and x_2 = y_2^2 12 = 4y_1^2 12 = y_1^2 3 . The length of the chord PQ is 3 13 , so PQ^2 = 117. Using the distance formula: (x_2 - x_1)^2 + (y_2 - y_1)^2 = 117 ( y_1^2 3 - y_1^2 12 )^2 + (2y_1 - y_1)^2 = 117 ( y_1^2 4 )^2 + y_1^2 = 117 y_1^4 16 + y_1^2 - 117 = 0 y_1^4 + 16y_1^2 - 1872 = 0 (y_1^2 + 52)(y_1^2 - 36) = 0 Since y_1^2 > 0, we get y_1^2 = 36, which gives y_1 = 6 (taking the positive root by symmetry). Substituting y_1 = 6, we get x_1 = 36 12 = 3. Thus, P is (3, 6). For Q, y_2 = 12 and x_2 = 144 12 = 12. Thus, Q is (12, 12). The focus of the parabola y^2 = 12x is S(3, 0). The vectors from the focus S to points P and Q are: SP = (3 - 3) i + (6 - 0) j = 6 j SQ = (12 - 3) i + (12 - 0) j = 9 i + 12 j The angle subtended by PQ at the focus is the angle between SP and SQ . = SP SQ | SP | | SQ | = 0(9) + 6(12) 6 9^2 + 12^2 = 72 6 15 = 72 90 = 4 5 Therefore, = 1 - ^2 = 1 - ( 4 5 )^2 = 3 5 Answer: 3 5