JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

For the parabola P: y^2 = 8x, we have 4a = 8 a = 2. The equation of the directrix is x = -a x = -2. Since the directrix cuts the x-axis at A, the coordinates of A are (-2, 0). Let the coordinates of point B on the parabola be (2t_1^2, 4t_1). The slope of AB is given as 3 5 . 4t_1 - 0 2t_1^2 - (-2) = 3 5 2t_1 t_1^2 + 1 = 3 5 10t_1 = 3t_1^2 + 3 3t_1^2 - 10t_1 + 3 = 0 (3t_1 - 1)(t_1 - 3) = 0 t_1 = 1 3 or t_1 = 3. For t_1 = 1 3 , = 2 ( 1 3 )^2 = 2 9 , which is rejected since > 1. For t_1 = 3, = 2(3)^2 = 18 > 1. Thus, B is (18, 12). Since BC is a focal chord, the parameter for C is t_2 = - 1 t_1 = - 1 3 . The coordinates of C are (2 (- 1 3 )^2, 4 (- 1 3 ) ) = ( 2 9 , - 4 3 ). The area of ABC is given by: = 1 2 | x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) | = 1 2 | -2 (12 - (- 4 3 ) ) + 18 (- 4 3 - 0 ) + 2 9 (0 - 12) | = 1 2 | -2 ( 40 3 ) - 24 - 24 9 | = 1 2 | - 80 3 - 24 - 8 3 | = 1 2 | - 88 3 - 72 3 | = 1 2 | - 160 3 | = 80 3 Six times the area of ABC = 6 80 3 = 160. Answer: 160