JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let the other end of the chord be (x_1, y_1). Since the chord is bisected by the y-axis, the midpoint of the chord lies on the y-axis. The midpoint of the chord joining (1, 2) and (x_1, y_1) is ( x_1 + 1 2 , y_1 + 2 2 ). Since it lies on the y-axis, its x-coordinate must be zero: x_1 + 1 2 = 0 x_1 = -1 Since the point (x_1, y_1) lies on the given circle x^2 + y^2 + x - 3y = 0, we substitute x_1 = -1 into the equation: (-1)^2 + y_1^2 + (-1) - 3y_1 = 0 1 + y_1^2 - 1 - 3y_1 = 0 y_1^2 - 3y_1 = 0 y_1(y_1 - 3) = 0 y_1 = 0 or y_1 = 3 Thus, the other ends of the two chords are R(-1, 0) and S(-1, 3). The midpoint of the line segment RS is given as ( , ). = -1 + (-1) 2 = -1 = 0 + 3 2 = 3 2 We need to find the value of 6( + ): 6( + ) = 6 (-1 + 3 2 ) = 6 ( 1 2 ) = 3 Answer: 3