JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given the equation of the parabola y = x^2 + px + q. Since it passes through (1, -1), we have: -1 = 1 + p + q q = -p - 2 The vertex of the parabola y = x^2 + px + q occurs at x = - p 2 . The y-coordinate of the vertex is: y_v = (- p 2 )^2 + p (- p 2 ) + q = - p^2 4 + q The distance between the vertex and the x-axis is |y_v|. Substituting q = -p - 2 into the expression for y_v: |y_v| = |- p^2 4 - p - 2 | = 1 4 |p^2 + 4p + 8| = 1 4 |(p+2)^2 + 4| To minimize this distance, the term (p+2)^2 must be minimum. Since (p+2)^2 0 for all real p, the minimum value occurs at p = -2. Substituting p = -2 into q = -p - 2, we get: q = -(-2) - 2 = 0 Therefore, the value of p^2 + q^2 is: p^2 + q^2 = (-2)^2 + 0^2 = 4 Answer: 4