JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The equation of the parabola is y = x^2 - 6x + 12, which can be rewritten as y - 3 = (x - 3)^2. The vertex of the parabola is P(3, 3). The equation of the circle is x^2 + y^2 - 2x - 4y + 3 = 0, which can be rewritten as (x - 1)^2 + (y - 2)^2 = 2. The center of the circle is C(1, 2) and its radius is r = 2 . The distance between the point P and the center C is: PC = (3 - 1)^2 + (3 - 2)^2 = 4 + 1 = 5 Since PC = 5 > r = 2 , the point P lies outside the circle. Let the line through P intersect the circle at R and S, and let M be the midpoint of the chord RS. The perpendicular distance from the center C to the chord is CM. Using the right-angled triangle PCM, the distance from P to the midpoint M is PM = PC^2 - CM^2 . The lengths of the segments are PR = PM - RM and PS = PM + RM. Therefore, the sum of the distances is: PR + PS = (PM - RM) + (PM + RM) = 2PM = 2 PC^2 - CM^2 To maximize (PR + PS)^2, we need to maximize PC^2 - CM^2. This occurs when CM is minimized, which means CM = 0. This happens when the line passes through the center C of the circle. Substituting CM = 0, we get the maximum sum: PR + PS = 2PC = 2 5 Thus, the maximum value of (PR + PS)^2 is: (2 5 )^2 = 4 5 = 20 Answer: 20