JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The latus rectum of the parabola P: y^2 = 4kx is the line segment x = k. The endpoints of this latus rectum are (k, 2k) and (k, -2k). The latus rectum of the ellipse E: x^2 a^2 + y^2 b^2 = 1 is the line segment x = ae (taking the one in the positive x-axis). The endpoints of this latus rectum are (ae, b^2 a ) and (ae, - b^2 a ). Since the line segment joining the points of intersection of P and E is the latus rectum for both curves, their endpoints must coincide. Comparing the coordinates, we get: k = ae 2k = b^2 a Substituting k = ae into the second equation: 2ae = b^2 a 2a^2e = b^2 For an ellipse, the relation between a, b, and e is b^2 = a^2(1 - e^2). Substituting this into the equation above: 2a^2e = a^2(1 - e^2) Since a 0, we can divide by a^2: 2e = 1 - e^2 e^2 + 2e - 1 = 0 Solving for e using the quadratic formula: e = -2 4 - 4(1)(-1) 2 = -2 2 2 2 = -1 2 Since the eccentricity of an ellipse must satisfy 0 e = 2 - 1 We need to find the value of e^2 + 2 2 . First, calculate e^2: e^2 = ( 2 - 1)^2 = 2 - 2 2 + 1 = 3 - 2 2 Therefore: e^2 + 2 2 = (3 - 2 2 ) + 2 2 = 3 Answer: 3