JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

For the ellipse E: x^2 18 + y^2 12 = 1, we have a_E^2 = 18 and b_E^2 = 12. The eccentricity e_E is given by: e_E = 1 - b_E^2 a_E^2 = 1 - 12 18 = 1 3 The foci of the ellipse E are ( a_E e_E, 0) = ( 18 1 3 , 0 ) = ( 6 , 0). For the hyperbola H, the eccentricity e_H is the reciprocal of e_E, so e_H = 3 . Since H has the same foci as E, its foci are ( 6 , 0). Let the equation of H be x^2 A^2 - y^2 B^2 = 1. The foci are ( A e_H, 0), which gives A 3 = 6 A = 2 A^2 = 2. Also, B^2 = A^2(e_H^2 - 1) = 2(3 - 1) = 4. Thus, the equation of the hyperbola H is x^2 2 - y^2 4 = 1. To find the points of intersection of H and the parabola 5 y = x^2, we substitute x^2 = 5 y into the equation of H: 5 y 2 - y^2 4 = 1 Multiplying by 4, we get: 2 5 y - y^2 = 4 y^2 - 2 5 y + 4 = 0 Solving for y using the quadratic formula: y = 2 5 20 - 16 2 = 5 1 Let y_1 = 5 - 1 and y_2 = 5 + 1. Both are positive. The corresponding x^2 values are: x_1^2 = 5 ( 5 - 1) = 5 - 5 x_2^2 = 5 ( 5 + 1) = 5 + 5 Since the points P and Q lie in the first quadrant, x > 0. Thus, x_1 = 5 - 5 and x_2 = 5 + 5 . The square of the distance d between P and Q is: d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 First, we evaluate (x_2 - x_1)^2: (x_2 - x_1)^2 = x_2^2 + x_1^2 - 2x_1x_2 x_2^2 + x_1^2 = (5 + 5 ) + (5 - 5 ) = 10 x_1x_2 = (5 - 5 )(5 + 5 ) = 25 - 5 = 20 = 2 5 So, (x_2 - x_1)^2 = 10 - 4 5 . Next, we evaluate (y_2 - y_1)^2: (y_2 - y_1)^2 = (( 5 + 1) - ( 5 - 1))^2 = 2^2 = 4 Adding these together gives d^2: d^2 = 10 - 4 5 + 4 = 14 - 4 5 Comparing this with d^2 = a + b 5 , we get a = 14 and b = -4. Therefore, a - b = 14 - (-4) = 18. Answer: 18