JEE Main 2024 Mathematics question
The sum of squares of all possible values of k, for which area of the region bounded by the parabolas 2y² = kx and ky² = 2(y − x) is maximum, is equal to .
Answer: 8
Solution
The parabolas meet at y = 2k/(k² + 4) and the enclosed area is 2k²/(3(k² + 4)²), which is maximum for k = ±2; 4 + 4 = 8.