Quadratic Equations — Medium JEE math MCQ
**[JEE Mains 2026]** If 4x² + y² = 52 where x, y ∈ ℤ (integers), then the number of ordered pairs (x, y) is:
- A. 4
- B. 6
- C. 8
- D. 12
Solution
4x² + y² = 52.
For integer solutions, 4x² ≤ 52, so x² ≤ 13, giving x ∈ {-3, -2, -1, 0, 1, 2, 3}.
x = 0: y² = 52 (not perfect square).
x = ±1: y² = 48 (not perfect square).
x = ±2: y² = 52 - 16 = 36, y = ±6 ✓
x = ±3: y² = 52 - 36 = 16, y = ±4 ✓
Pairs: (2, 6), (2, -6), (-2, 6), (-2, -6), (3, 4), (3, -4), (-3, 4), (-3, -4).
Total = 8 ordered pairs.