JEE Main 2026 — 22 January, Evening Shift — Question 3

We need the number of integer pairs $(x,y)$ such that $4x^2 + y^2 < 52$. For each integer $x$, $y^2 < 52 - 4x^2$, so $|y| < \sqrt{52 - 4x^2}$. $x$ can be $0, \pm1, \pm2, \pm3$ because for $|x| \ge 4$, $4x^2 \ge 64 > 52$. For $x=0$: $y^2 < 52 \Rightarrow |y| \le 7$ (since $7^2=49<52$, $8^2=64>52$).

So $y = 0, \pm1, \ldots, \pm7$: 15 values. For $x=\pm1$: $y^2 < 48 \Rightarrow |y| \le 6$ (since $6^2=36<48$, $7^2=49>48$).

So $y = 0, \pm1, \ldots, \pm6$: 13 values each, total $2 \times 13 = 26$. For $x=\pm2$: $y^2 < 36 \Rightarrow |y| \le 5$ (since $5^2=25<36$, $6^2=36$ not <).

So $y = 0, \pm1, \ldots, \pm5$: 11 values each, total $2 \times 11 = 22$. For $x=\pm3$: $y^2 < 16 \Rightarrow |y| \le 3$ (since $3^2=9<16$, $4^2=16$ not <).

So $y = 0, \pm1, \pm2, \pm3$: 7 values each, total $2 \times 7 = 14$. Total number of elements = $15 + 26 + 22 + 14 = 77$.