The number of elements in the relation
$$R = \{(x,y) : 4x^2 + y^2 < 52,\; x,y \in \mathbb{Z}\}$$
is
A) 86
B) 67
C) 89
D) 77 ✓
✓
Solution Steps
1
Find valid range of x
For any integer pair to exist, we need $4x^2 < 52$:
$x^2 < 13 \implies |x| \leq 3$ (since $3^2=9 < 13$ but $4^2=16 > 13$)
Valid $x$: $\{-3, -2, -1, 0, 1, 2, 3\}$
2
For each x, find valid y range
Condition: $y^2 < 52 - 4x^2$. Find largest $|y|$ satisfying this.
x = 0: $y^2 < 52$ → $|y| \leq 7$ (since $7^2=49 < 52,\; 8^2=64 \geq 52$)
$y \in \{-7,\ldots,7\}$ → 15 values
x = ±1: $y^2 < 48$ → $|y| \leq 6$ (since $7^2=49 \geq 48$)
$y \in \{-6,\ldots,6\}$ → $13 \times 2$ = 26 values
x = ±2: $y^2 < 36$ → $|y| \leq 5$ (since $6^2=36 \not< 36$)
$y \in \{-5,\ldots,5\}$ → $11 \times 2$ = 22 values
x = ±3: $y^2 < 16$ → $|y| \leq 3$ (since $4^2=16 \not< 16$)
$y \in \{-3,\ldots,3\}$ → $7 \times 2$ = 14 values
3
Summary table
| x | y² < | |y|max | y count | pairs |
|---|---|---|---|---|
| 0 | 52 | 7 | 15 | 15 |
| ±1 | 48 | 6 | 13 | 26 |
| ±2 | 36 | 5 | 11 | 22 |
| ±3 | 16 | 3 | 7 | 14 |
| Total | 77 | |||
4
Add all counts
$15 + 26 + 22 + 14$
$= 15 + 26 + 22 + 14$
$= 41 + 36 = \boxed{77}$
💡
Key Insight
Strict inequality excludes boundary. E.g., (±2, ±6): 4(4)+36=52 ✗. (±3, ±4): 36+16=52 ✗. Always check boundary cases!
📚
Theory
1. Lattice Points Inside an Ellipse
$4x^2+y^2 < 52$ is the interior of an ellipse $\dfrac{x^2}{13}+\dfrac{y^2}{52} = 1$ with semi-axes $a=\sqrt{13}\approx3.6$ (x-direction) and $b=\sqrt{52}\approx7.2$ (y-direction). Counting integer lattice points strictly inside is a classic combinatorics problem: fix each integer $x$ and count valid integer $y$ values.
2. Strict vs Non-Strict Inequality
$4x^2+y^2 < 52$ (strict) excludes points ON the ellipse boundary. For example, $(2,6)$: $4(4)+36=52$ — on the boundary, excluded. $(3,4)$: $36+16=52$ — excluded. This is why the count is 77, not some larger number. Always verify boundary points when counting!
3. Counting y for Each x
For fixed $x$, $y^2 < M$ has solutions $y \in \{-\lfloor\sqrt{M-1}\rfloor, \ldots, \lfloor\sqrt{M-1}\rfloor\}$ giving $2\lfloor\sqrt{M-1}\rfloor + 1$ values. Here $\lfloor\sqrt{M-1}\rfloor$ is the largest integer strictly less than $\sqrt{M}$. Be careful when $M$ is a perfect square: $y^2 < 36$ excludes $y=\pm6$.
4. Symmetry Shortcut
Since $4x^2+y^2$ depends on $x^2$ and $y^2$, the count is symmetric: positive and negative $x$ give same $y$-count. So for $x=\pm k$, total pairs $= 2 \times$(count for $|x|=k$). This halves the work. Only $x=0$ is counted once (no negative counterpart).
❓
FAQs
Q
Why can x only go up to ±3?
⌄
4x²<52 → x²<13. Since 3²=9<13 but 4²=16>13, x=±4 gives 4(16)=64>52 with no valid y.
Q
For x=0, why is |y|≤7 and not |y|≤6?
⌄
y²<52. 7²=49<52 ✓. 8²=64≥52 ✗. So |y|≤7, giving y∈{-7,...,7} = 15 values.
Q
For x=±2, why is |y|≤5 not |y|≤6?
⌄
y²<52-16=36. 6²=36, which is NOT less than 36 (strict inequality). So y=±6 excluded. Max |y|=5.
Q
For x=±3, why is |y|≤3?
⌄
y²<52-36=16. 4²=16, not <16. 3²=9<16 ✓. So max |y|=3, giving y∈{-3,...,3}=7 values.
Q
How many y values for |y|≤n?
⌄
y ∈ {-n, -n+1, ..., 0, ..., n-1, n} = 2n+1 values. E.g., |y|≤7: 15 values, |y|≤6: 13, |y|≤5: 11, |y|≤3: 7.
Q
What is the geometric shape?
⌄
4x²+y²=52 is an ellipse. Semi-axis along x: √13≈3.6, along y: √52≈7.2. We count integer points STRICTLY inside.
Q
Verify: 15+26+22+14 = 77?
⌄
15+26=41. 41+22=63. 63+14=77. ✓
Q
Why not 86 or 89?
⌄
86 or 89 would result from including boundary points or miscounting. Strict inequality <52 excludes pairs where 4x²+y²=52 exactly.
Q
Can we use 4x²+y²≤52 instead?
⌄
That would include boundary. Additional pairs: (±2,±6)×4=8 pts, (±3,±4)×4=8 pts, (0,±√52) not integer. So ≤52 gives 77+extra. Problem uses strict <.
Q
Is there a formula for lattice points inside an ellipse?
⌄
Approximately π·a·b = π√13·√52 = 13π ≈ 40.8... not quite 77 since that's area. Exact count needs enumeration as done above.
🔗
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