If 4x² + y² < 52 ,   x,y ∈ Z then the number of ordered pairs (x, y) is

We are given the inequality:
$4x^2+y^2<52$
where $x,y\in \mathbb{Z}$ (i.e., $x$ and $y$ are integers).

To find the number of ordered pairs $(x,y)$ satisfying this inequality, we can analyze the possible integer values of $x$ first.
Since $y^2\ge 0$ for all real numbers, we must have:
$4x^2<52$
Dividing by 4 on both sides gives:
$x^2<13$
Since $x$ must be an integer, the possible values for $x$ are:
$x\in \{-3,-2,-1,0,1,2,3\}$

Now, we can count the number of integer values of $y$ that satisfy the inequality for each possible value of $x$:

Case 1: When $x=0$
Substituting $x=0$ into the inequality:
$4\left(0\right)+y^2<52\Rightarrow y^2<52$
Since $y$ is an integer, $y$ can take any value from $-7$ to $7$ (since $7^2=49<52$ and $8^2=64>52$).
Number of values of $y$ = $2\times 7+1=15$.

Case 2: When $x=\pm 1$
Substituting $x^2=1$ into the inequality:
$4\left(1\right)+y^2<52\Rightarrow y^2<48$
Since $y$ is an integer, $y$ can take any value from $-6$ to $6$ (since $6^2=36<48$ and $7^2=49>48$).
Number of values of $y$ = $2\times 6+1=13$ for each value of $x$.
Total pairs for $x=\pm 1$ = $2\times 13=26$.

Case 3: When $x=\pm 2$
Substituting $x^2=4$ into the inequality:
$4\left(4\right)+y^2<52\Rightarrow 16+y^2<52\Rightarrow y^2<36$
Since $y$ is an integer, $y$ can take any value from $-5$ to $5$ (since $5^2=25<36$ and $6^2=36$).
Number of values of $y$ = $2\times 5+1=11$ for each value of $x$.
Total pairs for $x=\pm 2$ = $2\times 11=22$.

Case 4: When $x=\pm 3$
Substituting $x^2=9$ into the inequality:
$4\left(9\right)+y^2<52\Rightarrow 36+y^2<52\Rightarrow y^2<16$
Since $y$ is an integer, $y$ can take any value from $-3$ to $3$ (since $3^2=9<16$ and $4^2=16$).
Number of values of $y$ = $2\times 3+1=7$ for each value of $x$.
Total pairs for $x=\pm 3$ = $2\times 7=14$.

Total Number of Ordered Pairs:
Summing the ordered pairs from all the cases:
$\text{Total pairs}=15+26+22+14=77$

Thus, the number of ordered pairs $(x,y)$ is 77.