Animated Solution for Mathematics - Sets and Relations: The number of elements in the relation $R=\{(x,y):4x^2+y^2<52,x,y\in \mathbb{Z}\}$ is

To find the number of integer coordinates $(x,y)$ that satisfy the inequality $4x^2+y^2<52$, we treat this as a problem of counting lattice points within an elliptical boundary.

Since $y^2\ge 0$, we can establish a bound for $x$ by observing that $4x^2<52$. Dividing by $4$, we obtain the constraint:

The integers $x$ that satisfy this condition are $x\in \{0,\pm 1,\pm 2,\pm 3\}$. This reduces our search to seven distinct vertical lines.

We now evaluate the range of $y$ for each possible value of $x$ by substituting into the inequality $y^2<52-4x^2$.

Case 1: $x=0$ The inequality becomes $y^2<52$. The integers $y$ satisfying this are $\{-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7\}$. There are 15 points on this line.

Case 2: $x=\pm 1$ Substituting $x=\pm 1$ gives $4(1{)}^2+y^2<52$, which simplifies to $y^2<48$. The valid integers for $y$ are $\{-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6\}$. Each line contains 13 points. For both $x=1$ and $x=-1$, we have $2\times 13=$ 26 points.

Case 3: $x=\pm 2$ Substituting $x=\pm 2$ gives $4(4)+y^2<52$, which simplifies to $y^2<36$. Because the inequality is strict, $y$ cannot be $\pm 6$. The valid integers are $\{-5,-4,-3,-2,-1,0,1,2,3,4,5\}$. Each line contains 11 points. For both $x=2$ and $x=-2$, we have $2\times 11=$ 22 points.

Case 4: $x=\pm 3$ Substituting $x=\pm 3$ gives $4(9)+y^2<52$, which simplifies to $y^2<16$. The valid integers are $\{-3,-2,-1,0,1,2,3\}$. Each line contains 7 points. For both $x=3$ and $x=-3$, we have $2\times 7=$ 14 points.

To find the total number of lattice points, we sum the counts from all cases:

Performing the addition:

There are exactly 77 integer pairs $(x,y)$ that satisfy the given inequality.