The number of elements in the relation R= (x,y): 4x^2+y^2<52, x,y… — Solution | JEE Mathematics

MCQEasyJEE 2026Basics: Distance, Section Formula, Locus

The number of elements in the relation $R=\{(x,y): 4x^2+y^2<52,\; x,y\in\mathbb{Z}\}$ is

Answer

Correct answer:D

Standard Method

Given: We are required to count the number of integer ordered pairs $(x,y)$ satisfying

4x^2 + y^2 < 52

Find: The number of elements in the relation $R$ .

First restrict the possible integer values of $x$ .

4x^2 < 52

x^2 < 13

So,

x = -3,-2,-1,0,1,2,3

Now count the possible integer values of $y$ for each value of $x$ .

For $x=0$ ,

y^2 < 52

Hence $|y| \le 7$ , so there are $15$ values of $y$ .

For $x=\pm 1$ ,

y^2 < 48

Hence $|y| \le 6$ , so there are $13$ values of $y$ for each of these two cases.

For $x=\pm 2$ ,

y^2 < 36

Hence $|y| \le 5$ , so there are $11$ values of $y$ for each of these two cases.

For $x=\pm 3$ ,

y^2 < 16

Hence $|y| \le 3$ , so there are $7$ values of $y$ for each of these two cases.

Adding all valid ordered pairs,

15 + 2(13) + 2(11) + 2(7)

= 15 + 26 + 22 + 14

= 77

Therefore, the number of elements in the relation is $77$ . The correct option is D.

A common mistake is to treat $y^2<52$ as giving $|y|<\sqrt{52}$ and then count only positive values. This is wrong because integer values of $y$ include negative values and zero as well. Count symmetrically from $-7$ to $7$ , which gives $15$ values.
Another mistake is to use the boundary values as if the inequality were non-strict. For example, for $x=\pm 2$ , writing $y^2\le 36$ and allowing $y=\pm 6$ is wrong because the condition is $4x^2+y^2<52$ , not $\le 52$ . So here $y^2<36$ and hence $|y|\le 5$ only.
Students may count the values for $x=1,2,3$ and then forget to double for the corresponding negative values of $x$ . This is wrong because ordered pairs with $x$ and $-x$ are distinct. After counting for $x=1,2,3$ , include both signs separately.

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.