Let $\mathrm{S}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{N} \times \mathbb{N}: 9(\mathrm{x}-3)^{2}+16(\mathrm{y}-4)^{2} \leq 144\right\}$ and $\mathrm{T}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{R} \times \mathbb{R}:(\mathrm{x}-7)^{2}+(\mathrm{y}-4)^{2} \leq 36\right\}$. The $\mathrm{n}(\mathrm{S} \cap \mathrm{T})$ is equal to ____.

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Sets Relations and Functions

Let

S = {(x, y) \in N \times N : 9 (x - 3)^{2} + 16 (y - 4)^{2} \leq 144}

and

T = {(x, y) \in R \times R : (x - 7)^{2} + (y - 4)^{2} \leq 36}

.
The

n (S \cap T)

is equal to ____.

Answer: 27

Rewrite S in standard form:

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{9} \leq 1

with

x, y \in N

This is an ellipse centered at

(3, 4)

with semi-axes 4 and 3.

T is a circle centered at

(7, 4)

with radius 6.

Substitute

X = x - 3

Y = y - 4

S: $\frac{X ^{2}}{16} + \frac{Y ^{2}}{9} \leq 1$ with $X \in {- 2, - 1, 0, 1, 2, 3, 4}$ , $Y \in {- 3, - 2, - 1, 0, 1, 2, 3}$
T: $(X - 4)^{2} + Y^{2} \leq 36$

Counting lattice points in

S \cap T

Total:

7 + 10 + 8 + 2 = 27

points.

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer