Question

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

Answer 1

$S: \frac{(x-3)^2}{16}+\frac{(y-4)^2}{9} \leq 1$
$ x, y \in\{1,2,3, \ldots \ldots .\} $
$T:(x-7)^2+(y-4)^2 \leq 36 x, y \in R$
Let $x-3=x: y-4=y$
$ S: \frac{x^2}{16}+\frac{y^2}{9} \leq 1$
$ x \in\{-2,-1,0,1, \ldots \ldots\}$
$ T:(x-4)^2+y^2 \leq 36 ;$
$ y \in\{-3,-2,-1,0, \ldots \ldots\}$

$ S \cap T=(-2,0),(-1,0), \ldots \ldots(4,0) \rightarrow(7) $
$(-1,1),(0,1), \ldots \ldots . .(3,1) \rightarrow(5)$
$ (-1,-1),(0,-1), \ldots \ldots . .(3,-1) \rightarrow(5)$
$ (-1,2),(0,2),(1,2),(2,2) \rightarrow(4) $
$ (-1,-2),(0,-2),(1,-2),(2,-2) \rightarrow(4) $
$ (0,3)(0,-3) \rightarrow(2)$