Question

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

Answer 1

Answer: 27

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

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