Let S = \ (x, y) ∈ ℕ × ℕ: 9 (x - 3) ^ 2 + 16 (y - 4) ^ 2 ≤ 144\ and...

Given,

$S: 9 ( x - 3{)}^{2} + 16{\left(y - 4\right)}^{2} \le 144\text{where}\left(x, y\right)\inN \times N$

On rearranging we get,

$S: \dfrac{( x - 3{)}^{2}}{16} + \dfrac{{\left(y - 4\right)}^{2}}{9} \le 1\text{where}x, y \in \left\{1, 2, 3, \ldots \ldots\right\}$

Now $T: {\left(x - 7\right)}^{2} + {\left(y - 4\right)}^{2} \le 36\text{where}x, y \in R$

Now let $x - 3 = X: y - 4 = Y$

So, new equation and domain will become,

$S: \dfrac{{X}^{2}}{16} + \dfrac{{Y}^{2}}{9} \le 1; X \in \left\{- 2, - 1, 0, 1, \ldots \ldots\right\}$

$T: {\left(X - 4\right)}^{2} + {Y}^{2} \le 36; Y \in \left\{- 3, - 2, - 1, 0, \ldots \ldots\right\}$

Now on plotting the diagram we get,

Now number of point in $n\left(S \cap T\right)$ will be given by,

Take $Y = - 3, - 2, - 1, 0, 1, 2, 3$ and check for $X$

So, possible for $S \cap T$ will be,

When $Y = 0$$\left(- 2, 0\right), \left(- 1, 0\right), \ldots \ldots \left(4, 0\right) \to 7\text{possible case}$

When $Y = 1$

\left(- 1, 1\right), \left(0, 1\right), \ldots \ldots . \left(3, 1\right) \to 5\text{possible case}$</p><p>When$Y = - 1$</p><p>\left(- 1, - 1\right), \left(0, - 1\right), \ldots \ldots . \left(3, - 1\right) \to 5\text{possible case}$

When $Y = 2$

\left(- 1, 2\right), \left(0, 2\right), \left(1, 2\right), \left(2, 2\right) \to 4\text{possible case}$</p><p>When$Y = - 2$</p><p>\left(- 1, - 2\right), \left(0, - 2\right), \left(1, - 2\right), \left(2, - 2\right) \to 4\text{possible case}$

When $Y = 3 \text{ and } - 3$

\left(0, 3\right)\left(0, - 3\right) \to \text{Total}2\text{possible case}$</p><p>So, total cases will be$7 + 5 + 5 + 4 + 4 + 2 = 27