Two sets $A$ and $B$ are as under: $A=\left\{\left(a, b\right)\in R\times R :\left|a-5\right|<1 \text{and} \left|b-5\right|<1\right\};$

$B=\left\{\left(a, b\right)\in R\times R:4{\left(a-6\right)}^2+9{\left(b-5\right)}^2\le 36\right\}.$ Then :

Given sets, $A=\left\{\left(a, b\right)\in R\times R :\left|a-5\right|<1 \text{and} \left|b-5\right|<1\right\}$

$A=\left\{\left(a, b\right)\in R\times R :-1<a-5<1 \text{and} -1<b-5<1\right\}$

$A=\left\{\left(a, b\right)\in R\times R :4<a<6 \text{and} 4<b<6\right\}$

$B=\frac{{\left(a-6\right)}^2 }{9}+\frac{{\left(b-5\right)}^2}{4}\le 1$ 

For set $A,$

For corner coordinates of the set $A$is $\left(4,4\right),\left(4,6\right),\left(6,4\right),\left(6,3\right).$

For set $B,$

Now, Put $\left(4,4\right)$in the equation of the set $B,$

$\frac{4}{9}+\frac{1}{4}\le 1,$$\left(4,4\right)$lies in the set $B.$

Similarly, $\left(4,6\right),\left(6,4\right),\left(6,3\right)$also lies in the set $B.$

So, $A\subset B$.