Question

The set $S$ and $E$ are defined as given below:
$S=\left\{\right(x,y):|x-3|<1$ and $|y-3|<1\}$
$E=\left\{\right(x,y):4x^2+9y^2-32x-54y+109\le 0\}$
then $S\subset E$

Answer 1

Ans. $True$.
We first observe that $|x-3|<1$
$\Rightarrow -1<x-3<1$
$\Rightarrow 2<x<4$
Similarly $|y-3|<1\to 2<y<4$.
Thus $S$ consists of all points inside the square bounded by the lines $x=2$, $x=4$, $y=2$ and $y=4$.
This square region is shown in the figure by vertical lines.
Again $4x^2+9y^2-32x-54y+109$
$=4(x^2-8x+16)+9(y^2-6y+9)-36$
$=4(x-4{)}^2+9(y-3{)}^2-36$.
Hence $4x^2+9y^2-32x-54y+109\le 0$
$\Rightarrow 4(x-4{)}^2+9(y-3{)}^2-36\le 0$
$\Rightarrow \frac{{(x-4)}^2}{9}+\frac{{(y-3)}^2}{4}\le 1$.
Thus the set $E$ consists of all points within and on the ellipse whose centre is $(4,3)$ and semi major and minor axes are $3$ and $2$ respectively. This region is shown by dots in the diagram.
We now show that $S\subset E$.