Question

Let $A$ and $B$ be two sets defined as given below: $A=\{(x,y):|x-3|<1and|y-3|<1\}$ $B=\{(x,y):4x^2+9y^2-32x-54y+109\le 0\}$ Then,

• A. $A\subset B$
• B. $B\subset A$
• C. $A=B$
• D. None of these

Answer 1

Answer: (A)

We have, $|x-3|<1$ and $|y-3|<1$
$\Rightarrow$ $2<x<4$ and $2<y<4$
Thus, $A$ is the set of all points $(x,y)$
lying inside the square formed by the lines
$x=2,x=4,y=2$ and $y=4$ .
Now, $4x^2+9y^2-32x-54y+109\le 0$
$\Rightarrow$ $4(x^2-8x)+9(y^2-6y)+109\le 0$
$\Rightarrow$ $4{(x-4)}^2+9{(y-3)}^2\le 36$
$\Rightarrow$ $\frac{{(x-4)}^2}{3^2}+\frac{{(y-3)}^2}{2^2}\le 1$
Thus, $B$ is the set of all points lying inside the ellipse having its centre at $(4,3)$ and of lengths major and minor axes are 3 and 2 units. It can be easily seen by drawing the graphs of two regions that $A\subset B$ .