Question

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