Two sets A and B are as under: [A=(a,b) R R:|a-5|

Question

Two sets $A$ and $B$ are as under:
$$A=\{(a,b)\in\mathbb R\times\mathbb R:|a-5|<1\text{ and }|b-5|<1\}$$
$$B=\{(a,b)\in\mathbb R\times\mathbb R:4(a-6)^2+9(b-5)^2\le36\}.$$

Then

Accepted Answer

Given,

$$4{\left( {a - 6} \right)^2} + 9{\left( {b - 5} \right)^2} \le 36$$

Let $a - 6 = x$ and $b - 5 = y$

$$\therefore\,\,\,\,$$ $$4{x^2} + 9{y^2} \le 36$$

$$ \Rightarrow \,\,\,\,{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$

This is a equation of ellipse.

This ellipse will look like this,

According to set A,

$$\left| {a - 5} \right| < 1$$

as $a - 6 = x$ then $a - 5 = x + 1$

$$\therefore\,\,\,$$ $$\left| {x + 1} \right| < 1$$

$$ \Rightarrow \,\,\, - 1 < x + 1 < 1$$

$$ \Rightarrow \,\,\, - 2 < x < 0$$

$$\left| {b - 5} \right| < 1$$

as $b - 5 = y$

$$\therefore\,\,\,$$ $$\left| y \right| < 1$$

$$ \Rightarrow \,\,\,\, - 1 < y < 1$$

This will look like this,

By combining both those graphs it will look like this,

To check entire set A is inside of B or not put any paint of set A on the equation $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$ and  if this inequality satisfies, then it is inside B.

By looking at the graph you can surely say  (0, 1)  or (0, -1) inside the graph. But to check point ($-2$, 1) or (-2, -1) is inside of the ellipse or not, put on the $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1.$$

Putting (-2, 1) on the inequality,

LHS $$ = {{\left( { - 2} \right){}^2} \over 9} + {{{1^2}} \over 4}$$

$$ = {4 \over 9} + {1 \over 4}$$

$$ = {{25} \over {36}} < 1$$

$$\therefore\,\,\,$$ Inequality holds.

So,  $$\left( { - 2,1} \right)$$ is inside the ellipse.

Similarly by checking we can see $$\left( { - 2, - 1} \right)$$ is also inside the ellipse.

Hence,  we can say entire set A is inside of the set B.

$$\therefore\,\,\,\,$$ A $$ \subset $$ B

Solutions · 1