Let A = (x,y) : |x-1|4 and |y-5|6 and B = (x,y) : 16(x-2)^2+9(y-6)^2144.

Question

Let $A = \{(x,y) : |x-1|\leq4 	ext{ and } |y-5|\leq6\}$ and $B = \{(x,y) : 16(x-2)^2+9(y-6)^2\leq144\}$. Then:

Accepted Answer

Set $A$: $|x-1|\leq4 \Rightarrow -3\leq x\leq5$; $|y-5|\leq6 \Rightarrow -1\leq y\leq11$. So $A$ is the rectangle $[-3,5]\times[-1,11]$.

Set $B$: $16(x-2)^2+9(y-6)^2\leq144$, i.e.
$$\frac{(x-2)^2}{9}+\frac{(y-6)^2}{16}\leq1$$
This is a filled ellipse centered at $(2,6)$ with semi-axis 3 in the $x$-direction and 4 in the $y$-direction, so its bounding box is $[-1,5]\times[2,10]$.

Since $[-1,5]\times[2,10] \subseteq [-3,5]\times[-1,11]$, and the ellipse lies within its own bounding box, every point of $B$ lies in $A$:
$$B \subseteq A$$

Solutions · 1