Animated Solution for Mathematics - Conic Sections: Let $A=\{(\alpha ,\beta )\in \mathbb{R}\times \mathbb{R}:|\alpha -1|\le 4\text{ and }|\beta -5|\le 6\}$ and $B=\{(\alpha ,\beta )\in \mathbb{R}\times \mathbb{R}:16(\alpha -2{)}^2+9(\beta -6{)}^2\le 144\}$. Then

The Rectangular Boundary

We are given the conditions $|\alpha -1|\le 4$ and $|\beta -5|\le 6$. These absolute value inequalities represent distance constraints on the coordinate plane.

For the variable $\alpha$, the inequality $|\alpha -1|\le 4$ expands to:

Adding 1 to all parts, we find that $\alpha$ is constrained to the interval:

Similarly, for the variable $\beta$, the condition $|\beta -5|\le 6$ expands to:

Adding 5 to all parts, we find that $\beta$ is constrained to the interval:

Geometrically, Set $A$ defines a rectangle in the $\alpha$-$\beta$ plane where $\alpha \in [-3,5]$ and $\beta \in [-1,11]$.

The Elliptical Boundary

Set $B$ is defined by the inequality:

To identify the shape, we divide the entire inequality by 144:

This simplifies to the standard form of an ellipse:

The ellipse is centered at $(2,6)$ with semi-axes $a=3$ and $b=4$.

To determine if $B\subset A$, we examine the extreme reach of the ellipse. The horizontal reach is determined by the center $\alpha =2$ plus or minus the semi-axis $a=3$:

The vertical reach is determined by the center $\beta =6$ plus or minus the semi-axis $b=4$:

We now compare these ranges to the boundaries of rectangle $A$: 1. The horizontal range $[-1,5]$ is a subset of $[-3,5]$. 2. The vertical range $[2,10]$ is a subset of $[-1,11]$.

Because the ellipse is a convex shape, if its bounding box fits entirely within the rectangle, the entire ellipse must also be contained within the rectangle.