Animated Solution for Mathematics - Conic Sections: Two sets $A$ and $B$ are as under : $A=\{(a,b)\in \mathbb{R}\times \mathbb{R}:|a-5|<1\text{ and }|b-5|<1\}$; $B=\{(a,b)\in \mathbb{R}\times \mathbb{R}:4(a-6{)}^2+9(b-5{)}^2\le 36\}$. Then :

We are tasked with determining the relationship between two sets, $A$ and $B$, in the Cartesian plane. Set $A$ is defined as:

The condition $|a-5|<1$ is equivalent to $4<a<6$. Similarly, $|b-5|<1$ implies $4<b<6$.

Geometrically, this describes an open square region centered at $(5,5)$ with a side length of $2$. The vertices of this square are $(4,4)$, $(4,6)$, $(6,4)$, and $(6,6)$.

Set $B$ is defined by the inequality:

To visualize this, we divide the entire inequality by $36$:

This simplifies to the standard form of an ellipse:

This ellipse is centered at $(6,5)$ with a semi-major axis of $3$ and a semi-minor axis of $2$.

Convexity and Inclusion

We must determine if $A\subset B$. Since both the square $A$ and the ellipse $B$ are convex sets, we can utilize a powerful geometric property.

A convex set is entirely contained within another convex set if and only if all its extreme points (vertices) lie within the second set. Therefore, we only need to verify the four vertices of the square: $(4,4)$, $(4,6)$, $(6,4)$, and $(6,6)$.

We test each vertex against the inequality $4(a-6{)}^2+9(b-5{)}^2\le 36$:

For vertex $(4,4)$: Since $25\le 36$, this vertex is inside the ellipse.

For vertex $(4,6)$: Since $25\le 36$, this vertex is also inside.

For vertices $(6,4)$ and $(6,6)$: Since $9\le 36$, these vertices are also contained within the ellipse.

We have rigorously verified that all four vertices of the square $A$ lie within the ellipse $B$. Because both sets are convex, this confirms that the entire square $A$ is a subset of the ellipse $B$.

The final result is $A\subset B$.