Let A = \(alpha ,beta)in mathbfRtimes mathbfR:|alpha -1|leq 4 text and |beta -5|leq 6\$A = \{(\alpha ,\beta)\in \mathbf{R}\times \mathbf{R}:|\alpha -1|\leq 4 \text{ and } |\beta -5|\leq 6\}$ and B = \(alpha , beta) in mathbfR times mathbfR: 16 (alpha - 2)^2 + 9 (beta - 6)^2 leq 144\$B = \{(\alpha , \beta) \in \mathbf{R} \times \mathbf{R}: 16 (\alpha - 2)^{2} + 9 (\beta - 6)^{2} \leq 144\}$. Then

### Related Formula An ellipse equation is structured as: frac(x-h)^2a^2 + frac(y-k)^2b^2 leq 1$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} \leq 1$$ ### Core Logic Analyzing set A$A$: |alpha - 1| le 4 implies -4 le alpha - 1 le 4 implies -3 le alpha le 5$$|\alpha - 1| \le 4 \implies -4 \le \alpha - 1 \le 4 \implies -3 \le \alpha \le 5$$ |beta - 5| le 6 implies -6 le beta - 5 le 6 implies -1 le beta le 11$$|\beta - 5| \le 6 \implies -6 \le \beta - 5 \le 6 \implies -1 \le \beta \le 11$$ Thus, region A$A$ forms a rectangle bounded between x in [-3, 5]$x \in [-3, 5]$ and y in [-1, 11]$y \in [-1, 11]$. Analyzing set B$B$: 16(alpha - 2)^2 + 9(beta - 6)^2 le 144$$16(\alpha - 2)^2 + 9(\beta - 6)^2 \le 144$$ Dividing by 144: frac(alpha - 2)^29 + frac(beta - 6)^216 le 1$$\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \le 1$$ This represents the interior and boundary of an ellipse centered at (2, 6)$(2, 6)$ with semi-minor axis a = 3$a = 3$ and semi-major axis b = 4$b = 4$. ### Step 1: Spatial Inclusion Check Let's check the extreme horizontal and vertical extents of the ellipse B$B$: Horizontal extent: x in [2 - 3, 2 + 3] = [-1, 5]$x \in [2 - 3, 2 + 3] = [-1, 5]$ Vertical extent: y in [6 - 4, 6 + 4] = [2, 10]$y \in [6 - 4, 6 + 4] = [2, 10]$ Comparing with the boundaries of rectangle A$A$ (x in [-3, 5]$x \in [-3, 5]$ and y in [-1, 11]$y \in [-1, 11]$): [-1, 5] subseteq [-3, 5]$$[-1, 5] \subseteq [-3, 5]$$ [2, 10] subseteq [-1, 11]$$[2, 10] \subseteq [-1, 11]$$

Since all points of the ellipse lie perfectly inside the rectangular region, we conclusively find that B subset A$B \subset A$. ### Pattern Recognition A bounding box check (finding h pm a$h \pm a$ and k pm b$k \pm b$) for conics is the fastest analytical shortcut to verify set inclusion without plotting extensive coordinates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sets, Relations and Functions Class 11 Mathematics: Conic Sections