Question
Let $A = \{(\alpha, \beta) \varepsilon R\times R: |\alpha-1|\leq 4$ and $|\beta-5|\leq6\}$ and $B = \{(\alpha, \beta) \in R \times R : 16(\alpha-2)^2+9(\beta-6)^2 \leq 144 \}$. Then
The correct answer is
B ⊂ A
Analyzing Set A Definition
Set A is defined as $A = \{(\alpha, \beta) \varepsilon R\times R: |\alpha-1|\leq 4 \text{ and } |\beta-5|\leq6\}$.
- The condition $|\alpha-1|\leq 4$ implies: $-4 \leq \alpha-1 \leq 4$ Adding 1 to all parts gives: $1-4 \leq \alpha \leq 1+4$ $ -3 \leq \alpha \leq 5$
- The condition $|\beta-5|\leq6$ implies: $-6 \leq \beta-5 \leq 6$ Adding 5 to all parts gives: $5-6 \leq \beta \leq 5+6$ $ -1 \leq \beta \leq 11$
- Therefore, Set A represents a rectangular region defined by $-3 \leq \alpha \leq 5$ and $-1 \leq \beta \leq 11$.
Analyzing Set B Definition
Set B is defined as $B = \{(\alpha, \beta) \in R \times R : 16(\alpha-2)^2+9(\beta-6)^2 \leq 144 \}$.
To understand the shape of Set B, we can rewrite the inequality:
$ \frac{16(\alpha-2)^2}{144} + \frac{9(\beta-6)^2}{144} \leq 1 $ $ \frac{(\alpha-2)^2}{9} + \frac{(\beta-6)^2}{16} \leq 1 $This is the standard inequality for an ellipse centered at $(2, 6)$.
- The semi-axes are $a = \sqrt{9} = 3$ (in the $\alpha$ direction) and $b = \sqrt{16} = 4$ (in the $\beta$ direction).
- The range of $\alpha$ values within the ellipse is determined by the center plus/minus the semi-axis in the $\alpha$ direction: $ 2 - 3 \leq \alpha \leq 2 + 3 $ $ -1 \leq \alpha \leq 5 $
- The range of $\beta$ values within the ellipse is determined by the center plus/minus the semi-axis in the $\beta$ direction: $ 6 - 4 \leq \beta \leq 6 + 4 $ $ 2 \leq \beta \leq 10 $
- Therefore, Set B represents the region inside and on the boundary of an ellipse with $\alpha$ values ranging from -1 to 5 and $\beta$ values ranging from 2 to 10.
Comparing Sets A and B
We need to determine the relationship between Set A (a rectangle) and Set B (an ellipse).
- Check if B ⊂ A: Does every point in B also lie in A?
- Range of $\alpha$ in B: $[-1, 5]$. Range of $\alpha$ in A: $[-3, 5]$. Since $[-1, 5]$ is contained within $[-3, 5]$, the $\alpha$ condition for B is satisfied within A.
- Range of $\beta$ in B: $[2, 10]$. Range of $\beta$ in A: $[-1, 11]$. Since $[2, 10]$ is contained within $[-1, 11]$, the $\beta$ condition for B is satisfied within A.
- Check if A ⊂ B: Does every point in A also lie in B?
The rectangle A extends from $\alpha=-3$ to $\alpha=5$ and $\beta=-1$ to $\beta=11$. The ellipse B covers $\alpha$ from -1 to 5 and $\beta$ from 2 to 10. Points in A but outside these ranges of B (e.g., $\alpha=-3$) are not in B. Therefore, A is not a subset of B.
Conclusion
Based on the analysis, Set B is entirely contained within Set A.
Final Answer: The final answer is $\boxed{B \subset A}$
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