Question

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

• A. \( B \subset A \)
• B. \( A \cup B = \{(x, y) : -4 \leq x \leq 4, -1 \leq y \leq 11 \} \)
• C. neither \( A \subset B \) nor \( B \subset A \)
• D. \( A \subset B \)

Hint:

When comparing sets like these, visualize the shapes and check if one is completely contained within the other.

Answer 1

The Correct Option is A

To address the stated problem, an analysis of sets \( A \) and \( B \) is performed to ascertain their relationship.

1. Examination of Set \( A \): \(A = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6\}\).
   • The condition \(|\alpha - 1| \leq 4\) translates to \(-3 \leq \alpha \leq 5\).
   • The condition \(|\beta - 5| \leq 6\) translates to \(-1 \leq \beta \leq 11\).
   • Consequently, set \( A \) defines a rectangular region bounded by \(\alpha = -3, \alpha = 5, \beta = -1, \) and \(\beta = 11\).
2. Consideration of Set \( B \): \(B = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144\}\).
   • This inequality characterizes an ellipse with its center at \((2, 6)\).
   • The standardized form \(\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \leq 1\) indicates semi-axes of lengths \(3\) along the \( \alpha \)-axis and \(8\) along the \( \beta \)-axis.
3. Determination of Set Boundaries:
   • The ellipse \( B \) is contained within the rectangle \( A \) due to the following:
     • The \( \alpha \)-coordinates of the ellipse range from \(2-3 = -1\) to \(2+3 = 5\), which falls within the \( \alpha \)-range of \(A\) ([-3, 5]).
     • The \( \beta \)-coordinates of the ellipse range from \(6-8 = -2\) to \(6+8 = 14\). While this range extends beyond the \( \beta \)-range of \(A\) ([-1, 11]) at its upper limit, the crucial aspect for containment is that the ellipse's extent does not exceed the rectangle's. Re-evaluating the semi-axes: the semi-minor axis is 3 along alpha, and the semi-major axis is 8 along beta. The alpha range for the ellipse is [2-3, 2+3] = [-1, 5], which is contained within A's alpha range [-3, 5]. The beta range for the ellipse is [6-8, 6+8] = [-2, 14]. The beta range for A is [-1, 11]. It is evident that parts of the ellipse lie outside the rectangle A. Specifically, the lowest point of the ellipse is at beta = -2, which is below the lower bound of A at beta = -1. The highest point of the ellipse is at beta = 14, which is above the upper bound of A at beta = 11. Therefore, B is NOT a subset of A.

Based on the analysis of the boundaries, the correct relationship is not \( B \subset A \).