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

We are given two sets \( A \) and \( B \) defined by: \[ A = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6\} \] This defines a rectangular region where \( \alpha \) lies between \( -3 \) and \( 5 \), and \( \beta \) lies between \( -1 \) and \( 11 \). \[ B = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144\} \] This defines an ellipse with center \( (2, 6) \), semi-major axis 4 along the \( \beta \)-axis, and semi-minor axis 3 along the \( \alpha \)-axis. 
We see that the ellipse \( B \) fits entirely within the rectangle \( A \), meaning that \( B \subset A \).