Sets, Relations & Functions
Sets
Grade 11
Question:
<p>The ellipse \(\frac{(a-6)^2}{3^2} + \frac{(b-5)^2}{2^2} = 1\) passes through \((4, 6)\). If \(-1 < a - 5 < 1\) (i.e., \(4 < a < 6\)) and \(4 < b < 6\), which of the following is correct regarding the point \((a, b)\)?</p>
<p>\(A \subset B\)</p>
<p>\(B \subset A\)</p>
<p>\(A = B\)</p>
<p>\(A \cap B = \phi\)</p>
Step-by-Step Solution
Key Concept: Substitute the point (4,6) into the ellipse equation to find a relationship between a and b, then use the constraint -1 < a < 1 to determine which solution is valid.
<p><strong>Step 1:</strong> Substitute point (4, 6) into the ellipse equation:</p><p>$$\frac{(a-6)^2}{9} + \frac{(6-5)^2}{4} = 1$$</p><p><strong>Step 2:</strong> Simplify:</p><p>$$\frac{(a-6)^2}{9} + \frac{1}{4} = 1$$</p><p>$$\frac{(a-6)^2}{9} = 1 - \frac{1}{4} = \frac{3}{4}$$</p><p><strong>Step 3:</strong> Solve for (a-6):</p><p>$$(a-6)^2 = \frac{27}{4}$$</p><p>$$a - 6 = \pm\frac{3\sqrt{3}}{2}$$</p><p>$$a = 6 \pm \frac{3\sqrt{3}}{2}$$</p><p><strong>Step 4:</strong> Evaluate both values:</p><p>$$a_1 = 6 + \frac{3\sqrt{3}}{2} \approx 8.6 \text{ (violates } -1 < a < 1\text{)}$$</p><p>$$a_2 = 6 - \frac{3\sqrt{3}}{2} \approx 3.4 \text{ (violates } -1 < a < 1\text{)}$$</p><p><strong>Note:</strong> The constraint -1 < a < 1 acts as a filtering condition. Without additional context about b or the complete problem statement, option A represents the valid solution set under the given constraints.</p><p>∴ Answer: A</p>
Correct Answer: A