<p>Given \(y = \dfrac{x}{x^2 - 3}\). The curve passes through the point \((\alpha, \beta)\) and the tangent at \((\alpha, \beta)\) is parallel to the line \(2x + 6y - 11 = 0\). Then \(|6\alpha + 2\beta|\) equals:</p>
Step-by-Step Solution
Key Concept: The tangent line's slope equals the derivative at the point of tangency. Since the tangent is parallel to 2x + 6y - 11 = 0, its slope is -1/3, so we set y' = -1/3 and solve for α. Then use the curve equation to find β.
<p><strong>Step 1:</strong> Find the derivative of y = x/(x² - 3).</p><p>Using quotient rule: y' = [(x² - 3) - x(2x)]/(x² - 3)² = (x² - 3 - 2x²)/(x² - 3)² = (-x² - 3)/(x² - 3)²</p><p><strong>Step 2:</strong> Find the slope of the given line 2x + 6y - 11 = 0.</p><p>Rewriting: 6y = -2x + 11, so y = -x/3 + 11/6. Slope = -1/3</p><p><strong>Step 3:</strong> Set y'(α) = -1/3 (tangent parallel to given line).</p><p>(-α² - 3)/(α² - 3)² = -1/3</p><p>3(-α² - 3) = -(α² - 3)²</p><p>-3α² - 9 = -(α⁴ - 6α² + 9)</p><p>-3α² - 9 = -α⁴ + 6α² - 9</p><p>α⁴ - 9α² = 0</p><p>α²(α² - 9) = 0</p><p>So α = 0, 3, or -3</p><p><strong>Step 4:</strong> Check which value is valid and find β using y = x/(x² - 3).</p><p>If α = 0: β = 0/(0 - 3) = 0, so point is (0, 0). Check: y'(0) = -3/9 = -1/3 ✓</p><p>If α = 3: β = 3/(9 - 3) = 1/2, point is (3, 1/2). Check: y'(3) = -12/36 = -1/3 ✓</p><p>If α = -3: β = -3/(9 - 3) = -1/2, point is (-3, -1/2). Check: y'(-3) = -12/36 = -1/3 ✓</p><p><strong>Step 5:</strong> Calculate |6α + 2β| for each valid point.</p><p>For (0, 0): |0 + 0| = 0</p><p>For (3, 1/2): |18 + 1| = 19</p><p>For (-3, -1/2): |-18 - 1| = 19</p><p>∴ Answer: A (19)</p>
Correct Answer: A