Applications of Derivatives
Tangent and Normal
Grade 12
Question:
<p>If the tangent to the curve \(y = \dfrac{x}{x^2 - 3}\), \(x \in \mathbb{R}\), \((x \neq \pm\sqrt{3})\) at a point \((\alpha, \beta) \neq (0, 0)\) on it is parallel to the line \(2x + 6y - 11 = 0\), then</p>
<p>\(|6\alpha + 2\beta| = 19\)</p>
<p>\(|6\alpha + 2\beta| = 9\)</p>
<p>\(|2\alpha + 6\beta| = 19\)</p>
<p>\(|2\alpha + 6\beta| = 11\)</p>
Step-by-Step Solution
Key Concept: Find the derivative to get the slope of the tangent, set it equal to the slope of the given line, then solve for the point (α, β) where the tangent is parallel to 2x + 6y - 11 = 0.
<p><strong>Step 1:</strong> Find the derivative of y = x/(x² - 3).</p><p>Using quotient rule: y' = [(x² - 3)(1) - x(2x)]/(x² - 3)²</p><p>y' = (x² - 3 - 2x²)/(x² - 3)² = (-x² - 3)/(x² - 3)²</p><p><strong>Step 2:</strong> The line 2x + 6y - 11 = 0 has slope m = -2/6 = -1/3.</p><p>For the tangent to be parallel: (-α² - 3)/(α² - 3)² = -1/3</p><p><strong>Step 3:</strong> Cross multiply: 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>α = 0, ±3</p><p><strong>Step 4:</strong> Since (α, β) ≠ (0, 0), we reject α = 0. For α = ±3, note that x ≠ ±√3 is required, so α = ±3 is valid.</p><p>When α = 3: β = 3/(9 - 3) = 3/6 = 1/2</p><p>When α = -3: β = -3/(9 - 3) = -3/6 = -1/2</p><p>∴ The points are (3, 1/2) and (-3, -1/2)</p>
Correct Answer: C