Applications of Derivatives
Tangents and Normals
Grade 12
Question:
<p>Given curve \(y = x^3 + ax - b\). The slope of the tangent to this curve at \((1, -5)\) is perpendicular to the line \(x - y + 5 = 0\). Find the values of \(a\) and \(b\).</p>
<p>\(a = -4,\ b = 0\)</p>
<p>\(a = 2,\ b = 8\)</p>
<p>\(a = -4,\ b = 2\)</p>
<p>\(a = 2,\ b = -8\)</p>
Step-by-Step Solution
Key Concept: The slope of the tangent at a point equals the derivative evaluated at that point. Use the perpendicularity condition (product of slopes = -1) to find the derivative value, then use the point condition to find constants.
<p><strong>Step 1: Find the slope of the given line.</strong></p><p>Line: x - y + 5 = 0 ⟹ y = x + 5</p><p>Slope of given line = 1</p><p><strong>Step 2: Find slope of tangent at (1, -5).</strong></p><p>Since tangent is perpendicular to the given line:</p><p>m₁ × m₂ = -1</p><p>m_tangent × 1 = -1</p><p>m_tangent = -1</p><p><strong>Step 3: Use derivative condition.</strong></p><p>y = x³ + ax - b</p><p>dy/dx = 3x² + a</p><p>At x = 1: dy/dx = 3(1)² + a = 3 + a</p><p>Setting this equal to -1:</p><p>3 + a = -1</p><p>a = -4</p><p><strong>Step 4: Use point condition.</strong></p><p>Point (1, -5) lies on the curve:</p><p>-5 = (1)³ + a(1) - b</p><p>-5 = 1 + (-4) - b</p><p>-5 = -3 - b</p><p>b = 2</p><p><strong>∴ Answer: a = -4, b = 2</strong></p>
Correct Answer: A