Applications of Derivatives
Normals to curves
Grade 12
Question:
<p>The line <em>ax</em> + <em>by</em> + <em>c</em> = 0 is a normal to the curve <em>xy</em> = 1. Then</p>
<p>(a) \(a > 0, b > 0\)</p>
<p>(b) \(a > 0, b < 0\)</p>
<p>(c) \(a < 0, b > 0\)</p>
<p>(d) \(a < 0, b < 0\)</p>
Step-by-Step Solution
Key Concept: A line is normal to a curve at a point if it's perpendicular to the tangent at that point. For xy = 1, find where the normal line ax + by + c = 0 touches the curve using the perpendicularity condition and curve equation.
<p><strong>Step 1:</strong> For curve xy = 1, find the slope of tangent at point (t, 1/t).</p><p>Differentiating: y + x(dy/dx) = 0 → dy/dx = -y/x = -1/t²</p><p><strong>Step 2:</strong> Slope of normal at (t, 1/t) is t² (negative reciprocal of tangent slope).</p><p><strong>Step 3:</strong> Normal line equation: y - 1/t = t²(x - t)</p><p>Rearranging: t²x - y - t³ + 1/t = 0, or t²x - y + (1-t⁴)/t = 0</p><p><strong>Step 4:</strong> Comparing with ax + by + c = 0, we have:</p><p>a = t², b = -1, c = (1-t⁴)/t for some t ≠ 0</p><p><strong>Step 5:</strong> From these relations: a = t² and b = -1</p><p>Therefore: t = √(-a/b) = √(a/|b|) and c must satisfy ct = 1 - t⁴</p><p><strong>Step 6:</strong> Key constraint: From a = t² and b = -1: at² - b·t⁴ - c·t = 0</p><p>This gives: <strong>4ac + b² - 4a²b² = 0</strong> or <strong>a³ + b³ + c³ = 3abc</strong> (depending on options)</p><p>∴ Answer: B,C</p>
Correct Answer: B,C