<p>A common tangent is drawn to the parabolas \(y = x^2 - x + 1\) and \(y = x^2 - 3x + 1\). If the slope of the common tangent is \(m\), find \(|m|\).</p>
Step-by-Step Solution
Key Concept: A common tangent to two curves must have the same slope at its point of contact with each curve. Use the condition that the tangent line from a point satisfies both the curve equation and its derivative simultaneously for both parabolas.
<p><strong>Step 1:</strong> For parabola y = x² - x + 1, tangent at point (a, a² - a + 1) has slope m = 2a - 1.</p><p>Tangent line: y - (a² - a + 1) = (2a - 1)(x - a)</p><p>Simplifying: y = (2a - 1)x - a² + 1</p><p><strong>Step 2:</strong> For parabola y = x² - 3x + 1, tangent at point (b, b² - 3b + 1) has slope m = 2b - 3.</p><p>Tangent line: y = (2b - 3)x - b² + 1</p><p><strong>Step 3:</strong> For a common tangent, both slopes and y-intercepts must be equal:</p><p>2a - 1 = 2b - 3 → a = b - 1</p><p>-a² + 1 = -b² + 1 → a² = b²</p><p><strong>Step 4:</strong> From a² = b² and a = b - 1:</p><p>(b - 1)² = b²</p><p>b² - 2b + 1 = b²</p><p>-2b + 1 = 0 → b = 1/2</p><p><strong>Step 5:</strong> Then a = 1/2 - 1 = -1/2</p><p>Slope m = 2a - 1 = 2(-1/2) - 1 = -2</p><p>∴ |m| = 2</p>
Correct Answer: 2