<p>The shortest distance between the line \(y = x\) and the curve \(y^2 = x - 2\) is __________ (up to four decimal places).</p>
Step-by-Step Solution
Key Concept: The shortest distance occurs along the normal to the curve, which must be perpendicular to the line y=x. Find the point on the parabola where the tangent is parallel to y=x, then calculate the perpendicular distance from that point to the line.
<p><strong>Step 1:</strong> Parametrize the curve. Let y² = x - 2, so x = y² + 2. A point on the curve is P(y² + 2, y).</p><p><strong>Step 2:</strong> Find dy/dx. From y² = x - 2: 2y(dy/dx) = 1, so dy/dx = 1/(2y). For the tangent to be parallel to y = x (slope = 1), we need 1/(2y) = 1, giving y = 1/2.</p><p><strong>Step 3:</strong> When y = 1/2, x = (1/2)² + 2 = 1/4 + 2 = 9/4. The critical point is P(9/4, 1/2).</p><p><strong>Step 4:</strong> Distance from point (9/4, 1/2) to line x - y = 0 is: d = |9/4 - 1/2|/√2 = |9/4 - 2/4|/√2 = (7/4)/√2 = 7/(4√2) = 7√2/8.</p><p><strong>Step 5:</strong> Calculate: 7√2/8 = 7(1.41421356...)/8 ≈ 9.89949.../8 ≈ 1.2374.</p><p>∴ Answer: <strong>1.2374</strong></p>
Correct Answer: 1