<p>Let line (1) be \(x - y + 1 = 0\) and line (2) be the tangent to the curve \(x = y^2\). The shortest distance between the line and the curve is:</p>
<p>\(\dfrac{3\sqrt{2}}{8}\)</p>
<p>\(\dfrac{3}{4\sqrt{2}}\)</p>
<p>\(\dfrac{\sqrt{2}}{4}\)</p>
<p>\(\dfrac{1}{2\sqrt{2}}\)</p>
Step-by-Step Solution
Key Concept: The shortest distance from a line to a curve occurs along the perpendicular from the line to the curve. For a parabola, find the point where the tangent is parallel to the given line, then calculate the perpendicular distance from that point to the line.
<p><strong>Step 1:</strong> The parabola is x = y². For the shortest distance, the tangent to the parabola must be parallel to the given line x - y + 1 = 0 (or y = x + 1).</p><p><strong>Step 2:</strong> The slope of line (1) is 1. For the parabola x = y², we have dx/dy = 2y, so dy/dx = 1/(2y). Setting dy/dx = 1 gives 1/(2y) = 1, so y = 1/2.</p><p><strong>Step 3:</strong> At y = 1/2, we get x = (1/2)² = 1/4. The point on the parabola is P(1/4, 1/2).</p><p><strong>Step 4:</strong> The perpendicular distance from point P(1/4, 1/2) to line x - y + 1 = 0 is:</p><p>d = |1/4 - 1/2 + 1|/√(1² + (-1)²) = |1/4 - 1/2 + 1|/√2 = |3/4|/√2 = 3/(4√2) = 3√2/8</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A