Applications of Derivatives
Shortest distance between line and curve
Grade 12
Question:
<p>The shortest distance between the line <em>y</em> − <em>x</em> = 1 and the curve <em>x</em> = <em>y</em><sup>2</sup> is</p>
<p>\(\dfrac{3\sqrt{2}}{8}\)</p>
<p>\(\dfrac{2\sqrt{3}}{8}\)</p>
<p>\(\dfrac{3\sqrt{2}}{5}\)</p>
<p>\(\dfrac{\sqrt{3}}{4}\)</p>
Step-by-Step Solution
Key Concept: The shortest distance occurs along the common normal to both the curve and line. For a curve y² = x and line y - x = 1, find the point on the parabola where the normal is parallel to the perpendicular direction of the given line.
<p><strong>Step 1:</strong> For curve x = y², we have dx/dy = 2y, so dy/dx = 1/(2y). The slope of the normal at point (y₀², y₀) is -2y₀.</p><p><strong>Step 2:</strong> The line y - x = 1 has slope 1. For shortest distance, the normal to the curve must be parallel to the perpendicular from the line. The perpendicular to the line has slope -1.</p><p><strong>Step 3:</strong> Set -2y₀ = -1, giving y₀ = 1/2. The point on the curve is (1/4, 1/2).</p><p><strong>Step 4:</strong> Distance from point (1/4, 1/2) to line x - y + 1 = 0 is: d = |1/4 - 1/2 + 1|/√(1² + (-1)²) = |3/4|/√2 = 3/(4√2) = 3√2/8</p><p>∴ Answer: A</p>
Correct Answer: A