Parabola
Shortest distance from a line to a parabola
Grade 11
Question:
<p>The shortest distance between line <em>y</em> − <em>x</em> = 1 and curve <em>x</em> = <em>y</em><sup>2</sup> is</p>
<p>\(\dfrac{3\sqrt{2}}{8}\)</p>
<p>\(\dfrac{8}{3\sqrt{2}}\)</p>
<p>\(\dfrac{4}{\sqrt{3}}\)</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 the line. Find the point on the parabola where the tangent is parallel to the given line (slope = 1), then calculate the perpendicular distance from that point to the line.
<p><strong>Step 1:</strong> For the parabola x = y², the tangent at point (t², t) has slope dy/dx = 1/(2y) = 1/(2t).</p><p><strong>Step 2:</strong> For the shortest distance, the normal at the point must be parallel to the given line y − x = 1 (slope = 1). So the tangent slope = −1, giving 1/(2t) = −1, thus t = −1/2.</p><p><strong>Step 3:</strong> The point on the parabola is (1/4, −1/2).</p><p><strong>Step 4:</strong> The perpendicular distance from point (1/4, −1/2) to line x − y − 1 = 0 is:</p><p>d = |1/4 − (−1/2) − 1|/√(1² + 1²) = |1/4 + 1/2 − 1|/√2 = |−1/4|/√2 = 1/(4√2) = √2/8</p><p>∴ Answer: A</p>
Correct Answer: A