Applications of Derivatives
Minimum Distance from Origin to Curve
Grade 12
Question:
<p>From a point on the curve \(y = x^2 - 4\), the minimum distance from the origin is:</p>
<p>\(\dfrac{\sqrt{15}}{2}\)</p>
<p>\(\dfrac{\sqrt{13}}{2}\)</p>
<p>\(\dfrac{\sqrt{17}}{2}\)</p>
<p>\(\dfrac{\sqrt{11}}{2}\)</p>
Step-by-Step Solution
Key Concept: The minimum distance from origin to a point (x, x²-4) on the curve occurs when the line from origin to that point is perpendicular to the tangent line at that point. This means the position vector must be parallel to the normal vector.
<p><strong>Step 1:</strong> Let point P(x, x²-4) lie on the curve y = x² - 4. Distance from origin: d² = x² + (x²-4)²</p><p><strong>Step 2:</strong> For minimum distance, the line OP must be perpendicular to the tangent at P. Slope of tangent = dy/dx = 2x. Slope of OP = (x²-4)/x</p><p><strong>Step 3:</strong> Perpendicularity condition: (2x) · (x²-4)/x = -1 → 2(x²-4) = -1 → 2x² - 8 = -1 → x² = 7/2</p><p><strong>Step 4:</strong> At x² = 7/2: y = 7/2 - 4 = -1/2. So d² = 7/2 + 1/4 = 14/4 + 1/4 = 15/4</p><p><strong>Step 5:</strong> d = √(15/4) = √15/2</p><p>∴ Answer: A (√15/2)</p>
Correct Answer: A