Applications of Derivatives
Orthogonal intersection of curves
Grade 12
Question:
<p>Given \(y^2 = 6x\) and \(9x^2 + by^2 = 16\). If both curves intersect each other at right angles, find the value of \(b\).</p>
<p>\(\dfrac{3}{2}\)</p>
<p>\(\dfrac{9}{2}\)</p>
<p>\(\dfrac{27}{4}\)</p>
<p>\(6\)</p>
Step-by-Step Solution
Key Concept: Two curves intersect at right angles when the product of their slopes at the intersection point equals -1. Use implicit differentiation to find dy/dx for both curves, then apply the orthogonality condition m₁·m₂ = -1.
<p><strong>Step 1:</strong> Find dy/dx for curve y² = 6x using implicit differentiation.</p><p>2y(dy/dx) = 6 → <strong>dy/dx = 3/y</strong></p><p><strong>Step 2:</strong> Find dy/dx for curve 9x² + by² = 16 using implicit differentiation.</p><p>18x + 2by(dy/dx) = 0 → <strong>dy/dx = -9x/(by)</strong></p><p><strong>Step 3:</strong> Apply orthogonality condition m₁·m₂ = -1 at intersection point.</p><p>(3/y) · (-9x/by) = -1</p><p>-27x/(by²) = -1</p><p><strong>27x = by²</strong> ... (i)</p><p><strong>Step 4:</strong> From y² = 6x, substitute into equation (i).</p><p>27x = b(6x)</p><p>27x = 6bx</p><p><strong>27 = 6b</strong></p><p><strong>b = 27/6 = 9/2 = 4.5</strong></p><p>∴ Answer: <strong>b = 9/2</strong> (or 4.5)</p>
Correct Answer: B