<p>The equation of common tangent to the curves \(y^2 = 16x\) and \(xy = -4\) is:</p>
Step-by-Step Solution
Key Concept: A common tangent must satisfy the tangency condition for both curves simultaneously. For the parabola y² = 16x, use the standard tangent form y = mx + 4/m, and verify it's also tangent to the rectangular hyperbola xy = -4 by checking discriminant = 0.
<p><strong>Step 1:</strong> For parabola y² = 16x, the tangent with slope m is: y = mx + 4/m (using standard form y = mx + a/m where a = 4)</p><p><strong>Step 2:</strong> For this line to be tangent to xy = -4, substitute y = mx + 4/m into xy = -4:</p><p>x(mx + 4/m) = -4</p><p>mx² + (4/m)x + 4 = 0</p><p><strong>Step 3:</strong> For tangency, discriminant = 0:</p><p>(4/m)² - 4(m)(4) = 0</p><p>16/m² - 16m = 0</p><p>16/m² = 16m</p><p>1/m² = m</p><p>m³ = 1, so m = 1</p><p><strong>Step 4:</strong> Substituting m = 1 into tangent equation:</p><p>y = x + 4</p><p><strong>Verification:</strong> For y² = 16x: (x+4)² = 16x → x² - 8x + 16 = 0 → (x-4)² = 0 ✓</p><p>For xy = -4: x(x+4) = -4 → x² + 4x + 4 = 0 → (x+2)² = 0 ✓</p><p>∴ Answer: y = x + 4</p>
Correct Answer: A