Parabola
Common Tangent
Grade 11
Question:
<p>The equation of a common tangent to the curves, \(y^2 = 16x\) and \(xy = -4\), is</p>
<p>\(x - y + 4 = 0\)</p>
<p>\(x + y + 4 = 0\)</p>
<p>\(x - 2y + 16 = 0\)</p>
<p>\(2x - y + 2 = 0\)</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, then substitute into xy=-4 to find which value of m makes the line tangent to the hyperbola.
<p><strong>Step 1:</strong> Write the tangent to parabola y²=16x (where 4a=16, so a=4) as: y = mx + 4/m</p><p><strong>Step 2:</strong> Substitute y = mx + 4/m into xy = -4:<br/>x(mx + 4/m) = -4<br/>mx² + (4x/m) = -4<br/>mx² + (4x/m) + 4 = 0<br/>m²x² + 4x + 4m = 0</p><p><strong>Step 3:</strong> For tangency to xy = -4, the discriminant must be zero:<br/>Δ = 16 - 4(m²)(4m) = 0<br/>16 - 16m³ = 0<br/>m³ = 1<br/>m = 1</p><p><strong>Step 4:</strong> Substitute m = 1 into the tangent equation:<br/>y = (1)x + 4/1<br/>y = x + 4<br/>or x - y + 4 = 0</p><p><strong>Verification:</strong> For y²=16x: (x+4)²=16x ⟹ x²+8x+16=16x ⟹ x²-8x+16=0 ⟹ (x-4)²=0 ✓<br/>For xy=-4: x(x+4)=-4 ⟹ x²+4x+4=0 ⟹ (x+2)²=0 ✓</p><p>∴ Answer: A (x - y + 4 = 0)</p>
Correct Answer: A