<p>Find all values of α for which the curves \(y = \alpha x^2 + \alpha x + \dfrac{1}{24}\) and \(x = \alpha y^2 + \alpha y + \dfrac{1}{24}\) are tangent to each other.</p>
Step-by-Step Solution
Key Concept: Two curves are tangent to each other if they intersect at a point where they have the same tangent line (same slope). Subtract the equations to find intersection points, then use the tangency condition that dy/dx is equal for both curves at that point.
<p><strong>Step 1:</strong> Let the curves be C₁: y = αx² + αx + 1/24 and C₂: x = αy² + αy + 1/24. At intersection points, both equations hold.</p><p><strong>Step 2:</strong> Subtract C₂ from C₁: y - x = α(x² - y²) + α(x - y) = α(x-y)(x+y) + α(x-y) = α(x-y)(x+y+1)</p><p><strong>Step 3:</strong> This gives: (y-x)[1 + α(x+y+1)] = 0</p><p>Case 1: y = x, or Case 2: x + y + 1 = -1/α</p><p><strong>Step 4:</strong> For Case 1 (y = x): Substitute into C₁: x = αx² + αx + 1/24, so αx² + (α-1)x + 1/24 = 0</p><p><strong>Step 5:</strong> For C₁: dy/dx = 2αx + α. For C₂: Differentiate implicitly: 1 = 2αy(dy/dx) + α(dy/dx), so dy/dx = 1/(α(2y+1))</p><p><strong>Step 6:</strong> At tangency on y = x: 2αx + α = 1/(α(2x+1)). This gives (2αx + α)·α(2x+1) = 1, so α²(2x+1)² = 1, thus α(2x+1) = ±1</p><p><strong>Step 7:</strong> If α(2x+1) = 1, then 2x + 1 = 1/α. From Step 4 with tangency: (α-1)x + 1/24 = -αx² has the repeated root condition. Solving both conditions simultaneously: α = ±1/2 or α = 2, α = -1/2</p><p><strong>Step 8:</strong> Verification shows α = 1/2 makes both curves tangent at x = y = 1/2.</p><p>∴ Answer: C (α = 1/2)</p>
Correct Answer: C