<p>The angle between the curves \(x^3 - 3xy^2 + 2 = 0\) and \(3x^2y - y^3 - 2 = 0\) is:</p>
Step-by-Step Solution
Key Concept: The angle between two curves equals the angle between their tangent lines at the intersection point, found using the formula tan θ = |(m₁ - m₂)/(1 + m₁m₂)| where m₁ and m₂ are slopes from implicit differentiation. Notice the two equations are related by swapping x and y variables—they're reflections across y = x.
<p><strong>Step 1:</strong> Find intersection point. From x³ - 3xy² + 2 = 0 and 3x²y - y³ - 2 = 0, add the equations: x³ - 3xy² + 3x²y - y³ = 0, which factors as (x - y)(x² + y² + 4xy) = 0. For real intersection, use x = y: x³ - 3x³ + 2 = 0 → -2x³ + 2 = 0 → x = 1. Intersection point is (1, 1).</p><p><strong>Step 2:</strong> Find slopes using implicit differentiation. For C₁: x³ - 3xy² + 2 = 0: 3x² - 3y² - 6xy(dy/dx) = 0 → dy/dx = (x² - y²)/(2xy). At (1,1): m₁ = 0/2 = 0.</p><p><strong>Step 3:</strong> For C₂: 3x²y - y³ - 2 = 0: 6xy + 3x²(dy/dx) - 3y²(dy/dx) = 0 → dy/dx = -6xy/(3x² - 3y²) = -2xy/(x² - y²). At (1,1): m₂ = -2/0 → undefined (vertical tangent).</p><p><strong>Step 4:</strong> One tangent is horizontal (m₁ = 0) and the other is vertical (m₂ = ∞), so they are perpendicular. The angle between the curves is <strong>90° or π/2 radians</strong>.</p><p>∴ Answer: A</p>
Correct Answer: A