An angle of intersection of the curves, x²/a² + y²/b² = 1 and x² + y² = ab, a > b, is :

Solution

Maths sir · 4 steps

Find where the curves meet, take both tangent slopes there, and use the angle-between-lines formula.

1. Meeting point. Put y² = ab - x² into the ellipse. This gives x² = (a² b)/(a+b) and y² = ab²/(a+b), so x²/y² = a/b. By symmetry, the point in the first quadrant is enough, and there x/y = √(a/b).
2. Two tangent slopes. For the ellipse, m₁ = -(b² x)/(a² y) = -b²/a²√(a/b). For the circle, m₂ = -x/y = -√(a/b). Their product is m₁ m₂ = b²/a² × a/b = b/a.

   tan θ = |(m₁ - m₂)/(1 + m₁ m₂)| = √(a/b)(1 - b²/a²)/(1 + b/a) = (√(a/b) × (a-b)(a+b)/a²)/((a+b)/a)

3. Simplify. The factor (a+b) cancels, leaving tan θ = √(a/b) × (a-b)/a = (a-b)/√ab. Since a > b, this is positive, so θ is acute.
4. So. The angle of intersection is tan⁻¹((a-b)/√ab), which is option (c).

The idea to keep. The angle between two curves is the angle between their tangents at a common point. Find that point, get both slopes, then use tan θ = |(m₁ - m₂)/(1 + m₁ m₂)|.

Answer from the NTA final answer key. Solution by PrepWiser.