Let θ be the acute angle between the tangents to the ellipse x²/9 + y²/1 = 1 and the circle x² + y² = 3 at their point of intersection in the first quadrant. Then tanθ is equal to :

Solution

Maths sir · 4 steps

Find the meeting point first, then take both slopes there and use the angle-between-lines formula.

1. Point of intersection. From the circle, y² = 3 - x². Putting this in the ellipse gives x²/9 + 3 - x² = 1, so 8x²/9 = 2 and x² = 9/4. In the first quadrant the point is (3/2, √3/2).
2. Slope on ellipse. Differentiating x²/9 + y² = 1 gives dy/dx = -x/9y. At the point, m₁ = -(3/2)/(9 × √3/2) = -1/3√3.
3. Slope on circle. Differentiating x² + y² = 3 gives dy/dx = -x/y. At the point, m₂ = -(3/2)/(√3/2) = -√3.

   tan θ = |(m₁ - m₂)/(1 + m₁ m₂)| = (√3 - 1/3√3)/(1 + 1/3) = (8/3√3)/(4/3) = 2/√3

4. So. The acute angle between the tangents has tan θ = 2/√3, which is option (b).

The idea to keep. Angle between two curves means angle between their tangents at the common point. Get the point, get both slopes by implicit differentiation, then use |(m₁ - m₂)/(1 + m₁ m₂)|.

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