The angle between the tangents drawn from the point (1,4) to the parabola is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
The chord of contact from (1,4) to y² = 4x is T = 0: 4y = 2(x+1), i.e. x = 2y−1.
Substituting into y² = 4x: y² = 4(2y−1) → y²−8y+4 = 0, giving slopes m₁, m₂ of the two tangents via y = mx + 1/m. From the tangent condition, the two values of m satisfy m₁+m₂ = 4 and m₁m₂ = 1 (using Vieta's on the tangent equation m²·1 − 4m + 1... let me redo cleanly).
Tangent to y²=4x is y = mx + 1/m. Passing through (1,4): 4 = m + 1/m → m² − 4m + 1 = 0. So m₁+m₂ = 4, m₁m₂ = 1.
tan θ = |m₁−m₂|/|1+m₁m₂| = √((m₁+m₂)²−4m₁m₂) / (1+m₁m₂) = √(16−4)/(1+1) = √12/2 = √3.
Therefore θ = π/3.
Substituting into y² = 4x: y² = 4(2y−1) → y²−8y+4 = 0, giving slopes m₁, m₂ of the two tangents via y = mx + 1/m. From the tangent condition, the two values of m satisfy m₁+m₂ = 4 and m₁m₂ = 1 (using Vieta's on the tangent equation m²·1 − 4m + 1... let me redo cleanly).
Tangent to y²=4x is y = mx + 1/m. Passing through (1,4): 4 = m + 1/m → m² − 4m + 1 = 0. So m₁+m₂ = 4, m₁m₂ = 1.
tan θ = |m₁−m₂|/|1+m₁m₂| = √((m₁+m₂)²−4m₁m₂) / (1+m₁m₂) = √(16−4)/(1+1) = √12/2 = √3.
Therefore θ = π/3.
[AIEEE 2004]