From a point on the line x - y + 2 = 0 tangents are drawn to the hyperbola x 2 6 - y 2 2 = 1 such that the chord of contact passes through a fixed point λ , μ . Then, μ - λ is equal to

Step-by-step solution

Let the point be α , β ⇒ β = α + 2 Chord of contact of the hyperbola is T = 0 ⇒ α x 6 - β y 2 = 1 ⇒ α x 6 - α + 2 y 2 = 1 ⇒ α x 6 - y 2 - 2 y 2 + 1 = 0 Since, this passes through the fixed point λ , μ ∴ λ 6 - μ 2 = 0 and 2 μ 2 + 1 = 0 So, μ = - 1 and λ = - 3 ⇒ μ - λ = 2