If the eccentricity of the standard hyperbola passing through the point ( 4,6 ) is 2 , then the equation of the tangent to the hyperbola at ( 4,6 ) is:

Step-by-step solution

Let us suppose equation of hyperbola is x 2 a 2 - y 2 b 2 = 1 e = 2 ∵ b 2 = e 2 - 1 a 2 ⇒ b 2 = 3 a 2 It is passing through 4,6 16 a 2 - 36 b 2 = 1 ⇒ a 2 = 4 ,   b 2 = 12 So, equation of tangent is T = 0 x x 1 a 2 - y y 1 b 2 = 1 Tangent at point ( 4 , 6 ) is ⇒ x - y 2 = 1 ⇒ 2 x - y - 2 = 0