A tangent drawn to the hyperbola x 2 a 2 - y 2 b 2 = 1 at P a sec ⁡ π 6 , b tan ⁡ π 6 form a triangle of area 3 a 2 sq. units with the coordinate axes. The eccentricity of the conjugate hyperbola of x 2 a 2 - y 2 b 2 = 1 is

Step-by-step solution

P = 2 a 3 , b 3 Equation of the tangent at P is 2 x 3 a - y 3 b = 1 x intercept & y intercept of the tangent are 3 a 2 & - 3 b respectively Area of the triangle formed by the tangent with the coordinate axes is 1 2 × 3 a 2 × 3 b = 3 a 2 ⇒ b a = 4 ⇒ a b = 1 4 The eccentricity of the conjugate hyperbola equals to 1 + 1 16 = 17 4