A tangent drawn to hyperbola x 2 a 2 - y 2 b 2 = 1 at P π 6 forms a triangle of area 3 a 2 square units, with coordinate axes. If the eccentricity of hyperbola is e , then the value of e 2 - 9 is

Step-by-step solution

The point P π 6 is a sec π 6 , b tan π 6 or P 2 a 3 , b 3 ∴ Equation of tangent at P is x 3 a 2 - y 3 b = 1 1 ∴ Area of the triangle = 1 2 × 3 a 2 × 3 b = 3 a 2 ∴   b a = 4 e 2 = 1 + b 2 a 2 = 17 Now, e 2 - 9 = 17 - 9 = 8 Hence, option (d) is correct.