Let H be a hyperbola with its center at the origin and transverse axis along the x -axis. The eccentricity of H is 2 and the length of its latus rectum is 12 . Suppose a point P(x₁, 6) , where x₁ > 0 , lies on H . If S and S' are the foci of H , then the radius of the incircle of the triangle SPS' is equal to

Step-by-step solution

Let the equation of the hyperbola be x^2 a^2 - y^2 b^2 = 1 . Given eccentricity e = 2 and length of latus rectum 2b^2 a = 12 b^2 = 6a . Using the standard relation b^2 = a^2(e^2 - 1) : 6a = a^2(2^2 - 1) 6a = 3a^2 a = 2 (since a > 0 ). Thus, b^2 = 6(2) = 12 . The equation of the hyperbola is x^2 4 - y^2 12 = 1 . Point P(x₁, 6) lies on H : x₁^2 4 - 36 12 = 1 x₁^2 4 - 3 = 1 x₁^2 = 16 x₁ = 4 (since x₁ > 0 ). So, P is (4, 6) . The coordinates of the foci are ( ae, 0) = ( 4, 0) . Let S(4, 0) and S'(-4, 0) . The side leng