The eccentricity of the hyperbola whose length of its conjugate axis is equal to half of the distance between its foci, is

Step-by-step solution

For any standard hyperbola we know that eccentricity, e = 1 + b 2 a 2 , conjugate axis = 2 b and distance between the foci = 2 a e . Given, 2 b = 1 2 2 a e . ⇒ b a = e 2 Now, e =   1 + b 2 a 2 =   1 + e 2 4 ⇒     e 2 = 1 + e 2 4 ⇒     3 4 e 2 = 1 ⇒     e 2 = 4 3 ⇒     e = 2 3