Let X - axis be the transverse axis and Y - axis be the conjugate axis of a hyperbola H . Let the eccentricity of H be the reciprocal of the eccentricity of the ellipse x^2 4 + y^2 2 =1 . If (5,4) is a point on H , then the length of the transverse axis of H is

Step-by-step solution

Eccentricity of ellipse x^2 4 + y^2 2 is: e^ = 1- 2 4 = 1- 1 2 = 1 2 Now eccentricity of hyperbola: aligned & e = 1 e^ = 2 & 1+ b^2 a^2 = 2 1+ b^2 a^2 =2 & b^2 a^2 =1 b^2=a^2 aligned The equation of hyperbola is: x^2 a^2 - y^2 b^2 =1 x^2 a^2 - y^2 a^2 =1 ...(i) Since (5,4) lies on equation (i) So, 25 a^2 - 16 a^2 =1 9=a^2 a= 3 Now length of transverse axis =2|a|=2 3=6