A hyperbola passing through a focus of the ellipse x^2 169 + y^2 25 =1 . Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of eccentricities is 1 . Then, the equation of the hyperbola is

Step-by-step solution

Let the equation of hyperbola be x^2 a^2 - y^2 b^2 =1 Given equation of ellipse is x^2 (13)^2 + y^2 (5)^2 =1 Here, aligned & Here, a=13, b=5 & e= 1- b^2 a^2 & = 1- 25 169 = 144 169 = 12 13 & Focus ( a e, 0)= ( 13 12 13 , 0 ) & =( 12,0) aligned Since, Eq. (i) passes through ( 12,0) . array rrrl & 144 a^2 - 0 b^2 =1 & a^2=144 & a= 12 array Now eccentricity of hyperbola aligned e^ & = 1+ b^2 a^2 & = 1+ b^2 144 aligned According to the equation, array rlrl & e^ =1 & & 12 13 1+ b^2 144 & =1 & & 1+ b^2 144 & = 13 12 & &