1100. A pair of hyperbolae

    Two hyperbolae share their assymptotes in such a way that the two assymptotes separate the four hyperbola branches... Let  E  and  F  be the eccentricities of the hyperbolae. Prove that  

E² + F²  =  E² · F² 

(The two hyperbolae are not necessarily conjugated.)

    Yoyes Canasta did it easily...

    Dear Teacher:
    These two hyperbolae also share their axes. Both axes are perpendicular to each other and are bisectors of the angles between the assymptotes. Then...

E = sec α   and   F = sec β

    Since  α  and  β  are complementary...

F  =  csc α

    And...

E² + F²  =

=  sec²α + csc²α  =

=  1/cos²α + 1/sin²α  =

=  (sin²α + cos²α) / (cos²α · sin²α)  =

=  1/cos²α · 1/sin²α  =
=  E² · F²