martes, 17 de enero de 2017

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

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