Solution :
Since their vertices and their foci lie on the equal ordinate the general hyperbola equation should be :
(x-p)^2/a^2-(y-q)^2/b^2=1
where (p,q)=its asymptotes cross connect
a^2+b^2=c^2
their foci :
F1(p-c,q)
F2(p+c,q)
where q=3 (equal ordinate)
asymptotes cross connect :
its absisca=(2+8)/2=10/5
its ordinate=q=3
its asymptote cross connect should be(5,3)
a=distance between its asymptote cross connect to its vertices
a=8-5=3
c=distance between its foci to asymptote cross connect
c=11-5=6
since a^2+b^2=c^2
then
3^2+b^2=6^2
b^2=36-9
b^2=27
b=3*sqrt(3)
its solution should be :
(x-5)^2/9-(y-3)^2/27=1
