Help

i feel like i've seen this problem on AoPS Alcumus before, it was a tough one for me!

i went back to my problem logs (scrolled a while), and saw that the answer was 6 + sqrt6.

though i hope you are not just asking this question for the answer alone to get points. i'd recommend you to read the solution by AoPS thoroughly and understand it, as it can help you learn a lot.

hope this helped! 

Thanks! I already tried to do it my second attempt and I already got it wrong, I read the solution and it really helped. :D

b/a   is given as   sqrt 2/ 2

so b = sqrt 2

     a = 2

c^2 = a^2 + b^2

c^2 = 6

c = sqrt 6

then since c - a = 1    (given ...becuase the focus is at 3    the vertex is at 4)

         so c = a + 1

    find  a       a is to c    as    a is to    (a+1)   

                      2 / sqrt6  = a / (a+1)

              a sqrt 6 = 2a  + 2 

               a ( sqrt6 -2) = 2

               a = 2 / ( sqrt6-2)     rationalaize    multiply numerator and denominator by  - sqrt6 -2

                   = (-4 -2 sqrt 6 ) / ( 4 -6)

                  a = 2 +- sqrt 6                          this is the distanc from the center to the vertex at 4

                   vertex then becomes    4   +   ( 2+ sqrt6 )     = 6 + sqrt 6     <====== the x coordinate of the center  !