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My last repost showed melody saying that I just want answer, but there was never a solution..

 

 

https://web2.0calc.com/questions/hi-repost-d

 Mar 20, 2021
 #1
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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! 

 Mar 20, 2021
edited by idyllic  Mar 20, 2021
 #3
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But nevertheless thank you for having a good attitude!

Guest Mar 20, 2021
 #2
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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

 Mar 20, 2021
 #4
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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  !

 Mar 20, 2021
edited by ElectricPavlov  Mar 20, 2021

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