SOLUTION: Hi, how can I get this: [((x-3)^2)/4]-[((y+2)^2)/16]=1 from this: 4x^2-y^2-24x-4y+16=0 Thanks!

Algebra ->  Equations -> SOLUTION: Hi, how can I get this: [((x-3)^2)/4]-[((y+2)^2)/16]=1 from this: 4x^2-y^2-24x-4y+16=0 Thanks!      Log On


   

Question 861881: Hi, how can I get this:
[((x-3)^2)/4]-[((y+2)^2)/16]=1
from this: 4x^2-y^2-24x-4y+16=0
Thanks!
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
4x^2-y^2-24x-4y+16=0
4%28x-3%29%5E2-%28y%2B2%29%5E2=+-16+%2B+36+-+4 Completing the Square
4%28x-3%29%5E2-%28y%2B2%29%5E2=16
%28x-3%29%5E2%2F4-%28y%2B2%29%5E2%2F16=1