anonymous
  • anonymous
Find the center of the ellipse (x-4)^(2)+(y-1)^(2)/(16)=1
Mathematics
katieb
  • katieb
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watchmath
  • watchmath
The center simply the number inside your parenthesis, i.e. (4,1).
anonymous
  • anonymous
There are many numbers inside the parantheses
anonymous
  • anonymous
watchmath, did you do the ring problem, or were you just being challenging?

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watchmath
  • watchmath
I did it once, but I am not sure if I can remember the solution :).
anonymous
  • anonymous
Were you saying that (4,1) is the answer?
watchmath
  • watchmath
yes :)
anonymous
  • anonymous
thank you
watchmath
  • watchmath
Whenever you have \[\frac{(x-h)^2}{a^2}+\frac{(x-k)^2}{b^2}=1\] then the center is \((h,k)\).
anonymous
  • anonymous
then wouldn't it be (-4,-1)? or no, because of the squares?
watchmath
  • watchmath
Compare to \[\frac{(x-4)^2}{1^2}+\frac{(y-1)^2}{16}=1\] What is the \((h,k)\) here?
anonymous
  • anonymous
But it wouldn't be negative coordinates, it would be positive?
watchmath
  • watchmath
well if you compare (x-h) to (x-4) then the h is 4 in this case. Similarly if you compare (y-k) and (y-1) then k=1 here. So (4,1) is the center.
anonymous
  • anonymous
okay thanks

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