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anonymous
 5 years ago
Find the center of the ellipse
(x4)^(2)+(y1)^(2)/(16)=1
anonymous
 5 years ago
Find the center of the ellipse (x4)^(2)+(y1)^(2)/(16)=1

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0The center simply the number inside your parenthesis, i.e. (4,1).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There are many numbers inside the parantheses

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath, did you do the ring problem, or were you just being challenging?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0I did it once, but I am not sure if I can remember the solution :).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Were you saying that (4,1) is the answer?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Whenever you have \[\frac{(xh)^2}{a^2}+\frac{(xk)^2}{b^2}=1\] then the center is \((h,k)\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then wouldn't it be (4,1)? or no, because of the squares?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Compare to \[\frac{(x4)^2}{1^2}+\frac{(y1)^2}{16}=1\] What is the \((h,k)\) here?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But it wouldn't be negative coordinates, it would be positive?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0well if you compare (xh) to (x4) then the h is 4 in this case. Similarly if you compare (yk) and (y1) then k=1 here. So (4,1) is the center.
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