Question

Find the center of the ellipse
(x-4)^(2)+(y-1)^(2)/(16)=1

Answer 1

The center simply the number inside your parenthesis, i.e. (4,1).

Answer 2

There are many numbers inside the parantheses

Answer 3

watchmath, did you do the ring problem, or were you just being challenging?

Answer 4

I did it once, but I am not sure if I can remember the solution :).

Answer 5

Were you saying that (4,1) is the answer?

Answer 6

yes :)

Answer 7

thank you

Answer 8

Whenever you have
\[\frac{(x-h)^2}{a^2}+\frac{(x-k)^2}{b^2}=1\]
then the center is \((h,k)\).

Answer 9

then wouldn't it be (-4,-1)? or no, because of the squares?

Answer 10

Compare to
\[\frac{(x-4)^2}{1^2}+\frac{(y-1)^2}{16}=1\]
What is the \((h,k)\) here?

Answer 11

But it wouldn't be negative coordinates, it would be positive?

Answer 12

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.

Answer 13

okay thanks