## anonymous 5 years ago Find the center of the ellipse (x-4)^(2)+(y-1)^(2)/(16)=1

1. watchmath

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

2. anonymous

There are many numbers inside the parantheses

3. anonymous

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

4. watchmath

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

5. anonymous

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

6. watchmath

yes :)

7. anonymous

thank you

8. watchmath

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

9. anonymous

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

10. watchmath

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

11. anonymous

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

12. 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.

13. anonymous

okay thanks