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anonymous

  • 5 years ago

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

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  1. watchmath
    • 5 years ago
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    The center simply the number inside your parenthesis, i.e. (4,1).

  2. anonymous
    • 5 years ago
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    There are many numbers inside the parantheses

  3. anonymous
    • 5 years ago
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    watchmath, did you do the ring problem, or were you just being challenging?

  4. watchmath
    • 5 years ago
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    I did it once, but I am not sure if I can remember the solution :).

  5. anonymous
    • 5 years ago
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    Were you saying that (4,1) is the answer?

  6. watchmath
    • 5 years ago
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    yes :)

  7. anonymous
    • 5 years ago
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    thank you

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

  9. anonymous
    • 5 years ago
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    then wouldn't it be (-4,-1)? or no, because of the squares?

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

  11. anonymous
    • 5 years ago
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    But it wouldn't be negative coordinates, it would be positive?

  12. watchmath
    • 5 years ago
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    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
    • 5 years ago
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    okay thanks

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