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anonymous

  • one year ago

PLEASE HELP ME WRITE THE EQUATION FOR AN ELLIPSE: (MEDAL AND FAN!!!): Write the equation of an ellipse with center (2,1); vertex (7,1) ; and focus (5,1).

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  1. ganeshie8
    • one year ago
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    |dw:1439810883867:dw|

  2. anonymous
    • one year ago
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    Ok that makes sense, so how do I get a? I am just confused sorry.

  3. ganeshie8
    • one year ago
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    |dw:1439810946525:dw|

  4. ganeshie8
    • one year ago
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    so it seems it is going to be a horizontal ellipse ?

  5. anonymous
    • one year ago
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    Yes

  6. anonymous
    • one year ago
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    I know that the equation is going to look like this: \[\frac{ (x-2)^2 }{ ? } + \frac{ (y-1)^2 }{ ? } = 1\]

  7. ganeshie8
    • one year ago
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    Right, center = \((h,k)=(2,1)\) we need to find the values of \(a,b\) and plug in : \[\dfrac{(x-2)^2}{a^2}+\dfrac{(y-1)^2}{b^2}=1\]

  8. anonymous
    • one year ago
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    I just need a^2 and b^2...

  9. ganeshie8
    • one year ago
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    use the vertex formula : \((h+a, k)=(7,1)\) can you find the value of \(a\) ?

  10. anonymous
    • one year ago
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    is it 5?

  11. ganeshie8
    • one year ago
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    Correct. \(2+a=7\implies a=5\)

  12. anonymous
    • one year ago
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    ok and then b?

  13. ganeshie8
    • one year ago
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    Next use focus formula \((h+c,~k)=(5,1)\) can you find the value of \(c\) ?

  14. anonymous
    • one year ago
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    Is it 2?

  15. ganeshie8
    • one year ago
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    nope

  16. anonymous
    • one year ago
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    Not sure, sorry

  17. ganeshie8
    • one year ago
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    \(h+c=5\) you know that \(h=2\), therefore \(c=?\)

  18. anonymous
    • one year ago
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    3?

  19. ganeshie8
    • one year ago
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    Yes, \(a=5,~c=3\) we still need to find \(b\)

  20. anonymous
    • one year ago
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    ok so I use this right: c^2 = b^2 - a^2?

  21. ganeshie8
    • one year ago
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    it should be \(a^2 = b^2+c^2\)

  22. anonymous
    • one year ago
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    whoops ok, sorry: 5^2 = 3^2 + b^2 25 = 9 + b^2 16 = b^2 b = 4

  23. ganeshie8
    • one year ago
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    Excellent!

  24. anonymous
    • one year ago
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    Ok so my final answer is: \[\frac{ (x-2)^2}{ 25 } + \frac{ (y-2)^2 }{ }\]

  25. anonymous
    • one year ago
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    whoops i messed that up under the (y-2)^2 is 16. and it all equals 1 right?

  26. anonymous
    • one year ago
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    Ok so now that i got that, how do I approach this when I am given center, co-vertex, and focus? Can i post the question?

  27. ganeshie8
    • one year ago
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    Ok so my final answer is: \[\frac{ (x-2)^2}{ 25 } + \frac{ (y-\color{red}{1})^2 }{ \color{red}{16} }\]

  28. anonymous
    • one year ago
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    Sorry that is what i meant, lol i had that written down...

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