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yeval76

  • one year ago

***WILL FAN AND MEDAL*** There are two fruit trees located at (3,0) and (−3, 0) in the backyard plan. Maurice wants to use these two fruit trees as the focal points for an elliptical flowerbed. Johanna wants to use these two fruit trees as the focal points for some hyperbolic flowerbeds. Create the location of two vertices on the y-axis. Show your work creating the equations for both the horizontal ellipse and the horizontal hyperbola. Include the graph of both equations and the focal points on the same coordinate plane. @ChiefArnav

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  1. yeval76
    • one year ago
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    @nincompoop

  2. yeval76
    • one year ago
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    @Abhisar please help

  3. yeval76
    • one year ago
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    @Loser66

  4. Loser66
    • one year ago
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    |dw:1439645146687:dw|

  5. Loser66
    • one year ago
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    You are "Creating" the ellipse. Hence it is up to you to pick the a and b such that \(c^2= a^2-b^2\). Your c is 3, hence \(c^2 =9\) you can pick a, b satisfy that equation. Which pair do you want to pick?

  6. Loser66
    • one year ago
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    @Penguin7 This problem is horizontal, your problem is vertical. I work for him first, you convert everything to vertical. That is it.

  7. Loser66
    • one year ago
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    @yeval76 If you don't help yourself, you don't work for your own problem. You are better move to @Penguin7 post and convert to horizontal case from her case. I prioritize for the students who join me on their stuff.

  8. yeval76
    • one year ago
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    Im so sorry I didn't notice that you had responded. I will join in and work for my problem... Please help me @Loser66

  9. yeval76
    • one year ago
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    I pick that a=6

  10. Loser66
    • one year ago
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    Ok, answer me, pick a , b such that 9 = a^2 -b^2

  11. yeval76
    • one year ago
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    9 = 6^2 -b^2

  12. yeval76
    • one year ago
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    9 = 36- b^2

  13. Loser66
    • one year ago
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    a =6 , hence a^2 =36, so b =?

  14. yeval76
    • one year ago
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    b^2 = 27 but b = 5.196

  15. Loser66
    • one year ago
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    ok, how about a =5 and b=4? do they work?

  16. Loser66
    • one year ago
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    Your pick is PERFECTLY ok. I just suggest the easiest numbers to work with.

  17. yeval76
    • one year ago
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    If a = 5 and b = 4, it would be the following: 25-16=9

  18. Loser66
    • one year ago
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    Yes, hence your ellipse equation will be??? just plug into the standard form for HORIZONTAL one.

  19. Loser66
    • one year ago
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    http://www.math.binghamton.edu/grads/kaminski/math222_spring12/Sect10.5_ShiftedConics.pdf

  20. Loser66
    • one year ago
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    Print it out to use later. It is helpful

  21. Loser66
    • one year ago
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    Look at Horizontal Ellipse row. the formula right on the 3rd column.

  22. Loser66
    • one year ago
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    The net is so bad. I assume you cannot access. but you know how to do this part, right? Now, move to hyperbola.!!!

  23. Loser66
    • one year ago
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    The same token, but for hyperbola, \(c^2 = a^2 \color{red}{+} b^2\) You can create a and b such that \(9= a^2 +b^2\)

  24. Loser66
    • one year ago
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    On this case, no way for you to choose both a, b are integers. If you choose a = 1, then \(b= 2\sqrt2\), If you choose a=2, then \(b=\sqrt5\) Whatsoever, you pick 1 pair, a and b, then plug back to row HORIZONTAL HYPERBOLA on the paper to get the equation. Good luck.

  25. Loser66
    • one year ago
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    |dw:1439646772690:dw|

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