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anonymous

  • one year ago

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.

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  1. anonymous
    • one year ago
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    @Hero plz help

  2. anonymous
    • one year ago
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    i can help you :)

  3. anonymous
    • one year ago
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    OMG THANK YOU SO MUCH I NEED SLEEP OMG

  4. anonymous
    • one year ago
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    lol ok :)

  5. anonymous
    • one year ago
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    (3,0) and (–3,0) are to be the focal points of the ellipse right?

  6. anonymous
    • one year ago
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    yeah

  7. anonymous
    • one year ago
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    x^2/25+y^2/16=1 is that the equation for the elliptical flower bed?

  8. anonymous
    • one year ago
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    for this you want \[\frac{ x2 ^{} }{ a2 } + \frac{ y2 }{ b2 } = 1\] and you want \[a^{2} - b^{2} = 3^{2}\] the easiest way to do that is to use the famous 3−4−5 right triangle and make a=5,b=4 so c=3 and use \[\frac{ x^{2} }{ 5^{2} } + \frac{ y^{2} }{ 4^{2} } = 1\] that will make your foci (−3,0) and (3,0)

  9. anonymous
    • one year ago
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    ok I got that part but what about the hyperbolic flowerbed

  10. anonymous
    • one year ago
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    can you try

  11. anonymous
    • one year ago
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    I can't make up the value of A

  12. anonymous
    • one year ago
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    like I know x^2/a^2-y^2/b^2=1 and a^2+b^2=3^2

  13. anonymous
    • one year ago
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    for the hyperbola it will look like x2/ a2−y2/b2=1 and you want a2+b2=33 simplest way i can think to do it is to make a2=8,b2=1 and use x2/8−y2=1 but you have other choices

  14. anonymous
    • one year ago
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    how did you get 33?

  15. anonymous
    • one year ago
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    no its \[3^{3}\]

  16. anonymous
    • one year ago
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    ok

  17. anonymous
    • one year ago
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    medal and fan please

  18. anonymous
    • one year ago
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    yeah but one more question how do you graph the second one?

  19. anonymous
    • one year ago
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    here is a nice graph with both together if you need one http://www.wolframalpha.com/input/?i=+x^2%2F8-y^2%3D1%2Cx^2%2F25%2By^2%2F16%3D1

  20. anonymous
    • one year ago
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    ok thank you so much!!

  21. anonymous
    • one year ago
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    your welcome

  22. anonymous
    • one year ago
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    medal and fan?

  23. anonymous
    • one year ago
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    yeah i already did

  24. anonymous
    • one year ago
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    ok thanks :)

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