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anonymous

  • one year ago

Write the equation of an ellipse with vertices (0, 5) and (0, -5) and co-vertices (2, 0) and (-2, 0).

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  1. anonymous
    • one year ago
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    |dw:1439646942183:dw|

  2. anonymous
    • one year ago
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    a=5 b=2

  3. anonymous
    • one year ago
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    5^2=25 2^2=4

  4. anonymous
    • one year ago
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    y2/25 + x2/4 = 1? the horizontal get me mixed up sometimes but i feel thats right

  5. Loser66
    • one year ago
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    How to say!!! You are an EXPERT now. hehehe

  6. anonymous
    • one year ago
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    only bc u took the time to teach my slow mind lol thank you, now..can u teach me hyperbola? those confuess me more then anything @Loser66

  7. Loser66
    • one year ago
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    Let's look at it first. I am not sure whether I can or not. One more thing, you have only 15 more minutes. I have to go after then.

  8. anonymous
    • one year ago
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    one more thing! Ellipses have symmetry through only their major axis? true or false? lol

  9. anonymous
    • one year ago
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    i think its false?

  10. Loser66
    • one year ago
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    through both axis

  11. anonymous
    • one year ago
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    so its false

  12. Loser66
    • one year ago
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    yup

  13. Loser66
    • one year ago
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    To hyperbola, everything is the same with ellipse but the - sign

  14. anonymous
    • one year ago
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    okay. ill start a new post.

  15. Loser66
    • one year ago
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    a goes with main axis b goes with co-axis. and c^2 = a^2 + b^2 (opposite with ellipse, which is c^2 =a^2 -b^2\)

  16. Loser66
    • one year ago
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    why a new post. It is good here

  17. anonymous
    • one year ago
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    okay- What are the asymptotes of the following graph?

  18. Loser66
    • one year ago
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    |dw:1439647512939:dw|

  19. anonymous
    • one year ago
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    |dw:1439647585256:dw|

  20. Loser66
    • one year ago
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    ok, |dw:1439647624382:dw|

  21. Loser66
    • one year ago
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    Do you know how to find the slope of this line?

  22. anonymous
    • one year ago
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    one sec.

  23. anonymous
    • one year ago
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    m=y1-y2/x1-x2

  24. Loser66
    • one year ago
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    yes, =?

  25. anonymous
    • one year ago
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    m=-3-3/-1-1

  26. Loser66
    • one year ago
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    nope

  27. Loser66
    • one year ago
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    You plug in x1-x2 on top. it is not right,

  28. anonymous
    • one year ago
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    y=-1-1/-3-3

  29. Loser66
    • one year ago
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    yup, =?

  30. anonymous
    • one year ago
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    m=-2/-6 m=2/6 m=1/3

  31. Loser66
    • one year ago
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    Bingo

  32. Loser66
    • one year ago
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    Hence the asymptote of the hyperbola is just \(y =\pm\dfrac{1}{3}x\)

  33. anonymous
    • one year ago
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    okay i get it. can we get 1 more in really quickly?

  34. Loser66
    • one year ago
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    3 minutes for that.

  35. anonymous
    • one year ago
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    What is the center of the following graph?

  36. Loser66
    • one year ago
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    |dw:1439648000091:dw|

  37. anonymous
    • one year ago
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    (1,2)

  38. Loser66
    • one year ago
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    yup

  39. anonymous
    • one year ago
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    u think we can solve for a foci or not enough time?

  40. Loser66
    • one year ago
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    not enough

  41. Loser66
    • one year ago
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    but I can give you a guidance: find asymptote, it is the form of a/b, take that a, b to solve for c which is foci.

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