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Babynini

  • one year ago

Parabola help :)

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  1. Babynini
    • one year ago
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  2. Babynini
    • one year ago
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    \[16x^2-128x-9y^2+400=0\] \[16(x^2-8x)-9y^2+400=0\] \[16(x^2-8x+16-16)-9y^2=-288\] \[16(x-4)^2-256-9y^2=-288\] \[16(x-4)^2-9y^2=-32\]

  3. Babynini
    • one year ago
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    @jim_thompson5910 I know we just did one like this, but i'm not quite sure where to go next with 9y^2 because it has no match.

  4. jim_thompson5910
    • one year ago
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    you don't need to complete the square for y since it's already done think of -9y^2 as -9(y-0)^2

  5. jim_thompson5910
    • one year ago
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    how did you go from step 2 to step 3?

  6. Babynini
    • one year ago
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    factored out 16

  7. jim_thompson5910
    • one year ago
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    idk where that -288 comes from

  8. Babynini
    • one year ago
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    haha oh shoot, that was meant to be -400

  9. Babynini
    • one year ago
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    I don't know where it came from either 0.0

  10. Babynini
    • one year ago
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    wherever -288 is it's -400

  11. jim_thompson5910
    • one year ago
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    ok

  12. Babynini
    • one year ago
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    Yeah. So the thing i'm at right now is \[16(x-4)^2-9y^2=-144\]

  13. Babynini
    • one year ago
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    er so the right side is negative. So i divide everything by negative yeah?

  14. Babynini
    • one year ago
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    \[\frac{ (x-4)^2 }{ -9}+\frac{ y^2 }{ 16 }=1\]

  15. Babynini
    • one year ago
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    hrm that seems wrong?

  16. Babynini
    • one year ago
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    @ganeshie8 can you check this? :) it doesn't match up with the graph.

  17. ganeshie8
    • one year ago
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    Looks good, just rewrite it as \[\large \frac{y^2}{16}-\frac{(x-4)^2}{9}=1\]

  18. Babynini
    • one year ago
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    but that makes the center (-1,4)?

  19. ganeshie8
    • one year ago
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    center is still (4, 0)

  20. ganeshie8
    • one year ago
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    \[\large \frac{y^2}{16}-\frac{(x-4)^2}{9}=1\] is same as \[\large \frac{(y-0)^2}{16}-\frac{(x-4)^2}{9}=1\]

  21. Babynini
    • one year ago
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    http://www.wolframalpha.com/input/?i=16x%5E2+%E2%88%92+9y%5E2+%E2%88%92+128x+%2B+400+%3D+0 that's where im confused because the center should be (4,4)

  22. ganeshie8
    • one year ago
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    rearranging the equation wont change anything

  23. Babynini
    • one year ago
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    (plus or minus)

  24. Babynini
    • one year ago
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    oooh oh i see! that's the vertex, not center.

  25. Babynini
    • one year ago
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    right?

  26. ganeshie8
    • one year ago
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    Yes center is the midpoint of vertices

  27. ganeshie8
    • one year ago
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    |dw:1433392386227:dw|

  28. Babynini
    • one year ago
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    ahh, ok! I wasn't sure about that, so that's where I was confused. Thank you :)

  29. Babynini
    • one year ago
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    a=4, b=3, c=5

  30. Babynini
    • one year ago
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    now how do I find vertex, foci, asymptotes, dirtrex?

  31. Babynini
    • one year ago
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    asy: a/b(x-h)+k so for mine it's \[\pm \frac{ 4 }{ 3 }(x-4)\]

  32. Babynini
    • one year ago
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    (with a y = before that)

  33. ganeshie8
    • one year ago
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    looks good |dw:1433393241031:dw|

  34. ganeshie8
    • one year ago
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    scroll down to double check ur answers http://www.wolframalpha.com/input/?i=focus+16x%5E2+%E2%88%92+9y%5E2+%E2%88%92+128x+%2B+400+%3D+0

  35. Babynini
    • one year ago
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    is this correct so far? :)

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  36. Babynini
    • one year ago
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    (and DNE on the directrix as well)

  37. ganeshie8
    • one year ago
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    looks gud

  38. Babynini
    • one year ago
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  39. Babynini
    • one year ago
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    thank you so so much! ;)

  40. Babynini
    • one year ago
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    |dw:1433393976693:dw|

  41. ganeshie8
    • one year ago
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    Excellent!!!

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