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Finding points of intersection

  1. Jun 18, 2007 #1
    the parabola equation is:
    (x^2/25) - (y^2/9) = 1

    the line is y = 4-x

    according to my calculations, if i point y - 4-x into the equation, i get
    -16x^2 + 200x - 175 = 0. is that right so far?

    ~Amy
     
  2. jcsd
  3. Jun 18, 2007 #2

    VietDao29

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    No, it's not correct. =.="

    Can you show us how you get there?

    I get: -16x2 + 200x - 625 = 0.

    Btw, that's definitely not a parabola equation. Instead, it's a hyperbola equation. :smile:
     
    Last edited: Jun 18, 2007
  4. Jun 18, 2007 #3
    thanks

    i meant to say hyperbola :shy:


    i have 8 lines of work and my typing numbers is slow.. so here's part of my calculations:

    x^2/ 25 - (4-x^2)/ 9 = 1

    225(x^2/25) - 225 ((4-x)^2/9) = 225(1)
    9x^2 - 25(16 - 4x - 4x + x^2) = 225
    9x^2 - 25(16 - 8x + x^2) = 225
    9x^2 - 400 + 200x - 25x^2 = 225

    i see my mistake now :smile: thanks

    and then after i do the quadratic equation, i end up with x = 6.25, so y= -2.25. and that's my only intersection.

    ~Amy
     
  5. Jun 18, 2007 #4

    VietDao29

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    Yeah, looks good. Congratulations. :)
     
  6. Jun 18, 2007 #5
    if you dont mind.. one more hyperbola question:

    4x^2 - y^2 + 8x + 4y + 16 = 0
    what is this in standard form?

    in standard form:
    (x+1)^2/4 - (y-2)^2/16 = 1

    does that look accurate?

    ~Amy
     
  7. Jun 18, 2007 #6

    malawi_glenn

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    You can try to expand that and see if you get the same as the initial form :)
     
  8. Jun 18, 2007 #7
    thanks. i expanded it and got:
    4x^2 - y^2 + 8x + 4y + 4 = 0

    so.. either something went wrong when i was expanding, or the standard form i figured out is incorrect?

    ~Amy
     
  9. Jun 18, 2007 #8

    malawi_glenn

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    you are very close altough, if you do it again properly and careful :) i promise.
     
  10. Jun 18, 2007 #9
    so there was something wrong with my expanding?

    ~Amy
     
  11. Jun 18, 2007 #10

    malawi_glenn

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    "if you dont mind.. one more hyperbola question:

    4x^2 - y^2 + 8x + 4y + 16 = 0
    what is this in standard form?"

    You said, then you gave:

    " (x+1)^2/4 - (y-2)^2/16 = 1"

    as the normal form of the first equation you gave;
    and when this is expand, you get:

    4(x^2 + 2x + 1) - (y^2-4y + 4) = 16 ->
    4x^2 + 8x + 4 - y^2 + 4y - 4 = 16 ->
    4x^2 + 8x - y^2 + 4y - 16 = 0

    right?

    try:

    -[(x+1)^2]/4 + [(y-2)^2]/16 = 1

    expand:

    -4[x^2 +2x+1] + [y^2 - 4y + 4] = 16 ->
    4[x^2 +2x+1] - [y^2 - 4y + 4] +16 = 0 ->
    4x^2 + 8x + 4 - y^2 + 4y - 4 + 16 = 0 ->
    4x^2 - y^2 + 8x + 4y + 16 = 0
     
    Last edited: Jun 18, 2007
  12. Jun 18, 2007 #11
    k thanks :)

    but could i write that as:
    [(x+1)^2]/4 + [(y-2)^2]/16 = -1

    (i dont think its standard form to have a "-" infront of the equation)

    and for this one, b^2 = 16 ?
    (i have to do other calculations)
    ~Amy
     
  13. Jun 19, 2007 #12

    malawi_glenn

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    Well as you said, having a minus in front of the "1" is not standard form. So why do you want to do that?
     
  14. Jun 19, 2007 #13

    VietDao29

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    Nope, you can't. However, you "can" write it as:

    [(x+1)^2]/4 - [(y-2)^2]/16 = -1

    Yes, you can write it like that, but it's definitely not a standard form.

    [(y-2)^2]/16 - [(x+1)^2]/4 = 1

    Note that it's a North-south opening hyperbola, your b2 = 4. :)
     
    Last edited: Jun 19, 2007
  15. Jun 20, 2007 #14
    thanks for the help guys :eek:). im finished the course (just have to do the exam). :)

    ~Amy
     
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