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Help! Can't do conics!

  1. Jan 16, 2005 #1
    Ok I seem to be having problems with changing the general form of a conic to standard form. I'm mainly confused with how to factor, since I haven't done it in a while, as well as how to go about completing the square.

    Here's one of my problems:

    2x^2 + y^2 + 12x – 2y + 15=0

    I rearranged it to look like: 2x^2 + 12x + y^2 – 2y + 15=0

    Then I "supposedly" completed the square:

    (2x^2 + 12x +36) + (y^2 – 2y +1)= 22

    Factoring is where I got stuck: 2(x^2 + 6x +18) + (y-1)^2= 22

    I don't know what to do with what I got and the answer is supposed to be:

    (x+3)^2 / 2 + (y-1)^2 / 4 = 1
     
  2. jcsd
  3. Jan 16, 2005 #2

    Hurkyl

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    Your problem is that this isn't a square! (Though, x^2 + 12x + 36 is) Your problem is you need to factor out the two first, so that the coefficient on x^2 is a 1.
     
  4. Jan 16, 2005 #3
    I did that and I got 2(x^2 + 6x +18).
     
  5. Jan 16, 2005 #4
    1. 2x^2 + 12x + 18 which is equivalant to 2(x+3)(x+3) + y^2 - 2y + 1 which is equivalnt to (y-1)(y-1) = -15 + 18 +1
    2. Your equation is 2(x+3)^2 + (y-1)^2 = 4
    3. Divide each side by 4. Now you have:
    2(x+3)^2/4 + (y-1)^2/4 = 4/4
    4. Now, your final product is:
    (x+3)^2/2 + (y-1)^2/4 = 1

    Is that the needed answer?
     
  6. Jan 16, 2005 #5

    Hurkyl

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    You need to factor before you figure out the constant term. You picked 36, then factored, which is the wrong way around.
     
  7. Jan 16, 2005 #6
    How did you get 2x^2 + 12x + 18? I got 2x^2 + 12x + 36.
     
  8. Jan 16, 2005 #7
    All you know is 2x^2 + 12x + ?.

    Factor out the two to make it easy.

    Now you have 2(x^2 +6x + ?)

    Then you can fill in the square by making it 2(x^2 + 6x + 9) or 2(x+3)^2
     
  9. Jan 16, 2005 #8
    This should give you 2x^2 + 12x + 18.

    I hope I'm doing this right...
     
  10. Jan 16, 2005 #9

    dextercioby

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    If you know the answer,then u can cheat:
    1.Make in the initial quadratic form the 2 substitutions
    [tex] x\rightarrow u-3 [/tex]
    [tex] v\rightarrow v+1 [/tex]

    2.Show that the new quadratic form is
    [tex] \frac{u^{2}}{2}+\frac{v^{2}}{4}=1 [/tex]

    3.Reverse the substitution and find the answer.

    Daniel.
     
  11. Jan 16, 2005 #10
    Oh I see! Thanks for the help guys!
     
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