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Homework Help: Did I do this coversion to Standard correctly?

  1. Mar 16, 2005 #1
    Cover to standard form...

    [tex] 4x^2 - 9y^2 + 32x + 18y + 91 = 0 [/tex]

    [tex] 4x^2 + 32x [/tex]..................[tex]-9y^2 + 18y[/tex]..................[tex]+91 = 0[/tex]

    [tex] 4(x^2 + 8x + 16) [/tex]..........[tex]-9(y^2 - 2y + 1)[/tex]..........[tex]+91 -16 -1 = 0[/tex]

    [tex] 4(x + 4)^2 [/tex]..................[tex]-9(y - 1)^2[/tex]..................[tex]+74 = 0[/tex]

    therefore

    [tex]\frac{(y - 1)^2}{4} - \frac{(x+4)^2}{9} = 74[/tex]
     
  2. jcsd
  3. Mar 16, 2005 #2
    You made a few errors with the constant terms. See if you can figure out where.
     
  4. Mar 16, 2005 #3
    When the 74 goes over the = it becomes a negative, which is not possible. So, I'm thinking i did something wrong by dividing out -9. Somehow I need to get a balancing number that is larger that 91 so it won't be positive on the other side... :confused:
    But I do see something that might work...

    I have to change those balancing numbers. I only countered the -16 and -1. But those weren't multiplied by that factor.
    Ok, So, [tex]91 - 64 + 9 = 36[/tex]. 36?
    But I can't get a positive number...since when it goes over the = line it'll be negative.
     
    Last edited: Mar 16, 2005
  5. Mar 16, 2005 #4
    36 is correct. The negative is no problem: just multiply both sides by -1.

    You made one more mistake though. To get the 4 and 9 in the denominators you had to divide the equation by 36. Do you see anything that you missed?
     
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