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Conics with multiple squares

  1. Nov 6, 2008 #1
    1. The problem statement, all variables and given/known data

    graph the following

    2. Relevant equations


    3. The attempt at a solution

    I think I need to get it into [tex]\frac{(x-x0)^2}{a^2}+\frac{(y-y0)^2}{b^2}[/tex] but I'm not sure.
    I have [tex]\frac{9x^2}{-4}-8x+y^2-2y=1[/tex] and now I'm stuck
    Last edited: Nov 6, 2008
  2. jcsd
  3. Nov 6, 2008 #2
    Ok, update. Here's what I have so far:

    [tex]\frac{(x+2)^2}{2^2}+\frac{(y-1)^2}{3^2} = 1 [/tex]

    Does that look right?
  4. Nov 6, 2008 #3


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    It looks fine, if you meant 4y^2 in the original post and not 47^2.
  5. Nov 6, 2008 #4
    Thanks for catching that. I fixed it.
    Ok, so now I have the following:

    [tex]Center = (-2,1)[/tex]
    [tex]a = 2[/tex]
    [tex]b = 3[/tex]
    [tex]Verticies: (0,1),(-4,1),(-2,4),(-2,-2)[/tex]

    Does that look right?
  6. Nov 6, 2008 #5


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    Looks right to me. But I'm tired. You should double check.
  7. Nov 6, 2008 #6
    haha. you're supposed to be the double checker!

    I'm really stuck now. I'm trying to find 'c' and I get [tex]\sqrt{-5}[/tex]. Did I do something wrong? c is the square root of a^2 - b^2 right? Here, a = 2 and b = 3. I'm confused
  8. Nov 6, 2008 #7


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    No. YOU are supposed to be the double checker. It's your class. I'm just tossing off hints without being fully awake. I have no idea what 'c' is supposed to be. Could you just like say what it is supposed to be instead of dropping a cryptic letter? I'll take another guess and say 'distance from center to focus'? That's a lot better description than 'c'. Why don't you think it could be sqrt(3^2-2^2)? If you flip the x and y axes, do you think this distance should change from real to imaginary?
  9. Nov 7, 2008 #8
    hmm, I'm not sure. I didn't know you could swap them like that. The formula I was going by said "distance from center to focus" = sqrt(a^2 - b^2). If you flip them, you get the real answer?
  10. Nov 7, 2008 #9


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    Surely that formula was assuming a> b. That is, that a is the length of the longer semi-axis. Don't just memorize formulas. Learn what the mean.
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