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Graphing Conics help

  1. Nov 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Graph the following. Include center, verticies, foci, asymptote, and directrix as appropriate.

    2. Relevant equations

    [tex]x^2 + 8y - 2x = 7[/tex]

    3. The attempt at a solution

    So far I have:

    V = (1, -7/8)
    P = -2
    X = -1

    I have no clue where to go from here, or if I'm even right.
    Thanks
     
  2. jcsd
  3. Nov 5, 2008 #2

    Dick

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    Re: Conics

    Don't you have to decide what kind of conic it is first? Complete the square in x and try to write it in some kind of normal form.
     
  4. Nov 5, 2008 #3
    Re: Conics

    when i completed the square i got [tex](x-1)^2 = 8(-y+\frac{7}{8})[/tex]
     
  5. Nov 5, 2008 #4

    Dick

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    Re: Conics

    I think you mean (x-1)^2=8*(1-y). Try that once more. It's a parabola, isn't it?
     
  6. Nov 5, 2008 #5
    Re: Conics

    why would it be 8(y-1) ? What happened to the 7?
    How do you tell that it's a parabola? Because of the (x-1)^2?
     
  7. Nov 5, 2008 #6

    Dick

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    Re: Conics

    Yes, because it's quadratic in x and linear in y. The 7 changed to an 8 when you added the 1 to both sides to complete the square. You did do that, right?
     
  8. Nov 5, 2008 #7
    Re: Conics

    oh nooes. I didn't.

    ok i got it. Now what?
     
  9. Nov 5, 2008 #8

    Dick

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    Re: Conics

    Ok, now where is the vertex? And if you tell me what X and P are supposed to be I might be able to help you with those. Tomorrow. zzzzzzzzzzz.
     
  10. Nov 5, 2008 #9
    Re: Conics

    Ok thanks for all of your help!

    I got:

    v = (1, 1)
    p = 2
     
  11. Nov 6, 2008 #10
    Re: Conics

    So does that look right? If so, where do I go from here?
     
  12. Nov 6, 2008 #11
    Re: Conics

    really I think my problem is putting the initial equation in the [tex]\frac{(y-y0)^2}{a^2}-\frac{(x-x0)^2}{b^2}[/tex]
     
  13. Nov 6, 2008 #12
    Re: Conics

    Mistake. I know it's a parabola because it's in the form [tex](x-1)^2=8(1-y)[/tex]. And I know p is 2 and the vertex is at (1,1). But how do I know which direction from V to go two units? Up or down? Left or right?

    Also how do I find the asymptote?
     
  14. Nov 6, 2008 #13

    Dick

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    Re: Conics

    You never told me what P is. Is it the distance from the vertex to the focus? Does a parabola have any asymptotes?
     
  15. Nov 6, 2008 #14
    Re: Conics

    P = 2 right?
    Right now I have:

    [tex]v=(1,1)[/tex]
    [tex]p=2[/tex]
    [tex]F=(1,3)[/tex]
    [tex]Directrix=(1,-2)[/tex]

    Does that look right?
     
  16. Nov 6, 2008 #15

    Dick

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    Re: Conics

    You've got the vertex. Now does the parabola go up or down from the vertex? As x gets large does y increase to +infinity or -infinity? And the directrix is a line, not a point.
     
  17. Nov 6, 2008 #16
    Re: Conics

    the parabola opens up.
     
  18. Nov 6, 2008 #17

    Dick

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    Re: Conics

    I disagree. Why do you think so?
     
  19. Nov 6, 2008 #18
    Re: Conics

    oh wait. since it's (x-1)^2 that means it goes down right?
     
  20. Nov 6, 2008 #19

    Dick

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    Re: Conics

    You tell me. If x=101, what kind of number is y?
     
  21. Nov 6, 2008 #20
    Re: Conics

    a - number. I got -1249
     
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