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Homework Help: Change to standard form of conic PLz help

  1. Mar 28, 2005 #1
    Write [tex] y^2-4x-4y+8=0 [/tex] in standard form and state its key features, Identify the type of conic.

    I know to get into standard form I need to complete the square but with these numbers im not sure how to factor it should have been in the form [tex]ax^2 + by^2 + 2gx + 2fy + c = 0 [/tex] but the x^2 term is missing, so i dont know what to group in brackets and then what to factor.

    I think after I get this question into standard form I will be able to tell what type of conic it is and state the key features plz help me with the first step how to get this in to standard form.
    Last edited: Mar 28, 2005
  2. jcsd
  3. Mar 28, 2005 #2
    What type of conic has only one of the variables squared in its standard form?
  4. Mar 28, 2005 #3
    yeh thats what im thinking I think it is a MAJOR typo, you know what? I think they forgot to add an x^2

    If there is an x^2 then this is the equation of a circle right cuz the values of a and b are equal?

    this is what I did but somethin is wrong because I got the radius as 0

    after adding x^2 to the eqn I grouped the x and y terms and took the 8 to the RHS
    [tex] (x^2 - 4x)+ (y^2 -4y) =-8 [/tex]
    [tex] (x^2 -4x +4) + (y^2 -4y +4)=-8+4+4 [/tex]
    [tex] (x-2)^2 + (y-2)^2 = 0 [/tex]

    What is the problem?
  5. Mar 28, 2005 #4
    It's probably not a typo. Like I asked, can you think of a type of conic that has only one of the variables squared in its standard form (I was being serious!!! :smile:)? Review all the types of conics if you need to. There are four (well, really three, but you are probably counting circles separately from ellipses).
  6. Mar 29, 2005 #5
    opps lol sorry

    [tex] (x - h)^2 = 4p(y - k) [/tex] a parabola opening up or down

    [tex] (y - k)^2 = 4p(x - h) [/tex] a parabola opening left or right

    these two equations for parabolas in standard form only require one term to be squared.

    I dont know how to get the equation given into standard form at all plz help me, I have tried with ellipse but never parabola.
    Last edited: Mar 29, 2005
  7. Mar 29, 2005 #6
    Solve for [itex]x[/itex] in terms of [itex]y[/itex], then complete the square for the [itex]y[/itex] terms.
  8. Mar 29, 2005 #7
    Im not sure I understand what you mean.
  9. Mar 29, 2005 #8
    [tex] y^2-4x-4y+8=0 \Longrightarrow 4x = y^2 - 4y + 8[/tex]

    then complete the square on the right side.
  10. Mar 29, 2005 #9
    when I completed the square on the RHS I got

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

    what do I do now?
    Last edited: Mar 29, 2005
  11. Mar 29, 2005 #10
    That's fine. Now you just need to identify all the important characteristics and such. Is it opening left/right or up/down? Where is its vertex? etc.
  12. Mar 29, 2005 #11
    is that +4 supposed to stay on that side? How come this equation doesnt really look like one of the parabola standard form equations?

    I think the center is (2,4)

    How do you figure out the value of P (focal length)?

    and the point where the focus is and the equation of the direct x?

    The reason I am asking is because the example I have is on this equation
    State the key feature of (x -1)^2 = 12 (y -2) Is my equation like this one? Can you put my equation in this form so that I can follow the example to find all of the key features? :redface:
  13. Mar 29, 2005 #12
    [tex]4x = (y-2)^2+4 \Longrightarrow 4(x-1) = (y-2)^2[/tex]
  14. Mar 31, 2005 #13
    Does P=1 what are the points for the focus?
  15. Apr 1, 2005 #14
    i've just done stuff like this now

    you divide by 4 to isolate x, and you get an inverse of a parabola

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

    you can easily get the vertex and axis of symmetry from this, as well as y and x interecepts
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