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Math Help - parabolas

  1. #1
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    parabolas

    how do you find the focus and the directrix line if all you have is the equation in parabolic form?
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    Quote Originally Posted by danielle1 View Post
    how do you find the focus and the directrix line if all you have is the equation in parabolic form?
    see post #2 here
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    woah i have no idea where that p came from.
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    Quote Originally Posted by danielle1 View Post
    woah i have no idea where that p came from.
    it is just a general form. all the letters but x and y represent constants. your objective is to try and get your parabola into that form. then you can figure out what p and the other letters are
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  5. #5
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    ughh i wish my teacher would do his job. i have no idea what you mean. like if i have y=(x-2)+3, i know how to get the vertex but not the focus or the directrix
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    Quote Originally Posted by danielle1 View Post
    ughh i wish my teacher would do his job. i have no idea what you mean. like if i have y=(x-2)+3, i know how to get the vertex but not the focus or the directrix
    that's not a parabola
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  7. #7
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    wait.... what??? why not?
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    Quote Originally Posted by danielle1 View Post
    wait.... what??? why not?
    the x is not squared. i suppose you meant to type y = (x - 2)^2 + 3

    now, you should know that if a parabola is in the form y = a(x - h)^2 + k, the vertex is given by (h,k)...oh, you do know that. ok, moving on.

    now we want to get this parabola to look like the form i used in the post i directed you to.

    the form must look like (x - h)^2 = 4p(y - k)

    in this form, the focus is: (h, k + p)

    and the directrix is: y = k - p


    now, y = (x - 2)^2 + 3

    \Rightarrow (x - 2)^2 = y - 3

    the coefficient of y is 1, so where would the 4 come from? well, we can notice that 1 = \frac 44 = 4 \frac 14

    thus, we have (x - 2)^2 = 4 \frac 14y - 3 = 4 \frac 14(y - 3)

    so, we have (x - 2)^2 = 4 \cdot \frac 14(y - 3) ....look familiar?
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  9. #9
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    omg i think i actually get it now. thank you so much!
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    Quote Originally Posted by danielle1 View Post
    omg i think i actually get it now. thank you so much!
    ok, what is the answer then?

    (i will have to leave for a while, maybe a half hour or so, so don't get upset with me if i don't verify your answer right away)
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  11. #11
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    hmm.. what do i do with the p again?
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    Quote Originally Posted by danielle1 View Post
    hmm.. what do i do with the p again?
    you just said you got it.

    look back at my post. what did i say the focus was? what did i say te directrix was?
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  13. #13
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    k+p, but i don't know what to plug in for p
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    Quote Originally Posted by danielle1 View Post
    k+p, but i don't know what to plug in for p
    it seems you are not reading through my posts. first of all, i never said anything was k + p. neither the focus nor the directrix is k + p. the focus is a coordinate and the directrix is a horizontal line.

    i am sorry, but i really do not see the problem you are having. i went through the algebraic manipulations and tried to make it plain. but let's try again.


    (x - {\color{red}h})^2 = 4{\color{blue}p}(y - {\color{green}k}) ...........this is the form we want
    ....... \downarrow......... \downarrow....... \downarrow
    (x - {\color{red}2})^2 = 4 {\color{blue}\frac 14}(y - {\color{green}3}) ............this is the form we got our equation in

    now, the focus is: (h, k + p)

    and the directrix is: y = k - p

    can you tell me what h, p and k are?
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  15. #15
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    i meant k+p as the y coordinate. and i didn't realize that p is 1/4.... here i'll try again now
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