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  1. #1
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    parable

    Finding the equation of the parable and build it

    Axis parallel to y = 0 and goes through A(-2,4) B(-3,2) C(-11,-2)



    Answer
    y^2-8y+4x+24=0
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  2. #2
    Junior Member bleesdan's Avatar
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    I hope you mean "parabola"

    There are two forms of parabolas. What are they?

    Once you know which of the two you need, how can you determine the specifics of the equation?
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  3. #3
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    The equation is:

    (y-k)^2=2p(x-h)


    but with points and I think the equation
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  4. #4
    Junior Member bleesdan's Avatar
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    All right. We have an equation now. That's a step in the right direction.
    A parabola is defined by three points. Since we have three points here, we can define it.
    In our equation (y-k)^2=2p(x-h), we can use (x,y)=(-2,4) to get an equation in three unknowns: h,k,p. Doing the same with the other two points will give us three equations in three unknowns, and from there, we can solve for h,k,p.
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  5. #5
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    Thank you. Do you have any means of online communication, for example, msn
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  6. #6
    Junior Member bleesdan's Avatar
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    No, I really don't. My computer is so old, I can barely run Itunes and Firefox at the same time.
    However, I will be checking this site frequently, so a private message to me would probably be easiest.
    I'm glad to have been of assistance.
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  7. #7
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    I not find the answer and you?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    Finding the equation of the parable and build it

    Axis parallel to y = 0 and goes through A(-2,4) B(-3,2) C(-11,-2)



    Answer
    y^2-8y+4x+24=0
    Quote Originally Posted by Apprentice123 View Post
    The equation is:

    (y-k)^2=2p(x-h)


    but with points and I think the equation
    So
    (4 - k)^2 = 2p(-2 - h)

    (2 - k)^2 = 2p(-3 - h)

    (-2 - k)^2 = 2p(-11 - h)

    So we get
    k^2 - 8k + 16 = -4p - 2ph (1)

    k^2 - 4k + 4 = -6p - 2ph (2)

    k^2 + 4k + 4 = -22p - 2ph (3)

    Subtract equation (2) from equation (1) and then subtract equation (3) from equation (1)
    -4k + 12 = 2p

    -12k + 12 = 18p

    You can find values of k and p from these and then use one of the original equations to find h.

    Another (perhaps simpler in this case) way to approach this is to use the "standard" form for this kind of parabola:
    ay^2 + by + c = x
    and plug the points into that. This gives you a linear system in a, b, and c.

    -Dan
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  9. #9
    A riddle wrapped in an enigma
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    Quote Originally Posted by topsquark View Post


    Another (perhaps simpler in this case) way to approach this is to use the "standard" form for this kind of parabola:
    ay^2 + by + c = x
    and plug the points into that. This gives you a linear system in a, b, and c.

    -Dan
    (-2,4), (-3,2), (-11,-2)

    ay^2+by+c=x

    Substituting each ordered pair into the standard equation, we get:

    1.\ \ a(4)^2+b(4)+c=-2
    16a+4b+c=-2

    2. \ \ a(2)^2+b(2)+c=-3
    4a+2b+c=-3

    3. \ \ a(-2)^2+b(-2)+c=-11
    4a^2-2b+c=-11

    Solve the system using matrix equation (or any method you like):

    \left[\begin{array}{cccc}16 & 4 & 1 & \\ 4 & 2 & 1 & \\ 4 & -2 & 1 \\ \end{array}\right] \cdot \left[\begin{array}{c}a \\ b \\ c \end{array}\right] = \left[\begin{array}{c}-2 \\ -3 \\ -11 \end{array}\right]

    \left[\begin{array}{cccc}16 & 4 & 1 & \\ 4 & 2 & 1 & \\ 4 & -2 & 1 \\ \end{array}\right]^{-1} \cdot \left[\begin{array}{c}-2 \\ -3 \\ -11 \end{array}\right] = \left[\begin{array}{c}a \\ b \\ c \end{array}\right]

     \left[\begin{array}{c}a \\ b \\ c \end{array}\right]= \left[\begin{array}{c}-\frac{1}{4} \\ 2 \\ -6 \end{array}\right]

    Substituting back into your original equaton:

    ay^2+by+c=x

    -\frac{1}{4}y^2+2y-6=x

    \boxed{y^2-8y+4x+24=0}
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  10. #10
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    Thank you very much
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