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Math Help - Parabola(multiple choice)

  1. #1
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    Parabola(multiple choice)

    Find the equation of a parabola w/ axis parallel to the x-axis and passing through
    (3/2 , 1),(5,0) and (-1,2)?

    i tried solving this problem
    first substituting coordinate (5,0) to the general eqn giving me 2 unknowns
    substituting the rest of the two coords gives me 4 unknowns each

    ending up with 3 unknowns which is impossible to solve
    any other way to solve this?

    Choices:
    A. y^2 -2x -8y -10 = 0
    B. y^2 -2x -8y+10 = 0
    C. y^2 +2x -8y +10 = 0
    D. y^2 -2x +8y +10 = 0
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  2. #2
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    Re: Parabola(multiple choice)

    Hello, TechnicianEngineer!

    Find the equation of a parabola w/ axis parallel to the x-axis
    and passing through (3/2 , 1),(5,0) and (-1,2).

    . . \begin{array}{ccccccccc}(A) & y^2 -2x -8y -10 \:=\: 0 && (B) & y^2 -2x -8y+10 \:=\:0 \\ \\ (C) & y^2 +2x -8y +10 :=\: 0 && (D) & y^2 -2x +8y +10 \:=\: 0 \end{array}

    Since the axis is parallel to the x-axis, we have a "horizontal" parabola.
    . . The general form is: . x \:=\:ay^2 + by + c


    Substitute the three points:

    . . \begin{array}{cccccccccc}(\frac{3}{2},\,1): & \frac{3}{2} &=& a(1^2) + b(1) + c &\Rightarrow& a + b + c \:=\:\frac{3}{2}  \\ \\[-3mm](5,\,0): & 5 &=&  a(0^2) + b(0) + c &\Rightarrow& c\:=\:5 \\ \\[-3mm] (\text{-}1,\,2): & \text{-}1 &=& a(2^2) + b(2) + c &\Rightarrow& 4a + 2b + c \:=\:\text{-}1 \end{array}


    Solve the system of equations: . a = \tfrac{1}{2},\;b = \text{-}\:\!4,\;c = 5


    We have: . x \:=\:\tfrac{1}{2}y^2 - 4y + 5 \quad\Rightarrow\quad  2x \:=\:y^2 - 8y + 10

    Therefore: . y^2 - 2x - 8y + 10 \:=\:0\;\;\text{ Answer (B)}

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    Re: Parabola(multiple choice)

    thanks soroban

    just a little clarification since the general eqn of parabola is Cy^2 + Dx + Ey + F = 0 ,
    why does "x" variable not have an unknown constant coefficient Dx and just let D = 1
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  4. #4
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    Re: Parabola(multiple choice)

    Quote Originally Posted by TechnicianEngineer View Post
    thanks soroban

    just a little clarification since the general eqn of parabola is Cy^2 + Dx + Ey + F = 0 ,
    why does "x" variable not have an unknown constant coefficient Dx and just let D = 1
    1. To determine the coefficients of the variables you have to use 4 equations, that means you must know 4 points of the parabola.

    2. Cy^2 + Dx + Ey + F = 0~\implies~x=-\frac CD y^2-\frac ED y - \frac FD
    With a = -\frac CD,\ b= -\frac ED,\ c= - \frac FD you'll get Soroban's equation. Then you have

    • separated the variables
    • you only need 3 points to determine the coefficients.
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