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Math Help - Proving y^2=-4x

  1. #1
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    Proving y^2=-4x

    I am given equations with coefficients related by a common ratio, thus a geometric sequence, such as

    2x+4y=8
    3x+12y=48

    I plotted numerous lines which obey this geometric pattern and discovered that there is a curve which through trial and error I discovered to be y=(-4x)^(1/2). I also discovered that each line was tangential to the curve.

    I proceeded to create a general proof of the solutions of equations obeying this general form:

    ax+ary=ar^2
    bx+bdy=bd^2

    I discovered the solution of x and y using simultaneous equations (I am not posting the answer here because this is a portfolio task and direct answers are not allowed).

    I am now stuck on proving the curve; can anybody tell me if my approach above is correct in working the relations and to go where from this point.
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  2. #2
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    Hello, chaosier!

    I am given equations with coefficients related by a common ratio,
    thus a geometric sequence, such as: . \begin{array}{ccc}2x+4y&=&8 \\ 3x+12y&=&48 \end{array}

    I plotted numerous lines which obey this geometric pattern
    . . and discovered that there is a curve which,
    . . through trial and error, I found to be: . y\:=\:(-4x)^{\frac{1}{2}} . . . . I agree!
    I also found that each line was tangential to the curve.

    I proceeded to create a general proof of the solutions
    . . of equations with this general form: . \begin{array}{cccc}ax+apy &=&ap^2 & [1] \\ bx+bqy &=& bq^2 & [2] \end{array}

    I found the solution of x and y using simultaneous equations.

    I am now stuck on proving the curve.
    Can anybody tell me if my approach above is correct
    . . and where to go from this point?

    \begin{array}{ccccccc}\text{Divide [1] by }a\!: & x + py &=& p^2 & [3] \\ \text{Divide [2] by }b\!: & x + qy &=& q^2 & [4]\end{array}

    Subtract [3] - [4]: . py - qy \:=\:p^2-q^2 \quad\Rightarrow\quad (p-q)y \:=\:(p-q)(p+q)

    . . Hence: . y \:=\:p+q

    Substitute into [3]: . x + p(p+q) \:=\:p^2 \quad\Rightarrow\quad x \:=\:-pq


    We have parametric equations for the intersection
    . . of two members of this family of curves.

    . . . . . \begin{Bmatrix} x &=& -pq \\ y &=& p+q \end{Bmatrix}


    To determine the "envelope" of this family of curves, let q \to p.

    . . Hence, we have: . \begin{array}{ccccc}x &=& -p^2 & [5]\\ y &=& 2p & [6]\end{array}


    Eliminate the parameter:
    . . From [6]: . y \:=\:2p \quad\Rightarrow\quad p \:=\:\tfrac{y}{2}
    . . Substitute into [5]: . x \:=\:-\left(\tfrac{y}{2}\right)^2 \quad\Rightarrow\quad \boxed{y^2 \:=\:-4x}

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  3. #3
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    Soroban, you are a genius!
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  4. #4
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    Soroban could you please elaborate a bit on q->p?

    Is it because q and p only represent two ratios when there could be any ratio from q to p, and from letting q tend to p you are proving that the graphical pattern that manifests from these results are true for all ratios and by proving this generate a general form for the graphical pattern.
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