ゼロから放物線まで14シンボル

Post up, weekend analytic geometers, get ready to race for the
parabola
and approach the starting line with
0 and 7 other symbols:

      0   =  
( x −1) 2

And they’re off!   Three symbols ( +,
y  and 2 ) assume the lead at an
early point:

            0   =  
( x −1) 2 +
y 2

Prospects don’t yet seem good for a parabola.
This is one race not meant to end in a “winner’s
circle.”

Whoa!   A cloud of dust trails an impossibly
complex stagger through the far turn as another
two symbols ( + and 2 ) charge
ahead:

                      0   =  
( x −1) 2 +
y 2 + 2

Hold on.   Two more symbols come out
of nowhere, gain the homestretch,
and append themselves just in time.
Could it be?
Yes, at the finish line (curve, that is) it’s a real
parabola!

最後の2つのシンボルは何ですか?

And how did these boldfaced puns relate to nearby equations?
     starting line
     early point
     winner’s circle
     impossibly complex

ベストアンサー

最終的な解決策の試みは、それぞれの bare-faced 太字のパンを順に説明します。

  • starting line

    The equation $0=(x-1)^2$ describes a line in the $(x,y)$ plane,
    namely the vertical line $x=1$.

  • early point

    The equation $0=(x-1)^2+y^2$ describes a single point in the
    $(x,y)$ plane, namely the point $x=1,y=0$.

  • winner’s circle

    Any equation of the form $0=(x-1)^2+y^2-c$ with $c$ a positive
    constant would describe a circle in the $(x,y)$ plane, namely with
    centre $(1,0)$ and radius $sqrt{c}$.

  • impossibly complex

    The equation $0=(x-1)^2+y^2+2$ has no real solutions for $x$ and
    $y$, since squares of real numbers are non-negative and so the
    right-hand side should be at least $2$. So the only way to solve
    this equation involves treating $x$ and $y$ as complex
    variables.

  • real parabola

    We need to stick two symbols on the end of the equation
    $0=(x-1)^2+y^2+2$ in order to form the equation of a parabola,
    which in general should take the form $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
    with coefficients satisfying $B^2=4AC$.

    The two last symbols (found by @TheGreatEscaper):

    $xy$,

    thus forming the final result

    $0=(x-1)^2+y^2+2xy$, which can be rewritten as $(x+y)^2=2x-1$ –
    definitely a real parabola.

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