# 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 &hairsp; = &hairsp;
(&hairsp;x&hairsp;−1)&hairsp;2

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

0 &hairsp; = &hairsp;
(&hairsp;x&hairsp;−1)&hairsp;2 +
y&hairsp;2

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

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

0 &hairsp; = &hairsp;
(&hairsp;x&hairsp;−1)&hairsp;2 +
y&hairsp;2 + 2

Hold on. &hairsp; 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!

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

• 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

• 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.