three theorems on parabolas
In the Cartesian plane, pick a point with coordinates $(0,2f)$ (subtle hint!) and construct (1) the set $S$ of segments $s$ joining $F=(0,2f)$ with the points $(x,0)$, and (2) the set $B$ of rightbisectors $b$ of the segments $s\in S$.
Theorem 1 :
Proof:
We’re lucky in that we don’t need a fancy definition of envelope; considering a line to be a set of points it’s just the boundary of the set $C={\cup}_{b\in B}b$. Strategy: fix an $x$ coordinate and find the max/minimum of possible $y$’s in C with that $x$. But first we’ll pick an $s$ from $S$ by picking a point $p=(w,0)$ on the $x$ axis. The midpoint^{} of the segment $s\in S$ through $p$ is $M=(\frac{w}{2},f)$. Also, the slope of this $s$ is $\frac{2f}{w}$. The corresponding rightbisector will also pass through $(\frac{w}{2},f)$ and will have slope $\frac{w}{2f}$. Its equation is therefore
$$\frac{2y2f}{2xw}=\frac{w}{2f}.$$ 
Equivalently,
$$y=f+\frac{wx}{2f}\frac{{w}^{2}}{4f}.$$ 
By any of many very famous theorems^{} (Euclid book II theorem twentysomething, CauchySchwarzBunyakovski (overkill), differential calculus, what you will) for fixed $x$, $y$ is an extremum^{} for $w=x$ only, and therefore the envelope has equation
$$y=f+\frac{{x}^{2}}{4f}.$$ 
I could say I’m done right now because we “know” that this is a parabola, with focal length $f$ and $x$axis as directrix. I don’t want to, though. The most popular definition of parabola I know of is “set of points equidistant from some line $d$ and some point $f$.” The line responsible for the point on the envelope with given ordinate $x$ was found to bisect the segment $s\in S$ through $H=(x,0)$. So pick an extra point $Q\in b\in B$ where $b$ is the perpendicular bisector^{} of $s$. We then have $\mathrm{\angle}FMQ=\mathrm{\angle}QMH$ because they’re both right angles^{}, lengths $FM=MH$, and $QM$ is common to both triangles^{} $FMQ$ and $HMQ$. Therefore two sides and the angles they contain are respectively equal in the triangles $FMQ$ and $HMQ$, and so respective angles and respective sides are all equal. In particular, $FQ=QH$. Also, since $Q$ and $H$ have the same $x$ coordinate, the line $QH$ is the perpendicular^{} to the $x$axis, and so $Q$, a general point on the envelope, is equidistant from $F$ and the $x$axis. Therefore etc.
QED.
Because of this construction, it is clear that the lines of $B$ are all tangent^{} to the parabola in question.
We’re not done yet. Pick a random point $P$ outside $C$ (“inside” the parabola), and call the parabola $\pi $ (just to be nasty). Here’s a nice quicky:
Theorem 2 The Reflector Law:
For $R\in \pi $, the length of the path $PRF$ is minimal^{} when $PR$ produced is perpendicular to the $x$axis.
Proof:
Quite simply, assume $PR$ produced is not necessarily perpendicular to the $x$axis. Because $\pi $ is a parabola, the segment from $R$ perpendicular to the $x$axis has the same length as $RF$. So let this perpendicular hit the $x$axis at $H$. We then have that the length of $PRH$ equals that of $PRF$. But $PRH$ (and hence $PRF$) is minimal when it’s a straight line; that is, when $PR$ produced is perpendicular to the $x$axis.
QED
Hey! I called that theorem the “reflector law”. Perhaps it didn’t look like one. (It is in the Lagrangian formulation), but it’s fairly easy to show (it’s a similar^{} argument^{}) that the shortest path from a point to a line to a point makes and “reflected” angles equal.
One last marvelous tidbit. This will take more time, though. Let $b$ be tangent to $\pi $ at $R$, and let $n$ be perpendicular to $b$ at $R$. We will call $n$ the to $\pi $ at $R$. Let $n$ meet the $x$axis at $G$.
Theorem 3 :
The radius of the “bestfit circle” to $\pi $ at $R$ is twice the length $RG$.
Proof:
(Note: the $\approx $’s need to be phrased in terms of upper and lower bounds, so I can use the sandwich theorem, but the proof schema is exactly what is required).
Take two points $R,{R}^{\prime}$ on $\pi $ some small distance^{} $\u03f5$ from each other (we don’t actually use $\u03f5$, it’s just a psychological trick). Construct the tangent $t$ and normal $n$ at R, normal ${n}^{\prime}$ at ${R}^{\prime}$. Let $n,{n}^{\prime}$ intersect at $O$, and $t$ intersect the $x$axis at $G$. $RF,{R}^{\prime}F$. Erect perpendiculars $g,{g}^{\prime}$ to the $x$axis through $R,{R}^{\prime}$ respectively. $R{R}^{\prime}$. Let $g$ intersect the $x$axis at $H$. Let $P,{P}^{\prime}$ be points on $g,{g}^{\prime}$ not in $C$. Construct $RE$ perpendicular to $RF$ with $E$ in ${R}^{\prime}F$. We now have

i)
$\mathrm{\angle}PRO=\mathrm{\angle}ORF=\mathrm{\angle}GRH\approx \mathrm{\angle}{P}^{\prime}{R}^{\prime}O=\mathrm{\angle}O{R}^{\prime}F$

iii)
$ER\approx FR\cdot \mathrm{\angle}EFR$

v)
$\mathrm{\angle}{R}^{\prime}RE+\mathrm{\angle}ERO\approx \frac{\pi}{2}$ (That’s the number $\pi $, not the parabola)

vii)
$\mathrm{\angle}ERO+\mathrm{\angle}ORF=\frac{\pi}{2}$

ix)
$\mathrm{\angle}{R}^{\prime}ER\approx \frac{\pi}{2}$

xi)
$\mathrm{\angle}{R}^{\prime}OR=\frac{1}{2}\mathrm{\angle}{R}^{\prime}FR$

xiii)
${R}^{\prime}R\approx OR\cdot \mathrm{\angle}{R}^{\prime}OR$

xv)
$FR=RH$
From (iii),(iv) and (i) we have $\mathrm{\angle}{R}^{\prime}RE\approx \mathrm{\angle}GRH$, and since ${R}^{\prime}$ is close to $R$, and if we let ${R}^{\prime}$ approach $R$, the approximations approach equality. Therefore, we have that triangle ${R}^{\prime}RE$ approaches similarity with $GRH$. Therefore we have $R{R}^{\prime}:ER\approx RG:RH$. Combining this with (ii),(vi),(vii), and (viii) it follows that $RO\approx 2RG$, and in the limit ${R}^{\prime}\to R$, $RO=2RG$.
QED
This last theorem is a very nice way of shortcutting all the messy calculus needed to derive the Schwarzschild “BlackHole” solution to Einstein’s equations, and that’s why I enjoy it so.
Title  three theorems on parabolas 

Canonical name  ThreeTheoremsOnParabolas 
Date of creation  20130322 12:40:51 
Last modified on  20130322 12:40:51 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  19 
Author  CWoo (3771) 
Entry type  Topic 
Classification  msc 51N20 
Related topic  PropertiesOfParabola 