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Homethree theorems on parabolas
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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 $\angle FMQ=\angle QMH$ 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:
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 “incident” 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 normal to $\pi$ at $R$. Let $n$ meet the $x$axis at $G$.
Theorem 3 :
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 $\epsilon$ from each other (we don’t actually use $\epsilon$, 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 $G$. Join $RF,R^{{\prime}}F$. Erect perpendiculars $g,g^{{\prime}}$ to the $x$axis through $R,R^{{\prime}}$ respectively. Join $RR^{{\prime}}$. 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

$\angle PRO=\angle ORF=\angle GRH\approx\angle P^{{\prime}}R^{{\prime}}O=\angle OR% ^{{\prime}}F$

$ER\approx FR\cdot\angle EFR$

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

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

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

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

$R^{{\prime}}R\approx OR\cdot\angle R^{{\prime}}OR$

$FR=RH$
From (iii),(iv) and (i) we have $\angle R^{{\prime}}RE\approx\angle GRH$, 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 $RR^{{\prime}}:ER\approx RG:RH$. Combining this with (ii),(vi),(vii), and (viii) it follows that $RO\approx 2RG$, and in the limit $R^{{\prime}}\rightarrow R$, $RO=2RG$.
QED
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Comments
little advice
when creating entries titles, it's wise to capitalize only proper nouns and its derivates. This is because the way system works.
If we don't do it this way, when automatically linking happens, text like
"The Hypotenuse of a Square is equal to the Sum of the Squares of its Legs".
f
G > H G
p \ /_  ~ f(G)
\ / f ker f
G/ker f
Re: little advice
Many thanks, but now I'm somewhat confused, because
I don't know how to parse your second paragraph.
I'll try again...
Re: little advice
What he means is that if you capitalize an entry name, the automatic linker will capitalize references to that entry.
So if you wrote an entry called "Parabola" (instead of "parabola"), and someone else wrote
"Consider the parabola $y=x^2$..."
in his entry, it would end up as
"Consider the Parabola $y=x^2$..."
which just looks silly.
So the accepted style is to only capitalize proper names in titles.
Kevin
I like this
I like your fresh writing style.
Have you considered changing this to "topic" or writing topickind entries? we're in need of them. There are lots of definitions and theorems, but we need some texts on general stuff.
also, if you write some stuff like an exposition or a paper we have special sections for posting them.
f
G > H G
p \ /_  ~ f(G)
\ / f ker f
G/ker f
Re: I like this
Yes, that would be fine. (That's the 1st
isomorphism thm, your signature, right?)
And I'll see what I can do. My grounding is
rather sketchy in most areas, but I like writing,
I like math, geometry.
Re: I like this
yes, the 1st iso thm :)
it's ok, we are not experts anyway. And when one of us really goes wrong, someone else notices and sends a correction
f
G > H G
p \ /_  ~ f(G)
\ / f ker f
G/ker f