# 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 right-bisectors $b$ of the segments $s\in S$.

## Theorem 1 :

The envelope described by the lines of the set $B$ is a parabola with $x$-axis as directrix and focal length $|f|$.

## 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 right-bisector will also pass through $(\frac{w}{2},f)$ and will have slope $\frac{w}{2f}$. Its equation is therefore

 $\frac{2y-2f}{2x-w}=\frac{w}{2f}.$

Equivalently,

 $y=f+\frac{wx}{2f}-\frac{w^{2}}{4f}.$

By any of many very famous theorems (Euclid book II theorem twenty-something, Cauchy-Schwarz-Bunyakovski (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:

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 “best-fit 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 $\epsilon$ 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 $G$. $RF,R^{\prime}F$. Erect perpendiculars $g,g^{\prime}$ to the $x$-axis through $R,R^{\prime}$ respectively. $RR^{\prime}$. 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

• i)

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

• iii)

$ER\approx FR\cdot\angle EFR$

• v)

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

• vii)

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

• ix)

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

• xi)

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

• xiii)

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

• xv)

$FR=RH$

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

This last theorem is a very nice way of short-cutting all the messy calculus needed to derive the Schwarzschild “Black-Hole” solution to Einstein’s equations, and that’s why I enjoy it so.

Title three theorems on parabolas ThreeTheoremsOnParabolas 2013-03-22 12:40:51 2013-03-22 12:40:51 CWoo (3771) CWoo (3771) 19 CWoo (3771) Topic msc 51N20 PropertiesOfParabola