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 thepoints $(x,0)$, 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, whatyou 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. Themost 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 onthe 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 $b$ isthe perpendicular bisector of $s$. We then have $\mathrm{\angle }FMQ=\mathrm{\angle }QMH$ because they’re both right angles, lengths $FM=MH$, and $QM$ iscommon 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 asimilar 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 isrequired).

Take two points $R,R^{\prime }$ on $\pi$ some small distance $ϵ$ from each other (we don’t actually use $ϵ$, 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

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

• $ER\approx FR\cdot \mathrm{\angle }EFR$

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

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

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

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

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

• $FR=RH$

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 approximationsapproach 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 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.