Harmonic ratio in a parabola and a peaked kissing circle

Question

Using the GeoGebra program a few days ago, I came up with a wonderful feature about the parabola and the circle kissing its peak. I don't know if it is new or previously discovered. Please, if it was discovered previously, put a reference mentioning it in the comments, and in any case, can anyone prove it I don't think a lot of words are needed, this picture includes intuition

Answer 1

This is not exactly a pretty solution, since it is entirely based on analytic geometry and it doesn't really offer any insight referring to conics and curvature. I am sure there is a much more clever solution involving some nice invariant of some cleverly chosen transformation of the plane. Alas, I am no geometer and couldn't get one to work. Also no idea if this is a classical problem. The proof is divided into some steps.

Step 1 (Choosing coordinates): By an appropriate combinations of rotations, scaling and translations, we can assume that the directrix $L$ is the horizontal line with equation $y=-1$ and that the focal point is $F(0,1)$. This implies that the parabola will be symmetric around the $y$ axis and that $V$ is the origin of the plane, $V(0,0)$. This means the parabola, viewed as a mapping, will take the form $x\mapsto \alpha x^2$. Solving the equation $|PF|=|PQ|$, where $Q$ is the orthogonal projection of $P$ onto the directrix and $P$ is an arbitrary point of the parabola, quickly gives us that $\alpha=\tfrac{1}{4}$. Therefore, a general point on the parabola will have coordinates $(x,\tfrac{x^2}{4})$.

Step 2 (Osculating circle): By symmetry, it is very clear that the osculating circle will have its center on the $y$ axis. Therefore the center will have coordinates $O(0,c)$ for some positive $c$. Since the circle is tangent to the parabola at $V$, its radius will also be $c$. So the osculating circle will have equation $x^2+(y-c)^2=c^2$ or $y^2-2cy+x^2=0$. Expressing $y$ in terms of $x$ we have $y=c\pm \sqrt{c^2-x^2}$. Fix some small $\varepsilon>0$ and define the function $f:[0,\varepsilon]\to\mathbb{R}$ given by $f(x)=c-\sqrt{c^2-x^2}$ to describe the local behavior of the osculating circle near $V$ (on the positive side, but again we have symmetry). This is clearly differentiable with $f'(x)=\tfrac{x}{\sqrt{c^2-x^2}}$. Define $p:[0,\varepsilon]\to\mathbb{R}$ by $p(x)=\tfrac{x^2}{4}$ to capture the behavior of the parabola on the same neighborhood. Again, this is differentiable with $p'(x)=\tfrac{x}{2}$. The idea now is that if we want to figure out $c$, we need to find the maximum $c$ such that $f'(x)\geq p'(x)$ on $[0,\varepsilon]$. The slope of the circle being at least that of the parabola will guarantee no additional intersections locally and the maximality of $c$ will guarantee that we are taking the largest such circle possible, i.e., the osculating circle. The condition on the derivatives reads $\tfrac{x}{\sqrt{c^2-x^2}}\geq \tfrac{x}{2}$ or $2\geq \sqrt{c^2-x^2}$. So the maximal $c$ is $c=2$. Therefore the osculating circle has center $(0,2)$ and radius $2$.

Step 3 (The easy case when $NM$ is horizontal): Due to symmetry, it is clear that in this case the relevant points are $S(0,4)$ (as the 'north pole' of the osculating circle) and therefore $N(-4,4)$ and $M(4,4)$. Clearly then, $|SN|=4=|SM|$ and the desired equality is easy to check as $|FV|=1$.

Step 4 (The general case of the tangent $NM$): We want to show that $\tfrac{1}{\sqrt{|SN|}}+\tfrac{1}{\sqrt{|SM|}}=1$. Figuring out the correct parametrization of the problem to simplify as much as possible the ugly computations took some time, but inspired by the trivial case above and the intuition that some semblance of a variational argument should be use to prove an invariant, I settled on the idea of regarding the $x$ coordinate of $N$ as a sub unit scalar times the $x$ coordinate in the horizontal case.

In other words, we will think of $N$ as having coordinates $N(-4a,4a^2)$ for some $0<a\leq 1$. In fact, intuition says that the correct interval for $a$ is $(\tfrac{1}{2},1]$ and we will confirm this later. Them $M$ will have coordinates $M(4b,4b^2)$ for some $b\geq 1$. We are therefore looking at the case where the tangent $NM$ has non-negative slope, but again this comes with no loss of generality due to symmetry. Again, on an intuitive level, we expect that the tangent will touch the osculating circle in the fourth quadrant (top left), and we will confirm this later.

A direct computation shows that the equation of the line $NM$ is given by $y=(b-a)x+4ab$. A point $(x,y)$ lying on the intersection of this line and the osculating circle must satisfy the equation $x^2+\big((b-a)x+4ab-2\big)^2=4$. Simplifying, we get \begin{equation} \big(1+(b-a)^2\big)x^2 + 4(b-a)(2ab-1)x+16ab(ab-1)=0.\quad(*) \end{equation}

Crucially, we want a unique intersection point, so the discriminant of $(*)$ must be $0$. After some computations, this leads to \begin{equation} (1-4a^2)b^2+2ab+a^2=0.\quad(**) \end{equation}

$(**)$ leads to two possible solutions for $b$, namely $b=-\tfrac{a}{1+2a}$ and $b=\tfrac{a}{2a-1}$. The first solution would imply $b\leq 0$, so we can discard it and we see that \begin{equation} b=\tfrac{a}{2a-1}. \quad (***) \end{equation} This confirms our prior intuitive remarks. Note that $a$ must be greater than $\tfrac{1}{2}$ and as $a$ approaches $\tfrac{1}{2}$, $b$ goes to infinity. The $y$ coordinate of the tangent point will always be at least $2$, meaning that the correct dependency of $y$ on $x$ for the tangent point $S$ is given by $y=2+\sqrt{4-x^2}$.

Since the discriminant of $(*)$ is $0$, we can find the solution $x=-2\tfrac{(b-a)(2ab-1)}{1+(b-a)^2}$ and plugging $(***)$ yields, after some tedious computations, that \begin{equation} x=-2\tfrac{2a-2a^2}{2a^2-2a+1} \quad \text{ and } \quad y=2\tfrac{2a^2}{2a^2-2a+1} \end{equation} are the coordinates of $S$.

After more tedious computations and some rather 'miraculous' simplifications that suggest a deeper meaning to the problem, we get $\sqrt{|SN|}=2a$ and $\sqrt{|SM|}=\tfrac{2a}{2a-1}$, from which the desired conclusion trivially follows.

Answer 2

I will establish the property, without loss of generality, for the standard parabola with equation $y=x^2$, knowing that in this case $F=(0,\frac14)$, $O=(0,\frac12)$, with a radius $r=\frac12$ for the circle (see figure).

As any (non vertical) tangent line to the circle intersects the parabola in a point $M$ with abscissa $>\frac12$, we will assume that :

$$M=(a,a^2) \ \text{with} \ a>\frac12.$$

The generic straight line passing through $M$ has equation :

$$y-a^2-m(x-a)=0\tag{1}$$

Let us express that this line is at distance $r=\frac12$ from point $O$ ; by a classical formula :

$$\frac{|\tfrac12-a^2+am|}{\sqrt{1+m^2}}=\frac12\tag{2}$$

Squaring (2) gives the following quadratic equation where the unknown is $m$:

$$1+2(am-a^2)=1+m^2$$

Its roots are

$$m_1=2\frac{a^2-a}{2a-1}, \ \ \ m_2=2 \frac{a^2+a}{2a+1}\tag{3}$$

We need only consider root $m_1$ (in fact, it is the "upper" tangent to the circle issued from $M$) ; it means that (1) gives :

$$y-\frac{2(a^2-a)}{2a-1}(x-a)-a^2=0\tag{4}$$

for the equation of line $MN$.

In order to get the abscissa of $N$ (whose coordinates verify $y=x^2$), using (4), we have to solve the following quadratic equation in variable $x$ :

$$x^2-\frac{2(a^2-a)}{2a-1}(x-a)-a^2=0\tag{5}$$

It has two roots :

$$x_1=a, \ \ \ x_2=\underbrace{\frac{a}{1-2a}}_{f(a)}\tag{6}$$

Of course, only $x_2$ is of interest.

Therefore : $N=(f(a),f(a)^2)$.

Let us compute :

$$OM^2=distance((a,a^2),(0,\tfrac12))^2=a^2+(a^2-\tfrac12)^2=a^4+\tfrac14$$

Applying Pythagoras theorem in right triangle $OMS$ :

$$SM^2+SO^2=OM^2 \ \ \iff \ \ SM^2+(\tfrac12)^2=a^4+\tfrac14,$$

giving the very simple result :

$$SM^2=a^4 \iff \frac{1}{\sqrt{SM}}=\frac{1}{|a|}$$

For a similar reason :

$$\frac{1}{\sqrt{SN}}=\frac{1}{|f(a)|}$$

$$\frac{1}{\sqrt{SM}}+\frac{1}{\sqrt{SN}}=\frac{1}{|a|}+\frac{1}{|f(a)|}=\frac{1}{a}-\frac{1}{f(a)}$$

(we have taken into account condition $a>\frac12 \implies f(a)<0$). We can conclude :

$$\frac{1}{\sqrt{SM}}+\frac{1}{\sqrt{SN}}=\frac{1}{a}-\frac{1-2a}{a}=2=\frac{1}{\sqrt{FV}}$$

as desired.

Remark : formula (6) defining $f(a)$ is very similar to formula (***) in the answer by @AnCar.

Answer 3

I have already posted an answer. Meanwhile, I have found a more direct solution, where the concept of power of a point with respect to a circle plays the central role. It is why I have decided to present it as a separate answer.

Fig. 1 : Animated version here ; point $M$ is moveable (GeoGebra plot).

Let us consider without loss of generality the particular case of curve $y=x^2$ with "osculating circle" $(C)$ at vertex $V$ (maximal curvature, minimal radius of curvature) with center $O=(0,\tfrac12)$, radius $r=\tfrac12$, therefore with equation :

$$x^2+y^2-y=0\tag{1}$$

with respect to cartesian axes where $V$ is the origin and the abscissa's axis taken along the tangent line $(L)$ to the parabola at point $V$.

Now, consider the identity :

$$x^2-y=\underbrace{(x^2+y^2-y)}_{P_{(C)}(x,y)}-(y^2)\tag{2}$$

where $P_{(C)}(x,y)$ is the power of a generic point in the plane with respect to circle $(C)$ (let us recall that its value is $>0$, resp. $<0$, when the point is outside, resp. inside, circle (C), and zero when $(x,y)$ is on (C) the circle). (2) implies :

$$P_{(C)}(x,y) = y^2 \ \ \iff \ \ y=x^2\tag{3}$$

With words : the parabola is the locus of points $(x,y)$ for which there is equality between $P_{(C)}(x,y)$ and $y^2$.

But the power of a point $M$ exterior to a circle is known to be $MS^2$ where $S$ is the point of tangency of any of the two tangents issued from point $M$ to the circle. Therefore, if we take a tangent line $MN$ (with $M=(a,a^2), \ N=(b,b^2)$ on the parabola) where we assume WLOG that :

$$b<0<a,\tag{4}$$

we just have to add the following constraint : the distance of $O=(\color{red}{0,\tfrac12})$ to line $MN$ is equal to $r=1/2$.

As line $MN$ has equation :

$$-(a+b)x+y+ab=0,\tag{5}$$

$$\text{distance}(O,MN)=\frac{|- (a+b)\color{red}{0} + \color{red}{\tfrac12} + ab |}{\sqrt{(a+b)^2+1}} = \frac12\tag{6}$$

Squaring (4) and simplifying, one gets the constraint :

$$(a-b)^2=4a^2b^2 \ \iff \ (2ab+(a-b))\underbrace{(2ab-(a-b))}_{<0, \text{using} \ (4)}=0 \ \iff \ 2ab+(a-b)=0$$

itself equivalent to :

$$\frac1a - \frac1b=2 \ \iff \ \frac{1}{a} + \frac{1}{|b|}=2 \ \iff \ \frac{1}{\sqrt{SM}} + \frac{1}{\sqrt{SN}}=\frac{1}{\sqrt{FV}}$$

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Remark 1 : I have had the idea of identity (2) by using the technique of pencil of conics of the form :

$$\underbrace{(x^2+y^2-y)}_{\text{circle (C)}}+\lambda \underbrace{(y^2)}_{\text{tangent (T)}}=0\tag{3}$$

and it appeared that the parabola belongs indeed to the pencil (with $\lambda=-1$ in (3)).

Taking $y^2=0$ instead of $y=0$ for the equation of line $(L)$ is motivated by the fact that it is only with a square that $(L)$ can be considered as a (degenerate) conic curve (twice the equation of the $x$-axis = a pair of lines).

Remark 2 : One can wonder why the center of the osculating circle in $V$ is $(0,1/2)$. Here is an explanation : it is known that the set of centers of curvature of a given curve (here the parabola), called the evolute of this curve, is the envelope of the normals to this curve.

Here, the normal line to the parabola at point $M(a,a^2)$ has equation :

$$y=\frac{1}{2a}x+a^2+\frac12\tag{4}$$

A classical technique (elimination of parameter $a$ between equation (4) and its derivative with respect to parameter $a$) gives the equation of the envelope which is :

$$y=3\left(\frac{x}{4}\right)^{2/3}+\frac12\tag{5}$$

As a consequence, the minimal point $(0,\tfrac12)$ of this curve is the center of the osculating circle in the apex $V$ of the parabola.

Fig. 2 : The locus of centers of curvature of the parabola (evolute curve) in red as envelope of (blue) normal lines to the parabola (GeoGebra plot).

Answer 4

The simplest solution possible is the one that utilizes elementary geometry. I am not currently available to write a full proof, but I can quickly clarify the idea: We know from a special case of Descartes' theorem that:

$\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=\frac{1}{\sqrt{z}}$

Therefore, to prove the theorem, all we need to show is that the two circles in the diagram touch the straight line tangent to the parabola at its vertex:

and then the three circles are clearly formed by defining the parabola.