Proving Frégier's Theorem for conics

Question

I am looking for a proof of Frégier's theorem for conics.

Pick any point $P$ on a conic section, and draw a series of right angles having this point as their vertices. Then the line segments connecting the rays of the right angles where they intersect the conic section concur in a point $P^\prime$.

In my work, I am mainly focusing on the "linear algebra" approach of conics in projective geometry, using mainly matrices. I am therefore looking for a proof of this theorem using mainly linear algebra, which I have not been thus far able to find.

Could anybody help?

Answer 1

This isn't a linear algebra proof, but you may be able to translate it into one.

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Consider the general equation of the conic: $$Ax^2 + 2 Bxy+ Cy^2+2Dx+2Ey+F=0 \tag{1}$$

Note that, if there is a "Frégier Point" corresponding to a point $P$, then it must lie on the normal at $P$. Let's assume that $P$ lies at the origin and its normal is vertical (that is, its tangent is horizontal). These conditions require that $(0,0)$ satisfies $(1)$ (making $F=0$), and that $(0,0)$ and $y^\prime=0$ satisfy $(1)$'s derivative: $$2Ax + 2B(x y'+y)+2Cyy'+2D+2Ey'=0 \quad\stackrel{(0,0),y'=0}{\to}\quad D=0 \tag{2}$$

So, this is our conic:

$$Ax^2 + 2Bxy+ Cy^2+2Ey=0 \tag{3}$$

(where we may assume $E\neq 0$, else there would also be a vertical tangent at the origin). Converting to polar coordinates via $(x,y)\to (r\cos\theta,r\sin\theta)$, we can write $$r = \frac{-2 E \sin\theta}{A\cos^2\theta + 2 B \sin\theta \cos\theta + C \sin^2\theta} \tag{4}$$ which in turn leads to a parametric equation for the conic. The points corresponding to $\theta$ and $\theta+\pi/2$ are joined by this line $$2 x ( A \cos 2\theta + B \sin 2\theta) + y ( A + C )\sin 2\theta + 2 E \sin 2\theta = 0 \tag{5}$$ This line crosses the $y$-axis (that is, the normal at $P$) when $x=0$: $$y ( A + C ) \sin 2\theta + 2 E \sin 2\theta = 0 \quad\to\quad y = -\frac{2E}{A+C} \tag{6}$$ That value is independent of $\theta$, and therefore corresponds to the point common to the lines for all $\theta$. That's the desired Frégier Point. $\square$

Answer 2

Start with a point $A$ on the conic, join it with $P$, we get a line $AP$. Now consider the line perpendicular to $PA$ through $P$, cutting the conic in a second point $A'$. To show that the lines $AA'$, as $A$ varies, pass through a fixed point.

The map $A\to A'$ is a projective transformation from the conic to itself ( a non-singular conic is isomorphic to the projective line, so the notion of "projective transformation" makes sense). Moreover, it is an involution.

Now, we will prove the fact above for an involution of the non-singular conic $\mathcal{C}$, that is not the identity).

So, consider $\phi\colon \mathcal{C} \to \mathcal{C}$ an involution. To show that there exists a point $P'$ such that $A \phi(A)$ passes through the point $P'$, for all $A$. In other works, we need to show that the involution $\phi$ is given by "projection through" $P'$.

Now, consider two points on $\mathcal{C}$, $A$, $B$, not in involution $\phi$. The lines $A \phi(A)$, and $B \phi(B)$ intersect in a point $P'$. Now the two involutions, $\phi$, and the projection through $P'$, coincide at $A$, and at $B$. So they must be the same map.

( Note: $\mathcal{C}$ is iso to $\mathbb{P}^1$. The involutions of $\mathbb{P}^1$ are of the form $t \leftrightarrow t'$, with $a t t' + b(t+t') + c = 0$, so are determined by the images at two points)

Obs: the Frégier map determined by $P$, being an involution of $\mathcal{C}$, will have fixed points, but those are not real.