Proving Frégier's Theorem for conics 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?
 A: This isn't a linear algebra proof, but you may be able to translate it into one.

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. $\square$
