A projective geometry problem about a set of lines envelope a curve of class 2 Let $P \in RP^2$ fixed. Let C be a conic. Consider all lines $L_{AB}$ such that there exist $A,B \in C$ such that $\measuredangle APB = \frac{\pi}{2}$.
Prove that the set of lines $L_{AB}$ envelope a curve of class 2.
I think Laguerre's Formula can be used to prove this, and I need help for some strategy.
 A: Here is a sketch of resolution that uses a projective method at the end.


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*1) Take point $M$ as the origin of coordinates.

*2) Describe your conical curve with polar coordinates $p=p(\theta)$. This is of course especialy simple when $M$ is a focus of the conic (in this case $p=\dfrac{p_0}{1+e \cos(\theta-\theta_0)}$).

*3) Write equation of $L_{AB}$ under the following determinant form (which, btw, is an alignment constraint of points $M,A,B$, interpreted in projective space !):
$$\begin{vmatrix}x&p(\theta)\cos(\theta)&p(\theta+\pi/2)\cos(\theta+\pi/2)\\
y&p(\theta)\sin(\theta)&p(\theta+\pi/2)\sin(\theta+\pi/2)\\ 1&1&1\end{vmatrix}=0 \ \ \iff$$
$$\begin{vmatrix}x&p(\theta)\cos(\theta)&-q(\theta)\sin(\theta)\\
y&p(\theta)\sin(\theta)&q(\theta)\cos(\theta)\\ 1&1&1\end{vmatrix}=0 \ \ \  \text{with} \ q(\theta):=p(\theta+\pi/2)\ \ \iff$$
$$\tag{1}x\underbrace{(p(\theta)\sin(\theta)-q(\theta)\cos(\theta))}_{r(\theta)}-y\underbrace{(p(\theta)\cos(\theta)+q(\theta)\sin(\theta))}_{s(\theta)}+\underbrace{p(\theta)q(\theta)}_{t(\theta)}=0$$


*

*4) Form the equation obtained by differentiating (1) with respect to $\theta$:


$$\tag{2}x r'(\theta)-y s'(\theta)+t'(\theta)=0$$


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*5) Solve the system obtained by grouping (1) and (2), in order to obtain a parametric representation of $x$ an $y$ with respect to parameter $\theta$ (this is the classical method for obtaining an envelope, see (Loci of intersection of lines with positive intercepts??) or (https://math.stackexchange.com/q/1553478)).


I haven't enough time to treat the general case. 
Nevertheless, I have validated the method in a particular case, with the following parabola with focus (point $M$) at the origin, represented by polar curve:
$$p(\theta):=\dfrac{1}{1+ \cos(\theta)}$$
(see graphics below, generated by a Matlab program: parabola in red, lines $L_{AB}$ in blue,  envelope curve in green).

One easily finds that the equation of $L_{AB}$ is:
$$x(\sin(\theta)-\cos(\theta)-1)+y(-\sin(\theta)-\cos(\theta))+1=0$$
The equation obtained by differentiation with respect to $\theta$ is:
$$x(\cos(\theta)+\sin(\theta))+y(-\cos(\theta)+\sin(\theta))=0$$
Solving the system formed by these two equations gives the parametric equations of the envelope:
$$\tag{3}X=\dfrac{\cos(\theta)-\sin(\theta)}{2+\cos(\theta)-\sin(\theta)}, \ \ \ Y=\dfrac{\cos(\theta)+\sin(\theta)}{2+\cos(\theta)-\sin(\theta)}$$
Important remark: (3) can be expressed as the image of unit circle $(x=\cos(\theta),y=\sin(\theta))$ by projective transformation 
$$\tag{3}X=\dfrac{x-y}{x-y+2}, \ \ \ Y=\dfrac{x+y}{x-y+2}$$
thus is a conic. We recognize here a full ellipse.
Finding the locus as the image of unit circle by a projective transformation should be the case in a  more general context (to be checked).
(the ellipse here has semi-major axis $\sqrt{2}$ and semi-minor axis $1$.)
