Establish that the set of centers of rectangular circum-hyperbolas to a triangle is a conic

Question

This is a "follow on" of this question which was dealing with a (single) rectangular circum-hyperbolas (RCH in short) to a certain triangle passing through a certain "fourth point".

Being in need of a general understanding of the set of all these RCH (in order in particular to understand the "degrees of freedom" of this issue), I wrote a computer program (reproduced at the bottom of this question) giving images like this one which has conducted me to a new issue about the set of centers of these RCH hyperbolas.

Fig. 1 : The $n$ RCH hyperbolas passing through points $P_k$ (black stars). Their centers are representd by blue stars. One checks that all RCH hyperbolas pass through the orthocenter.

Fig. 2 : Another case (isosceles triangle). Here, the $3$ pairs (altitudes, opposite sides) can be considered as degenerated cases (RCH hyperbolas reduced to their asymptotes).

Before that, let us give some explanations.

We can assume without loss of generality that vertices $A,B,C$ belong to the unit circle, which is henceforth the circumcircle. Fig. 1 displays a certain number $n$ of RCHs to a fixed triangle. For each of them, the "fourth point" has been constrained to be with cartesian coordinates

$$P_k=(\cos(2k \pi/n), \sin(2k \pi/n)), \ \ k=1\cdots n$$ It looks a thorough choice because in this way, the curves attached to generic point $P_{\theta}=(\cos(\theta),\sin(\theta))$ provide a "foliation" of the plane.

Circum-conics to a triangle have (with respect to this triangle) a barycentric equation of the form (see this article):

$$uyz+vzx+wxy=0 \ \ \begin{cases}\text{ellipse if u,v,w have the same sign}\\\text{hyperbola if u,v,w have not all the same sign}\\\text{parabola if } u+v+w=0\end{cases}\tag{1}$$

Restriction to RCH hyperbolas brings the supplementary condition ;

$$u \cos \hat{A}+v \cos \hat{B}+u \cos \hat{C}=0\tag{2}$$

giving rise (see p. 121 of Paul Yiu's notes here $\color{red}{\text{Take care : not secure site}}$) to the general equation (3), with the following notation : let $P$ with barycentric coordinates $(u:v:w)$ (think to $P$ as being the "fourth point" of the initial quesstion), the (unique) barycentric equation of the RCH to triangle $ABC$ passing through $P$ is :

$$u(S_Bv-S_Cw)yz+v(S_Cw-S_Au)zx+w(S_Au-S_Bv)xy=0\tag{3}$$

with Conway's conventions $S_A=\tfrac12(b^2+c^2-a^2), $ etc.

Experimental observation is that the hyperbolas' centers lie on a specific conic, more precisely an ellipse passing through the feet of altitudes. This ellipse attached to the triangle is almost surely a known one, but till now, I haven't been able to find which one...

So my main question is :

• How can we prove that the locus of centers of all > RCHs is a conic, more precisely an ellipse ? Besides, is it a known one ?

Another question :

• Is it possible to find a differential equation whose integral curves are these RCH hyperbolas ?

Remark : among the RCHs, there are two particular ones, the Kiepert hyperbola and the Jerabek hyperbola with resp. centers $X_{115}$ and $X_{125}.$

Matlab program for Fig. 1 :

 clear all;close all;
 L=1.6;n=40;
 plot(0,0,'sk');hold on;axis([-L L -L L]);axis equal; % origin=circumcenter
 aa=(4/9)*pi;bb=(9/8)*pi;cc=(15/8)*pi;
 A=exp(i*aa);B=exp(i*bb);C=exp(i*cc);T=[A,B,C]; % vertices
 a=abs(B-C);b=abs(C-A);c=abs(A-B); % sidelengths
 %aC=(bb-aa)/2;aA=(cc-bb)/2;aB=(aa-cc+2*pi)/2; % angles
 SA=(b^2+c^2-a^2)/2;SB=(c^2+a^2-b^2)/2;SC=(a^2+b^2-c^2)/2; % Conway
 M=[real(T);imag(T);[1,1,1]]; % direct matrix cart. -> baryc
 plot(M(1,[1,2,3,1]),M(2,[1,2,3,1]),'-or','LineWidth',3);hold on;axis equal
 N=inv(M); % inverse matrix cart (x,y) -> baryc. (aBC, bBC, cBC)
 aBC=@(x,y)(N(1,1)*x+N(1,2)*y+N(1,3));
 bBC=@(x,y)(N(2,1)*x+N(2,2)*y+N(2,3));
 cBC=@(x,y)(N(3,1)*x+N(3,2)*y+N(3,3));
 gr=@(u,v,w)((u*A+v*B+w*C)/(u+v+w)); % bary (u,v,w) -> cartesian
 for k=1:n;
    ang=(k/n)*2*pi;co=cos(ang);si=sin(ang);
    u=aBC(co,si);v=bBC(co,si);w=cBC(co,si);
    plot(gr(u,v,w),'pk');hold on; 
    U=u*(SB*v-SC*w);V=v*(SC*w-SA*u);W=w*(SA*u-SB*v); %  formula (3)
    fimplicit(@(x,y)(U.*bBC(x,y).*cBC(x,y)+...
       V*cBC(x,y).*aBC(x,y)+W*aBC(x,y).*bBC(x,y)),...
       [-2 2],'MeshDensity', 100); hold on;
    Uc=U*(-a*U+b*V+c*W);Vc=V*(a*U-b*V+c*W);Wc=W*(a*U+b*V-c*W); % center
    plot(gr(Uc,Vc,Wc),'pb');hold on;
 end;