An interesting family of intersecting self-orthogonal trajectories The ODE $$xy'^2-yy'-x=0$$
See:
Solve $xy'^{~2}-yy'-x=0$ for a new family of intersecting self-orthogonal trajectories
and the ODE $$xy'^2-2yy'-x=0$$
See: prove $x^2 = 4 c(y+c)$ is self orthogonal trajectory
are invariant under $y'\rightarrow -\frac{1}{y'}$ and give rise to two families of intersecting self-orthogonal trajectories.
Similarly, the simpler ODE $$y'^2-2xy'-1=0$$ can also give rise to such a family of trajectories but with an interesting difference with regard to the two examples referred here. The question is to find this family of curve(s) and their disparate feature.
 A: As noted by @MOO
$$y'^2-2xy'-1=0 \implies y'= x\pm \sqrt{1+x^2} \implies \int dy= \int(x \pm \sqrt{x^2+1}) dx$$ $$ \implies y= \frac{x^2}{2}\pm \left(\frac{x\sqrt{1+x^2}}{2}+\sinh^{-1} x \right)+C~~~(1)$$
So for a fixed value of $C$, these are two curves that cut each other orthogonally. Also for all real values of $C$, these curves represent a network of self-orthogonal curves.
This situation is different from that of earlier two examples mention in the OP where a single curve represents self-orthogonal trajectories for $C>0$ and $C<0$, for instance, the confocal parabolas: $y^2=4C(x+C).$
See the orhohonal trajecories for $C=1,2,3,4$ (blue) and $C=-1,-2,-3,-4$ (red) in the  fig. below
 below.
Similarly, teachers may also motivate students by the family
of intersecting self-orthogonal trajectories arising from $y'^2=1$, these are $y=\pm x+C$. It may be mentioned that $\frac{x^2}{a^2+C}+\frac{y^2}{b^2+C}=1$ for fixed values of $a$ and $b$ is a family of nonintersecting-self-orthogonal curves.
