Isogonal trajectory

A plane curve intersecting the curves of a given one-parameter family in the plane at one and the same angle. If

$$F(x,y,y')=0\label{1}\tag{1}$$

is the differential equation of the given family of curves, then an isogonal trajectory of this family intersecting it at an angle $\alpha$, where $0<\alpha<\pi$, $\alpha\neq\pi/2$, satisfies one of the following two equations:

$$F\left(x,z,\frac{z'-\tan\alpha}{1+z'\tan\alpha}\right)=0,\quad F\left(x,z\frac{z'+\tan\alpha}{1-z'\tan\alpha}\right)=0.$$

In particular, the equation

$$F\left(x,z,-\frac{1}{z'}\right)=0\label{2}\tag{2}$$

is satisfied by an orthogonal trajectory, that is, a plane curve that forms a right angle at each of its points with any curve of the family \eqref{1} passing through it. The orthogonal trajectories for the given system \eqref{1} form a one-parameter family of plane curves — the general integral of equation \eqref{1}. For example, if the family of lines of force of a plane electrostatic field is considered, then the family of orthogonal trajectories are the equipotential lines.

References

[1]	W.W. [V.V. Stepanov] Stepanow, "Lehrbuch der Differentialgleichungen" , Deutsch. Verlag Wissenschaft. (1956) (Translated from Russian)

References

[a1]	H.S.M. Coxeter, "Introduction to geometry" , Wiley (1961)
[a2]	D. Hilbert, S.E. Cohn-Vossen, "Geometry and the imagination" , Chelsea (1952) (Translated from German)