isogonal trajectory

Let a one-parametric family of plane curves  $\gamma$  have the differential equation

$F(x,y,{\displaystyle \frac{dy}{dx}})=\mathrm{\hspace{0.33em}0}.$(1)

We want to determine the isogonal trajectories of this family, i.e. the curves  $\iota$  intersecting all members of the family under a given angle, which is denoted by $\omega$.For this purpose, we denote the slope angle of any curve $\gamma$ at such an intersection point by $\alpha$ and the slope angle of $\iota$ at the same point by $\beta$.  Then

$$\beta -\alpha =\omega \mathit{\hspace{1em}}(\text{or alternatively }-\omega ),$$

and accordingly

$$\frac{dy}{dx}=\tan \alpha =\frac{\tan \beta -\tan \omega }{1+\tan \beta \tan \omega }=\frac{y^{\prime }-\tan \omega }{1+y^{\prime }\tan \omega },$$

where $y^{\prime }$ means the slope of $\iota$.  Thus the equation

$F(x,y,{\displaystyle \frac{y^{\prime }-\tan \omega }{1+y^{\prime }\tan \omega }})=\mathrm{\hspace{0.33em}0}$(2)

is satisfied by the derivative $y^{\prime }$ of the ordinate of $\iota$.  In other , (2) is the differential equation of all isogonal trajectories of the given family of curves.

Note.  In the special case  $\omega =\frac{\pi }{2}$,  it’s a question of orthogonal trajectories.