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# isogonal trajectory

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

$\displaystyle F(x,\,y,\,\frac{dy}{dx})\;=\;0.$ | (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

$\beta-\alpha\;=\;\omega\quad(\mbox{or alternatively\;\;}-\omega),$ |

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

$\displaystyle F(x,\,y,\,\frac{y^{{\prime}}-\tan\omega}{1+y^{{\prime}}\tan% \omega})\;=\;0$ | (2) |

is satisfied by the derivative $y^{{\prime}}$ of the ordinate of $\iota$. In other words, (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.

## Mathematics Subject Classification

51N20*no label found*34A26

*no label found*34A09

*no label found*

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