If a family of plane curves (with one free parameter) satisfies the differential equation $$F(x,\,y,\,y') \;=\; 0,$$ where $y' = \frac{dy}{dx}$ , then the family of curves intersecting orthogonally all the first curves satisfies the differential equation $$F\left(x,\,y,\,-\frac{1}{y'}\right) \;=\; 0.$$ Anyone of the latter curves is an orthogonal curve of the former ones.

Example. Let's consider the family of rectangular hyperbolas $$x^2-y^2 \;=\; c$$ with the parameter $c$ taking any real value. Derivating with respect to $x$ gives the differential equation of this family, $$x-yy' \;=\; 0,$$ and by replacing here $y'$ with $-\frac{1}{y'}$ we obtain the differential equation $$x+\frac{y}{y'} \;=\; 0$$ of the orthogonal curves. Integrating its form $$\frac{dy}{y} \;=\; -\frac{dx}{x}$$ gives the solution $$xy \;=\; C,$$ which represents another family of rectangular hyperbolas.

In the picture below (by drini), there are four hyperbolas of the first family (blue) given by the values $c = -1,\,-2,\,-4,\,-8$ and four hyperbolas of the orthogonal family (red) given by the values $C = 1,\,2,\,4,\,8$ .