Prove q = k times focal length

Meaning of focal length: the Casagrande base parabola has its focus F at the beginning of the horizontal filter. Let $y_0$ be the distance from the focus to the directrix (the "focal length" in this context, the parabola's parameter). The distance from focus to the vertex is $y_0/2$.

   y
   |   .  .           directrix
   |        .   .  <--- parabola
   |   y0 --->   .
   +------F-----------------> x   (filter at the toe)

Proof

Take the origin at F, $x$ positive towards the upstream side and $y$ vertical. Any point P(x, y) on the parabola is equidistant from the focus and from the directrix ($x = -y_0$):

$$\sqrt{x^2 + y^2} = x + y_0$$ $$x^2 + y^2 = x^2 + 2xy_0 + y_0^2 \quad\Rightarrow\quad y^2 = 2y_0\,x + y_0^2$$

Differentiate with respect to $x$:

$$2y\frac{dy}{dx} = 2y_0 \quad\Rightarrow\quad y\frac{dy}{dx} = y_0$$

By Dupuit's assumption, the hydraulic gradient on a vertical section at height $y$ is $i = dy/dx$ (slope of the free surface), and the flow area per unit width is $y \times 1$:

$$q = k\,i\,A = k\,\frac{dy}{dx}\,y = k\,y_0$$

Hence the discharge per unit width is equal to the coefficient of permeability times the focal length $y_0$. $\blacksquare$

At the focus ($x = 0$) the ordinate is $y = y_0$ and the slope is 1, so $q = k \times 1 \times y_0$, which agrees.

(If the focal length is taken as the focus-to-vertex distance $a = y_0/2$, then $q = 2ka$.)