Consider the parabolas $y=x^{2}$ and $y=-(x-1)^{2}$. Let's find a pair of parallel tangents, one to each parabola. Give the distance $d$ between the two tangents as a function of the tangent's slope $m$.
First, we determine the equations of the tangents with slope $m$ to the two parabolas. For the parabola $y=x^{2}$, the slope of the tangent at the point with abscissa $x_{0}$ is $2 x_{0}$, so we must choose $x_{0}-\mathrm{t} \frac{m}{2}$, and the equation of the tangent is

$$
y-\frac{m^{2}}{4}=m\left(x-\frac{m}{2}\right)
$$

For the second parabola, the derivative at $x_{0}$ is $-2\left(x_{0}-1\right)$, so now $x_{0}=-\frac{m}{2}+1$, and the equation of the line is

$$
y+\left(\frac{m}{2}\right)^{2}=m\left(x+\frac{m}{2}-1\right)
$$

The length $h$ of the segment on the $y$-axis between the two lines is equal to the absolute value of the difference of the $y$-values obtained by substituting $x=0$ into the equations:

$$
h=\left|\frac{m(m-2)}{2}\right|
$$

From this, we get the distance between the two lines by multiplying by the cosine of the common angle between the lines:

$$
d=h|\cos \alpha|=\frac{h}{\sqrt{1+\operatorname{tg}^{2} \alpha}}=\frac{|m(m-2)|}{2 \sqrt{1+m^{2}}}
$$