A road follows the parabola y² = 4x on a map, with lengths in kilometres…

Step 1. The normal. Here $a = 1$ and $t = 2$, so $P = (4, 4)$ and the normal is $y + 2x = 4 + 8 = 12$. At $y = 0$, $x = 6$.

Step 2. The second point. The normal meets the parabola where $y = 12 - 2x$ and $y^2 = 4x$. Putting $x = \dfrac{t'^2}{1}$ and $y = 2t'$ gives $2t' + 2t'^2 = 12$, so $t'^2 + t' - 6 = 0$ and $t' = 2$ or $-3$. The second root is $t' = -3$. This agrees with $-t - \dfrac{2}{t} = -2 - 1 = -3$. So $Q = (9, -6)$.

Step 3. The distance. $PQ = \sqrt{(9 - 4)^2 + (-6 - 4)^2} = \sqrt{125} \approx 11.18$ kilometres.