Show that the subtangent of the parabola y²=2px always

Show that the subtangent of the parabola $y^2=2px$ always equals twice the abscissa of the point of tangency.

Step 1: The subtangent formula

For $y=f(x)$, the subtangent at $(x_0,y_0)$ is the length $S_T=\dfrac{y_0}{y'(x_0)}$ — the projection onto the $x$-axis of the segment from the point of tangency to where the tangent crosses the axis.

Step 2: Differentiate

For $y_0\neq0$, solving $y=\sqrt{2px}$ and differentiating, \[y' = \frac{p}{\sqrt{2px}} = \frac{p}{y}. \tag{1}\]

At $(x_0,y_0)$, since $y_0^2=2px_0$, this reads $y'(x_0)=p/y_0$.

Step 3: Compute the subtangent

\[ S_T = \frac{y_0}{y'(x_0)} = \frac{y_0}{p/y_0} = \frac{y_0^2}{p}. \tag{2} \]

Since $(x_0,y_0)$ lies on the curve, $y_0^2=2px_0$; substituting into (2), \[S_T = \frac{2px_0}{p} = 2x_0. \tag{3}\]

Step 4: Verification

\[p=3,\ x_0=4\text{: } y_0=\sqrt{24},\ y'(x_0)=p/y_0\approx0.61237,\ S_T=y_0/y'(x_0)=8.00000=2x_0\ \checkmark\]