Find the locus of the feet of the perpendiculars dropped

Find the locus of the feet of the perpendiculars dropped from the focus of the parabola $y^2=2px$ onto its tangents.

Step 1: Setup

The focus is $F=(p/2,0)$. Parametrize the curve by $T=\left(\dfrac{pt^2}2,\,pt\right)$ (so $y_0=pt$ and $y_0^2=p^2t^2=2p\cdot\frac{pt^2}2$). Since $y'=p/y_0=1/t$, the tangent at $T$ is

\[ x - ty + \frac{pt^2}{2} = 0. \tag{1} \]

Step 2: The perpendicular from $F$

With slope $-t$ (perpendicular to slope $1/t$), through $F$: \[y = -t\!\left(x-\frac p2\right). \tag{2}\]

Step 3: The foot $M$

Substituting (2) into (1): \[x + t^2\!\left(x-\frac p2\right) + \frac{pt^2}2 = 0 \;\Longrightarrow\; x(1+t^2) = 0 \;\Longrightarrow\; x_M = 0, \tag{3}\] since $1+t^2>0$; then $y_M=-t(0-p/2)=pt/2$, which sweeps every real value as $t$ does.

Step 4: Verification

\[p=4,\ t=1.3\text{: tangent-perpendicular intersection computed directly }=(0,\,2.60000)=(0,\,pt/2)\ \checkmark\]