Answer to Question #202793 in Analytic Geometry for tanya

Question

Prove that the path traced by the foot of the perpendicular from the focus of a parabola on any tangent to the parabola is the tangent at its vertex.

Answer 1

$\space 1. \space Let \space F \space be \space the \space focus \space of \space the \space \\ \space parabola \space x^2=2py, \space DH \space be \space its \space directrix,\\ \space FO=OD=\frac{FD}{2} \space \space =\frac{p}{2}. \space \\ From \space the \space definition \space of \space parabola \space it \space follows \space that \space PF=PH, \space \\ \space \space where \space P \space is \space any \space point \space on \space the \space parabola \space and \space \\ \space H \space its \space projection \space on \space the \space directrix. \space \\ ----------------------\\ 2. \space The \space slope \space of \space the \space tangent \space at \space the \space point \space P \space is \space \\ \space determined \space by \space the \space formula: \\ k=tan(φ)=y'(x)=(\frac{x^2}{2p})'=\frac{x}{p}. \\ Because \space \\ \space tan(∠DFH) \space = \space \frac{DH}{DF} \space =\frac{ \space x}{p} \space = \space tan(φ), \space then \space ∠DFH=φ. \\ ----------------------\\ 3. \space Consider \space triangles \space FDH \space and \space NMH.\\ \space The \space angle \space H \space is \space common \space to \space them, \space \\ \space ∠DFH=∠MNH \space as \space shown \space above,\\ \space so \space the \space angle \space NMH \space is \space right: \\ ∠NMH=∠FDH=90°. \\ ----------------------\\ 4. \space In \space the \space isosceles \space triangle \space FPH \space the \space height \space PM \space is \space the \space median: \space FM=MH,\\ \space so \space the \space point \space M \space lies \space on \space the \space midline \space OM \space of \space the \space triangle \space FDH,\\ \space that \space is, \space on \space tangent \space OM \space of \space the \space parabola \space at \space its \space vertex. \space \\ \space It \space is \space obvious \space that \space the \space set \space of \space points \space M \space represent \space the \space tangent \space at \space its \space vertex,\\ \space what \space we \space wanted \space to \space prove. \space \\$