Problem
Given a parabola \ (C: ~ y ^ 2 = 2x \) focus as \ (F. \) , The quasi-line \ (L \) , the point \ (P \) is a parabola \ (C \) on a fixed point ( different from vertices), through \ (P \) do alignment \ (L \) perpendicular, to a foot \ (H \) , and \ (\ triangle PFH \) of gravity of \ (M \) , confirmation \ (MP \) and the parabola \ (C \) tangent.
Solution
Set \ (P (x_0, y_0) \) , there \ (H (- \ FRAC. 1} {2} {, y_0), ~ F. (\ FRAC. 1} {} {2, 0) \) .
Set \ (Q \) of \ (the PF \) midpoint, the \ (Q (\ frac {x_0 + \ frac {1} {2}} {2}, {\ frac {y_0} {2}}) \) .
\ (k_ {PF} = \ frac {y_0} {x_0- \ frac {1} {2}} \)
For convenience of description, with the following \ (BOT \ \) subscripts a vertical bisector of the side edges.
\(k_{PF_\bot}=\frac{-x_0+\frac{1}{2}}{y_0}\)
\(l_{PF_\bot}:~y=\frac{-x_0+\frac{1}{2}}{y_0}(x-\frac{x_0+\frac{1}{2}}{2})+\frac{y_0}{2}=\frac{-x_0+\frac{1}{2}}{y_0}(x-\frac{x_0}{2}-\frac{1}{4})+\frac{y_0}{2}\)
\(l_{PH_\bot}:~x=\frac{x_0-\frac{1}{2}}{2}=\frac{x_0}{2}-\frac{1}{4}\)
\(M=l_{PF_\bot}\cap l_{PH\bot}\rightarrow y=\frac{-x_0+\frac{1}{2}}{y_0}(\frac{x_0}{2}-\frac{1}{4}-\frac{x_0}{2}-\frac{1}{4})+\frac{y_0}{2}=\frac{x_0-\frac{1}{2}}{2y_0}+\frac{y_0}{2}=\frac{y_0^2+x_0-\frac{1}{2}}{2y_0}=\frac{3x_0-\frac{1}{2}}{2y_0}\)
\(k_{PM}=\frac{y_0-\frac{3x_0-\frac{1}{2}}{2y_0}}{x_0-\frac{2x_0-1}{4}}=\frac{\frac{2y_0^2-3x_0}{2y_0}+\frac{1}{2}}{\frac{4x_0-2x_0+1}{4}}=\frac{\frac{x_0+\frac{1}{2}}{2y_0}}{\frac{2x_0+1}{4}}=\frac{\frac{2x_0+1}{y_0}}{2x_0+1}=\frac{1}{y_0}\)
Implicit derivative to give \ (2Y \ FRAC {\ text {D} Y} {\ text {D} X} = 2 \) , i.e., \ (\ frac {\ text { d} y} {\ text {d } x} = \ frac {1 } {y} \)