Answer to Question #91739 in Analytic Geometry for Ra

Question

How to 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

1. Let F be the focus of the parabola x²=2py, DH be its directrix, FO=OD=FD/2=p/2. 

From the definition of parabola it follows that PF=PH, where P is any point on the parabola and H its projection on the directrix.

 

2. The slope of the tangent at the point P is determined by the formula:

k=tanφ=y'(x)=(x^2/2p)'=x/p.

Because tan∠DFH = DH/DF = x/p = tanφ, then ∠DFH=φ.

3. Consider triangles FDH and NMH. The angle H is common to them, ∠DFH=∠MNH as shown above, so the angle NMH is right:

∠NMH=∠FDH=90°.

4. In the isosceles triangle FPH the height PM is the median: FM=MH, so the point M lies on the midline OM of the triangle FDH, that is, on tangent OM of the parabola at its vertex. It is obvious that the set of points M represent the tangent at its vertex, what we wanted to prove.