Keplerian orbit, parabolic
The center of gravity is at 'C', the focus.
line 'a' is the directrix.
'c' is the perpendicular bisector of segment CD, where 'D' is any point on 'a'.
Line 'd' intersects 'a' perpendicularly at 'D'.
'E' is the intersection of 'c' and 'd'.
The locus of E(locus1) is a parabola, which is the trajectory of an object that falls freely from infinity, passes the center of gravity once and departs for good.
Circle 'g' is drawn with its center placed parallel to 'a' from the focus, C, and its perimeter passing through the focus.
The line segment CG, perpendicular to CD and cut by the circle 'g', is proportional to the speed of the orbiting object at each point, and is parallel to the tangent 'c' to the orbit at 'E', as the velocity vector should be. The red arrow is a parallel translation of CG with its tail at E.
Drag the point 'D' up and down to see how the point E and the velocity vector varies along 'locus1'.

Proofs are left to you(for your fun),
-- that the locus of the velocity vector is a circle(see 'Feynman's Lost Lecture' for reference),
-- that CG is proportional to the orbital speed (:use Kepler's 2nd Law and the properties of a circle),
-- and that the line c is tangent to the locus of E.