Start with an easy exercise:
Consider an isosceles right-angled triangle OXY. Project points P of the hypotenuse at Px, Py on its orthogonal sides and to point F' on the parallel to XY from O. The intersection point Z of PxPy and FF', F being the middle of XY, has the property ZF = ZF'.
[1] The circle (c) circumscribing the rectangle PPyOPx passes through F' and F (both view OP under a right angle).
[2] OFPF' is a rectangle, FF' is a diameter of (c), and is orthogonal to PxPy since angle(PyOF)=pi/4.
This property shows that Z lies on a parabola with focus F and directrix OF'. In addition shows that PxPy is tangent to this parabola at Z. The parabola passes through points X, Y and this identifies it with the Artzt parabola of the triangle OXY relative to side XY. This parabola is characterized by its property to be tangent to sides OX, OY at the points X, Y respectively.
It is interesting to transfer the previous construction to an arbitrary triangle via an affinity mapping OXY to that triangle.
The figure above shows the image parabola under the affinity P'=A(P) transforming OXY to the arbitrary triangle O'X'Y'. The map respects parallelity but no metrical relations and orthogonality. Px, Py map to P'x, P'y, so that P'P'x is parallel to O'X' and P'P'y parallel to O'Y'. P'xP'y is the diagonal of parallelogram P'P'xO'P'y and is again tangent to the parabola.
Thus, the Artzt parabola, in the general case, is the envelope of diagonals of parallelograms constructed from points P' by drawing parallels to the sides O'X', O'Y'.
The focus and the directrix are not preserved in general. Thus, F1=A(F) is no more the focus, but only the middle of X'Y'. The axis of the parabola, though, is parallel to the median O'F1. Interesting is also the ellipse c'=A(c), passing through the six points {O',F2,P'y,P',F1,P'x} and having its center on the middle of F1F2.