CircleBundleTransformationParabola3

Transforming lines to parabolas-III

Given two points {A(0,r),B(0,-r)}, taken on the y-axis symmetrically to the origin and a direction e(cos(u),sin(u)) define a transformation as follows:
- For each point X not lying on the x-axis consider the circle member c_X(x,y) of the circle bundle of all circles orthogonal to the circle (x²+y²)-r²=0 and the line x=0 (i.e. the bundle of non-intersecting type with limit-points {A(0,r),B(0,-r)}).
- Then construct Y to be the other intersection point of c_X with the line {X+te} through X and parallel to e.
The transformation Y=F(X) is well defined for every point of the plane except the x-axis. It is involutive (F² = 1) and has also the properties:

[1] F maps horizontal lines of the plane to parabolas.
[2] Let the line (v) be described through a vector equation X=a+tb, with (a=(0,a₂)) on the y-axis and (b) the unit vector (1,0). Then the parabola h=F(v) has axis parallel to (e).
[3] Let (c) be the circle of the bundle tangent at C to line v. Then c is tangent to hyperbola at its point K, which is the reflected of C with respect to the diameter of c parallel orthogonal to (e).
[4] For each fixed direction (e) every circle of the bundle intersecting line (v) at points {X,X'} has corresponding images {Y=F(X),Y'=F(X')} so that chords YY' of the parabola are parallel. The direction of YY' and the horizontal x-axis are equal inclined to the axis of the parabola.
[5] Line KC is the line of middles and conjugate direction of that of parallel chords YY'.
[6] The focus of the parabola can be determined geometrically from a tangential triangle whose one side is the tangent at K.

All the statements follow through an analogous reasoning to the one to be found in CircleBundleTransformationHyperbola2.html . Let us here analyse in some detail only [6].
From trapezium XYX'Y' follows that chord YY' is equal inclined to the direction of the directrix with the x-axis. Thus, the horizontal tangent to the hyperbola and the tangent at K intersect at M on the axis of the hyperbola and triangle MKL is isosceles, L being on the line KD orthogonal to the axis of the hyperbola. Let c' be the circle-bundle member passing through L, then drawing the parallel to e from L find R on (v). Quadrangle LDCR is cyclic implying that the intersection point O' of LR and DC satisfies O'D*O'C=O'L*O'R. Thus O' is on the radical axis of circles {c,c'} which implies that it coincides with O.
Thus, having K, L is easily constructed and from this triangle MNP is also easy to construct and find its circumcircle. The focus of the parabola is on the circumcircle of MNP and the medial line of KL. The vertex is found by the middle of the parallel to side KL of MKL from the middle of MK.
See also CircleBundleTransformationParabola.html and CircleBundleTransformationParabola2.html for the other cases of circle bundles (intersectin type and tangential respectively).

Transforming lines to parabolas-III

See Also