Consider an angle s = ABC and the lines defined through its sides. Take on these lines points X on BA and Y on BC such that BX = |BA|x, BY = a*|BC|(x²) . Construct the parallelogram p = XBYP and show that point P moves, for varying x, on a parabola c. Determine the focus (F) and directrix (d) of c.
Show that the parameter p of the parabola (distance |FF'| of focus to directrix) is equal to (|BA|*sin(w))2/(2*a*|BC|).
[1] Given the angle ABC, BA becomes tangent to the parabola and BC parallel to its axis.
[2] Thus the directrix is orthogonal to side BC.
[3] Consider a point X, the corresponding Y and P, defined through the parallelogram XBYP.
[4] Q is the symmetric w.r to Y and is also on the parabola.
[5] Besides, Z which is the symmetric of Y w.r. to B is the intersection point of the tangents to the parabola at P and Q. Thus ZP, ZQ are tangents to the parabola.
[6] The focus F is found by intersecting the symmetrics FP, FQ w.r to ZP and ZQ, of the parallels to BY from P and Q respectively.
[7] By taking QS = QF on the parallel to BY from Q we determine a point S on the directrix.
[8] Notice also that ZF is the symmedian from Z of the triangle ZPQ and also bisects the angle QFP, which is the double in measure of the angle QZP.
[9] Also triangles QFZ and ZFP are similar. Later implies that (ZF)² = (QF)*(PF).
[10] These properties of point F identify it with a vertex of the second Brocard triangle, and the parabola itself with an Artzt parabola of triangle ZPQ. These facts, concerning the triangle QZP, are also discussed in ParabolaChords.html .
[11] To find the parameter p of the parabola, find first point X' such that |BX'| = 2*|BY'|, show that then FY' is parallel to BX' and calculate |FF'| in terms of the data given.
[12] Consider the homothety with center at Z and ratio 1/2 and apply it to the circumcircle of ZPQ to get the circumcircle of ZIJ and the proof of the property: The circumcircles of triangles circumscribed to a parabola pass through its focus.
The figure shows also another property of the parabola.