Show that the parameter p of the parabola (distance |FF'| of focus to directrix) is equal to (|BA|*sin(w))

[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

[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

[11] To find the

[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.

AllParabolasCircumscribed.html

Artzt.html

BrocardSecond.html

MedialParabola.html

Miquel_Point.html

Parabola.html

ParabolaChords.html

ParabolaProperty.html

Thales_General.html

ThalesParabola.html

TrianglesCircumscribingParabolas.html

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