The parabola can be characterized as the locus (c) of points P which are at equal distances, PF = PA, from a fixed point F (focus) and a fixed line a (directrix).
It follows that given the focus F and the directrix a of a parabola, one can construct arbitrary many points by the following recipe: - Start with a point A on a, - Draw the orthogonal AP to a at A, - Join A with F and draw the middle-orthogonal to AF, line BF, - The intersection point P of the lines BP, AP is on the parabola.
The following properties are immediate consequences or each follows from the preceding ones:
[1] The tangent (t) to (c) at P is the middle-orthogonal line (medial line) of AF.
[2] The locus of centers of circles tangent to a fixed line (a) and passing through a fixed point F is the parabola with directrix a and focus F.
[3] Given a fixed line (a) and a fixed point F, the envelope of all medial lines AF, for points A on a, is the parabola with focus F and directrix (a).
[4] Line CF is orthogonal to PF and tangent to the circle with radius PF = PA.
[5] As A moves on line (a), the middles B of AF describe a line (b) parallel to (a) at p/2, where p (called parameter of the parabola) is the distance of the focus from the directrix (a) . The line (b) is tangent at the vertex O of the parabola. The vertex is the nearest to the directrix point of the parabola.
[6] Draw from F a parallel (d) to PC and from C a parallel (t') to FA. t' is again a tangent to the parabola.
[7] The tangents to the parabola CP, CP' from a point C on the directrix form a right angle at C.
[8] The line through the tangent points P, P' passes through the focus F.
[9] C is the middle of AD. Quadrangles APFC and DP'FC are similar. CF2 = AP*DP'.
[10] CF is the symmedian of the triangle CPP'. BB'P'P is a circular quadrangle.
[11] The circle with diameter PP' touches the directrix (a) at C.
[12] AR = DP' and APQD is a rectangle.
[13] Let G be the intersection point of the directrix (a) with PP'. Let I be the intersection point of CF with the parabola c. Triangles GIH and GIF are equal. GI is the bisector of the angle AGP. The tangent to c at I passes through G.
[14] All quadrangles APFC are bicircular i.e. they have inscribed and circumscribed circles. The center of the inscribed circle is the intersection point of PC and GI.
[15] Let J, E, K be correspondingly the middles of the sides of triangle CPP'. The middle M of CE is on the parabola, since it is on KJ, which is orthogonal to the middle of CF (CM=MF).
The previous remarks suggest the following way to construct a parabola: Consider a fixed line (a) and a fixed point F. For every point G on (a) let GI be the bisector of the angle between the line (a) and GF. This line coincides with the medial line of FH. The envelope of the lines GI (or the medial lines of FH) is the parabola with directrix (a) and focus F.
Selecting the appropriate coordinate axes we get the well known equation of the parabola. In fact, set the origin at the vertex O and choose for x-axis the tangent (b) to the parabola and y-axis the axis of the parabola. Then we have the relations: AP2 = PF2 <=> (y+p/2)2 = x2 + (y-p/2)2 =>
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1. Parabola ![[0_0]](Parabola_files/Parabola_0_0_0.png)
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