This is a supplement to the discussion in PowerGeneral.html . It deals with two secants of the parabola one of which is parallel to the parabola's axis, hence intersects the parabola at one point only. For a secant BLK of the parabola parallel to a fixed direction e from an arbitrary point B the following relation holds and is independent of the position of B, depending only on the direction of e: (BK*BL)/BA' = k (constant). Here A' is the unique intersection point of the parabola with the line parallel to the axis of the parabola and passing through B (the diameter of the parabola through B).
To prove it represent the parabola in coordinates and in the form: y = ax2, with a constant a. Then apply first the well known relation (see PowerGeneral.html ) for the two secants KBL, C'BD' by which the ratio r1= (BK*BL)/(BC'*BD') is independent of the position of B, depending only on the direction e of line KBL. Then calculate the ratio r2= (BC'*BD')/BA' = (x1-x)(x2-x)/(y-ax2), which by the symmetry x2 = -x1 becomes: -(x12-x2)/(y-ax2)=-(ax12-ax2) = -(1/a), therefore proving the claim using the product r1*r2.
Note that, by the aforementioned reference, the ratio r1 = (a*cos2(fi)-sin(fi))/a, where e=(cos(fi), sin(fi)) determines the direction of the secant. Hence the ratio in total is r = (sin(fi)-a*cos2(fi))/(a2).
A similar situation occuring in hyperbolas is discussed in PowerGeneralHyperbola.html . [SalmonConics, p. 151]
Given a trapezium ABCD and a point P on its side AD investigate the possibility to construct a parabola passing through {B,C}, having axis parallel to the parallel sides and having AD as its tangent at P. Locate the position P=L on AD for which such a construction is possible.
Assume L found and project {B,C} on the parallel e to AB from L. Then take the symmetrics {B',C'} of {B,C} with respect to {N,M} respectively. The parabola has to pass through points {B,C',L,C,B} hence can be constructed as a conic passing through five points. By the previous section the ratios s1 = MC2/LM = LD2/DC and s2 = BN2/LN = AL2/AB must be equal, thus implying that: LD2/LA2 = DC/AB. This is then a necessary condition and is satisfied by exactly one point L on AD. The condition is also sufficient, since setting the value of the common ratio s1=s2 = s we have: LD2 = s*DC, LA2 = s*AB. We find easily that the five points {B,C',L,C,B'} are on the parabola, which in skew axes coinciding with LD (x-axis) and LM (y-axis) is represented by ( see ParabolaSkew.html ): x2 = s*y.
Repeating the construction of points {B',C'} of the previous section, but this time for arbitrary positions of P on line AD we obtain conics passing through {P,B,C,B',C'}, tangent to AD at P and having the line eP which is parallel to the parallel sides and passes through P as a diameter of the conic.
By applying the results obtained in PowerGeneral.html we locate the other intersection point Q of the conic with line eP:
The case of the parabola is an exceptional one for P=L and for which point Q is at infinity. Another exceptional case occurs for P=E', where E' is the harmonic conjugate of E with respect to {A,D}. In this case Q is coincident with P. For all other cases we obtain conics, which depending on the position of P, relative to L are ellipses (as shown) or hyperbolas. The two cases of conics are separated by the location of L. For points on the half-line LE, containing the harmonic E' of E we obtain hyperbolas. For points on the half-line LA, not containing E', we obtain ellipses.