## 1. Power generalization parabola

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]

## 2. Parabola construction from trapezium

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.

## 3. The general case

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.