form of the quadratic equation

f(x,y) = ax

This follows by setting y=0 => ax

a=0.

the coefficients:

1) the quadratic terms remain the same a'=a, h'=h, b'=b.

2) the linear terms transform linearly: { g' = g + ax

3) the new constant term is: c' = f(x

Let e be a line parallel to an asymptote of the hyperbola c and intersecting it at the (unique) point A.

For a secant BLK of the hyperbola parallel to a fixed direction e' from an arbitrary point B of e the following

relation holds and is independent of the position of B (on e), depending only on the direction of e':

(BK*BL)/BA = k (constant).

The proof follows easily by taking the origin of coordinates at B, the axis of x equal to e and the y-axis equal to e'.

Then the equation has coefficient a = 0 (Remark-1), and BA = x satisfies x = c/(2g).

The coordinates {y

c/b, consequently the ratio of interest is r = (c/b)/(c/2g) = 2g/b.

By the transformation rule of Remark-2, translating B along e (x

(since a=0, y

Salmon [SalmonConics, p. 151] uses this argument to give a unified proof for hyperbolas and parabolas,

whereas for the case of parabolas I repeated essentially the same work using the *canonical form* of the

parabola in the file PowerGeneralParabola.html .

having one asymptote parallel to AB, where B is a point on KL?

The answer is a

resulting by drawing a parallel to KL from A.

In fact, taking the origin again at B and the axes as before and dividing by h we obtain the equation in the form:

2xy + by

The four coefficients result by introducing the coordinates of the three given points A(x

the equation and solving the system for the coefficients, giving:

b = k,

f = -k(y

c= k(y

g= -k(y

Assuming the coordinates A'(x

k = 2x

This shows that the conics are of the form:

2xy + k(b

Here the coefficients are the factors of k of the corresponding {b, g, ...} in the previous equalities.

The fact that all members of the pencil are hyperbolas follows from the

J

(tm,tn) and requiring from this line to have one root at infinity. This gives the equation:

2n + kb

This introducing the values for k an b

(x

First section gives another possibility to parameterize this pencil: Select an arbitrary parallel to BKL line

K'L'. Then select point K' on it arbitrary but L' so that it satisfies (B'K'*B'L')/B'A = (BK*BL)/BA. This

defines five points {A,K,L,K',L'} through which to pass the conic. The pencil depends on the parameter

fixing the positin of K'.

A further interesting question connected to the trapezium KLL'K' is discussed in PowerGeneralHyperbola2.html .

PowerGeneralHyperbola2.html

Produced with EucliDraw© |