[alogo] 1. Power generalization hyperbola

Here I follow closely Salmon, who proves the property in the frame of its exhaustive discussion of the general
form of the quadratic equation
                                                f(x,y) = ax2 + 2hxy + by2 + 2gx + 2fy + c = 0.
Remark-1 When the x-axis intersects this curve at  a point at infinity then a = 0.
This follows by setting y=0 => ax2 +2gx + c = 0 => a + 2g/x + c/x2 = 0. If x tends to infinity this is zero only when
Remark-2 Translation of the axes to another origin {x = x'+x0, y = y' + y0} has the following effect on
the coefficients:
1)  the quadratic terms remain the same a'=a, h'=h, b'=b.
2)  the linear terms transform linearly: { g' = g + ax0+hy0,   f ' = f + hx0 + by0 }.
3)  the new constant term is:  c' = f(x0, y0).

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).

[0_0] [0_1] [0_2]
[1_0] [1_1] [1_2]

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 {y1, y2} of {K,L} respectively satisfy the quadratic  by2+2fy+c = 0 hence their product is
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 (x0 arbitrary, y0=0) the corresponding ratio
(since a=0, y0=0 in g'=g+ax0+hy0) will remain equal to 2g/b i.e. constant.

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 .

[alogo] 2. Related pencil of hyperbolas

The above discussion rises the question: How many hyperbolas there are  passing through {A,K,L} and
having one asymptote parallel to AB, where B is a point on KL?

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]

The answer is a pencil of conics parameterized through the location of A', the fourth vertex of the trapezium
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 + by2 + 2gx + 2fy + c = 0.
The four coefficients result by introducing the coordinates of the three given points A(x0,0), K(0,y1), L(0,y2) into
the equation and solving the system for the coefficients, giving:
                                                  b = k,
                                                  f = -k(y1+y2),
                                                  c=  k(y1*y2),
                                                  g= -k(y1*y2)/(2x0).
Assuming the coordinates A'(x0,y3) and introducing to the equation we find that:
                                                  k = 2x0/(2(y1+y2)-y3).
This shows that the conics are of the form:
                                                  2xy + k(b0y2 + 2g0x + 2f0y + c0) = 0.
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 invariant J2<0, which is
J2 = ab-h2 = -4. One asymptote is y=0 (line AB). The other asymptote is determined by setting (x,y)=
(tm,tn) and requiring from this line to have one root at infinity. This gives the equation:
                                                  2n +  kb0m = 0 =>  (m,n) = s(2, -kb0).
This introducing the values for k an b0 shows that the second asymptote is:
                                                 (x0)x + (2(y1+y2)-y3)y = 0.

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 .

See Also



[SalmonConics] Salmon, G. A treatise on conic sections Longmans, London 1855

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