## 1. Rectangular hyperbolas defined by a line

Given the line  ax+by+c = 0, equation
xy + k(ax + by + c) = 0,
for variable parameter k defines a family (ck) of conics with the following properties.

[1] All conics of the family are rectangular hyperbolas with asymptotes parallel to the coordinate axes.
[2] All these conics pass through the intersection points {A,B} of the line with the axes.
[3] The focal points of these hyperbolas lie on two confocal conics with foci at {A,B}, which also pass through the origin O.
[4] The two previous conics are orthogonal.
[5] The centers of the conics ck lie on the diagonal OQ of the rectangle OAQB defined by {A,B,O}.
[6] This diagonal OQ is symmetric with respect to the line y=x of the normal from O to the line.

All of these properties follow easily from the simple form of this equation. First the fact that these are all rectangular hyperbolas
results from the vanishing of the square terms (see Conic_Equation.html , section-11). Then the coordinates of the center are easily
computed (same reference, section-8) to be
(x0, y0) = -k(b, a).
Then translating the coordinates-origin to that point via the new coordinates
x = x' - kb,
y = y' - ka,
reduces to the equation for (x', y'):
x' y' = k2(ab)-kc.
Introducing the new coordinates

leads to the equation in (x'', y''):
(x'')2 - (y'')2 = 2k(kab-c),
which for positive k(kab-c) has its focal points at

Back substitution x''--> x' --> x and ellimination of k from the equations of the two cases leads to the conics with matrices:

It is easily verified that the two conics have the stated properties. Among others their common center is at
-(c/2)(1/a, 1/b),
and their second invariant J2 (see Conic_Equation.html , section-3) is correspondingly  -2ab(b-a)2  and  2ab(a+b)2. Thus in the case ab>0,
as this is illustrated above, the first is a hyperbola and the second an ellipse.

## 2. Constructing a particular hyperbola

The construction of the particular hyperbola of the above family for k=1 can be done in several ways. One of them is to use the above
formulas to find the coordinates of the focal points in each case:

From these follows that the focal point is on the parallel to one of the main diagonals from point (a,b). The hyperbola can be constructed,
knowing its focal points, consequently its center and a point of it like one of {A,B} (see HyperbolaAsymptotics.html ). It is even simpler
to use the formula for the center M above locating it at  (-b, -a). Then, point {A,B} and their symmetrics {A',B'} are on the hyperbola,
as well as the orthocenter of one triangle constructed by selecting three out of these four points (see OrthoRectangular.html ). Then the conic can
be constructed as the one passing through these five points.
This particular cubic is needed in the process of graphically solving a cubic equation   x3 + ax2 + bx +c = 0 (see Cubic_Graphic_Solution.html ).

Conic_Equation.html
HyperbolaAsymptotics.html
OrthoRectangular.html
Cubic_Graphic_Solution.html