Intersection of Hyperbolas I have two hyperbolas, given in the form:
$$\tag{1}A_1x^2+2B_1xy+C_1y^2+2D_1x+2E_1y+F_1=0$$
$$\tag{2}A_2x^2+2B_2xy+C_2y^2+2D_2x+2E_2y+F_2=0$$
With $A_1=A_2=0$.
I wish to attain all intersection points. I saw this example, but my representation is different. In addition, somehow this transformation does not seem to work for me.
Thanks!
 A: The case $A_1=A_2=0$ leads to a cubic equation. They are at least one or up to three intersection points :

The general case, valid not only for two hyperbolas but for two quadratic curves, leads to a quartic equation, so from zero up to four intersection points.

A: The solution is voluminous, found by Cramer's Rule solving for 2 unknowns from 2 equations.  Roots are as quadratic equation depending on sign of quantity under the radical 
$$
(-2 B2^2 C1 D1 + 2 B1 B2 C2 D1 + A2 C1 C2 D1 - A1 C2^2 D1 + 
    2 B1 B2 C1 D2 - A2 C1^2 D2 - 2 B1^2 C2 D2 + A1 C1 C2 D2 + 
    A2 B2 C1 E1 - 2 A2 B1 C2 E1 + A1 B2 C2 E1 + A2 B1 C1 E2 - 
    2 A1 B2 C1 E2 + A1 B1 C2 E2)/(-4 A2 B1 B2 C1 + 4 A1 B2^2 C1 + 
    A2^2 C1^2 + 4 A2 B1^2 C2 - 4 A1 B1 B2 C2 - 2 A1 A2 C1 C2 + 
    A1^2 C2^2)  + ..... + \sqrt {...}
$$
EDIT 1:
Considering a single brach it may be like one of three situations:

But for  two branches, there is a situation of double the number of roots/intersections  as may be in a quadratic equation.
