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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!

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  • $\begingroup$ There can be 2 real, one double or a single complex root & conjugate. $\endgroup$
    – Narasimham
    Commented Nov 9, 2016 at 18:35
  • $\begingroup$ Are you looking for a general formula? Or do you have a specific problem? $\endgroup$
    – user307169
    Commented Nov 9, 2016 at 18:42
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    $\begingroup$ @Narasimham, that doesn't sound quite right unless I'm misreading what you said. There can be 1 real solution (e.g. $xy + y^2 + x + y + 1 = 0$ and $-2xy + y^2 + x + y + 1 = 0$) or 3 real solutions (e.g. $xy + y^2 + x + y + 1 = 0$ and $2xy -12y^2-3x-31y+37=0$), etc. $\endgroup$
    – user307169
    Commented Nov 9, 2016 at 18:59
  • $\begingroup$ @tilper I have a set of hyperbolas of this form ($A_i=0$), and I'm looking for a general formula to find all intersection points of any two hyperbolas in the set. $\endgroup$ Commented Nov 9, 2016 at 18:59
  • $\begingroup$ In the last example you gave in the question you chose axes rotated hyperbolas without an $xy$ term, $\endgroup$
    – Narasimham
    Commented Nov 9, 2016 at 19:28

2 Answers 2

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The case $A_1=A_2=0$ leads to a cubic equation. They are at least one or up to three intersection points :

enter image description here

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.

enter image description here

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  • $\begingroup$ Great, this is exactly what I looked for. Thanks! $\endgroup$ Commented Nov 10, 2016 at 9:23
  • $\begingroup$ Just a correction, the term $k_1$ should be $2(E_1 D_2 + F_1 B_2 - 2 E_2 D_1-F_2B_1)$ $\endgroup$ Commented Nov 10, 2016 at 9:55
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    $\begingroup$ Yes some terms are missing. I will correct it. Thanks. $\endgroup$
    – JJacquelin
    Commented Nov 10, 2016 at 9:58
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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:

Hyp ntrxn

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

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  • $\begingroup$ Do you mean Cramer's rule? Any point to a valuable explanation? $\endgroup$ Commented Nov 9, 2016 at 19:13
  • $\begingroup$ Changed spelling. $\endgroup$
    – Narasimham
    Commented Nov 9, 2016 at 19:24
  • $\begingroup$ My hope is that since $A_i=0$ this expression can be simplified.... $\endgroup$ Commented Nov 9, 2016 at 19:33
  • $\begingroup$ To an extent may be by dividing by $ F$ s , $A_i =0$ but depends on the discriminant basically. $\endgroup$
    – Narasimham
    Commented Nov 9, 2016 at 19:40
  • $\begingroup$ Do you refer to $D$ or $\Delta$ here? Anyway, it holds that $x_c=0$ (in the same notations) $\endgroup$ Commented Nov 9, 2016 at 19:44

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