Disjoint conic sections?

Question

is there any simple way to figure out whether two conic sections (e.g. two ellipses or an ellipse and a hyperbola) are disjoint or intersect each other? The conic sections are expected to be known employing the 6-parameter form Q(x,y)=0 where Q is a second-degree function of (x,y). Of course, one can go through the full procedure that derives the intersection points (see e.g. http://en.wikipedia.org/wiki/Conic_section#Intersecting_two_conics). However, I expect (at the first glance, at least), that the full derivation of the degenerated pencils are a bit bloat for this problem... but maybe i'm wrong and it cannot be skipped. thanks in advance!

Answer 1

This might allow you to quickly determine if they might be intersecting.

Each conic section can be modeled with an equation $Ax ^2 + B y^2 +C x y + D x + E y + F = 0 $. The conic can be decomposed along a principal coordinate system such that the equation along the rotated and translated coordinates is $A' x'^2 + B' y'^2 + F'=0$. To find this principal coordinates use the following:

1. Center of conic $$(x_c,y_c) = \left( \frac{2 B D-C E}{C^2-4 A B},\ \frac{2 A E - C D}{C^2-4 A B} \right)$$
2. Angle of conic $$\theta = \frac{1}{2} \tan^{-1} \left( \frac{C}{A-B} \right) $$
3. Transformation law $$(x,y) = (x_c + x' \cos\theta -y' \sin\theta, y_c + x'\sin\theta + y'\cos\theta)$$
4. Conic Radii (distance extrema, 1 real if hyperbola, 2 reals if ellipse) $$ r_x^2 = 2 \left( \frac{A E^2 + D ( B D - CE )}{\sqrt{C^2-4 A B}^3} \right) \left(1-\sqrt{1+\frac{C^2-4 A B}{(A+B)^2}} \right) \\ r_y^2 = 2 \left( \frac{A E^2 + D ( B D - CE )}{\sqrt{C^2-4 A B}^3} \right) \left(1+\sqrt{1+\frac{C^2-4 A B}{(A+B)^2}} \right)$$
5. Transformed conic equation $$ \frac{x'^2}{r_x^2} + \frac{y'^2}{r_y^2} + 1 = 0 $$

So if you at least transform to coordinates along one of the conics, it might be easier to check if points on the other conic intersect. For example if both are ellipses and the horizontal and vertical separation of the centers (in rotated axes) is more that the combined radii in that direction then they won't intersect.