# Intersection of two ellipses

Printable View

• Oct 18th 2008, 08:09 AM
againstthebes
Intersection of two ellipses
Hello all,

I have two ellipses, both share the same origin, one is rotated relative to the other which is aligned with the coordinate system. For the unrotated ellipse, it is in the form Ax^2+By^2+F=0. The rotated ellipse has a similar form, but includes a rotational component of course: Ax^2+By^2+Exy+F=0. I am looking for the intersection points (there should be four) of these two ellipses. It is where the two equations are equal to one another. Setting these two equations equal to one another and simplifying yields something like: (B1-B0)y^2=-E1xy-(A1-A0)x^2. My question is how can I solve for both x and y to find these intersections? Is there a simpler manner? Thank you for any input.
• Oct 19th 2008, 01:46 AM
CaptainBlack
Quote:

Originally Posted by againstthebes
Hello all,

I have two ellipses, both share the same origin, one is rotated relative to the other which is aligned with the coordinate system. For the unrotated ellipse, it is in the form Ax^2+By^2+F=0. The rotated ellipse has a similar form, but includes a rotational component of course: Ax^2+By^2+Exy+F=0. I am looking for the intersection points (there should be four) of these two ellipses. It is where the two equations are equal to one another. Setting these two equations equal to one another and simplifying yields something like: (B1-B0)y^2=-E1xy-(A1-A0)x^2. My question is how can I solve for both x and y to find these intersections? Is there a simpler manner? Thank you for any input.

Solve both as quadratics in x (so the solutions are in terms of y and the coefficients), and equate the solutions and solve for y.

CB
• Oct 19th 2008, 07:26 AM
awkward
I think you will end up having to solve a 4th degree equation (a.k.a. a biquadratic). This can be done analytically but is not very pleasant.