TITLE: Finding Intersection of an ellipse with another ellipse when both are rotated
QUESTION [8 upvotes]: Equation of first ellipse=>
$$\dfrac {((x-xFirstEllipseCenterPoint)\cdot \cos(A)+(y-yFirstEllipseCenterPoint)\cdot \sin(A))^2}{(a_1^2)}+\dfrac{((x-xFirstEllipseCenterPoint)\cdot \sin(A)-(y-yFirstEllipseCenterPoint)\cdot \cos(A))^2}{(b_1^2)}=1$$

Equation of the second ellipse=>
$$\dfrac {((x-xSecondEllipseCenterPoint)\cdot \cos(B)+(y-ySecondEllipseCenterPoint)\cdot \sin(B))^2}{(a_2^2)}+\dfrac{((x-xSecondEllipseCenterPoint)\cdot \sin(B)-(y-ySecondEllipseCenterPoint)\cdot \cos(B))^2}{(b_2^2)}=1$$

I know that the ellipse will intersect at 

One Point
Two Point 
Three Point 
Four Point
No intersection at all

Is there a general set of equation to solve the same.

REPLY [4 votes]: In addition to what @bubba said, you can ease the computation as follows: translate the first ellipse center to the origin, counter rotate by angle $A$, and apply a rescaling with factors $\frac1{a_1}$, $\frac1{b1}$. This will turn the first ellipse in the unit circle.
Then, you can rewrite the second equation in its implicit form
$$\frac{((a_1x-\delta x)\cos(B-A)+(b_1y-\delta y)\sin(B-A))^2}{a_2^2}+\frac{((a_1x-\delta x)\sin(B-A)-(b_1y-\delta y)\cos(B-A))^2}{b_2^2}
-1=0\\
=Ax^2+Bxy+Cy^2+Dx+Ey+F.$$
and plug the Weierstrass expressions for the circle, $x=\dfrac{1-t^2}{1+t^2}$ and $y=\dfrac{2t}{1+t^2}$, giving
$$(A+F-D)t^4+2(E-B)t^3+2(F+2C-A)t^2+2(B+E)t+A+D+F=0.$$

For an easy numerical solution, you can also use the parametric representation of the ellipse $2$ and discretize it using line segments or, better, circular arcs. This can be done by sampling points in triples or quads along the ellipse and bounding the deviation between the true curve and the linear or curvilinear approximation.
Then you reduce the problem to that of line/arc or arc/arc intersection, which is quite tractable.