*in space and two perpendicular orientation vectors*

**C***and*

**U***that define both the plane where the ellipse lays and the size of the axes. So you can store a 3D ellipse in just 8 floats. In the case of the ellipse being degenerated to a disk, then a single orientation vector is needed (perpendicular to the plane containing the disk) and a radious (so, 6 floats). Planar ellipses can become very useful for computer graphics. For example, they appear when you cut a cylinder with a plane. They also appear when rendering point clouds with a splatting algorithm, or when raytracing point clouds. They can also help in realtime ambient occlusion and indirect lighting computations, where occluders can be approximated by a pointcloud.*

**V**Here I will show how to do two of the most basic operations on planar ellipses: bounding box calculation and ray intersection (sounds like this is what you need for a fast kd-tree based raytracer, uh?). Let's see the bounding box first:

**Bounding Box:**

As we know any point in the ellipse boundary is described by the following parametric equation:

The bounding box of the ellipse has to be tangent to this boundary. This tangential points will be the maximum and minimum

**,**

*x***and**

*y***coordinates of the boundary equation. So, we just have to find the minima/maxima of the equation. From elementary school we know that we can get them by finding where the derivative annulates. So,**

*z*must equal zero for each of the three coordinates. Let's rename first and solve for the

**coordinate:**

*x*to get |

meaning that

Summarizing, we can calculate our bounding box corners like this:

Note that this is quite SIMD (SSE/Altivec) friendly, so you can do a single square root for the complete bounding box. Ready for building your acceleration kd-tree for point clouds?

**Ray-ellipse intersection:**

Similarly to the ellipse border, the interior can be defined by

and |

where the equality holds for the border. Now, if we define the ray with the equation

where

**is the ray origin and**

*ro***is the (not necessarily normalized) ray direction, then we must make both expressions equal in order to get the intersection point, thus we must solve the equation that actually is a system of three equations (one for each of the**

*rd***,**

*x***and**

*y***coordinates) with three unknowns. Re-arranging these three equations we get the following system:**

*z*We can solve this by Cramer's law:

Note that

**will be zero when the ray is parallel to the plane containing the ellipse, so that needs special care. Of course,**

*de***corresponds to the ray-plane intersection distance. Finally, to test that the intersection point is in the ellipse, check for**

*t*I hope all this is helpful for somebody :)