Note 2. The Apollonian ellipse in standard form

To see that the curve is an ellipse in standard form, notice that the gradient of at is




Let and assume that . If then is on the t-axis, the tangent plane is , and its intersection with the cone is a unit circle.

Now consider the pair of tangent vectors at the point on the hyperboloid :





For the Euclidean inner product, which I represent as a "dot" product, it is clear that





Now recall that the vector is normal (perpendicular) to the hyperboloid at . It is easy to see that:






Thus are tangent to the hyperboloid at , and, since they are perpendicular, they span the tangent plane.

I now establish that these conics, appropriately translated and rotated, have the familiar analytic form of central conics in a Euclidean plane -- for ,

For that, consider the curve of points of the form




That expands to


A straightforward verification shows that these points all belong to the cone




Therefore this curve is contained in an ellipse, by the plane-slicing-cone definition. Since the intersection of the plane with the cone is a connected curve, this ellipse is the entire intersection. This parametric characterization leads to the usual satellite concepts associated with the ellipse : the center, foci, axes, and so on.