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 taxis, 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:

and

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 planeslicingcone 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.