To see that the curve is an ellipse in standard form, notice that the gradient of at is
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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 :
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For the Euclidean inner product, which I represent as a "dot" product, it is clear that
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Now recall that the vector is normal (perpendicular) to the hyperboloid at . It is easy to see that:
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and
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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
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That expands to
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A straightforward verification shows that these points all belong to the cone
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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.