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In Spivak's chapter on differentiation, he asks the reader to prove that the tangent line to an ellipse or a hyperbola intersect the figure at exactly one point.

How is this done most elegantly? I can crunch the numbers and show that it comes out right, but I have a feeling there's a more elegant way. And why is it (is it?) that a smooth region in $\mathbb{R}^n$ is convex iff for every tangent hyperplane, that hyperplane intersects the set in only one point?

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    $\begingroup$ That's basically the definition of convex graphs as suprema of a family of linear functions. $\endgroup$ – Adam Hughes Jun 30 '14 at 21:26
  • $\begingroup$ By "convex graph", do you mean the graph of a convex function? I would be interested to know how to generalize this result to figures in the plane (e.g. conics) which are not graphs of functions. $\endgroup$ – Eric Auld Jun 30 '14 at 21:39
  • $\begingroup$ I'm not sure in what sense one can consider a hyperbola to be convex, unless one only takes the interior of one of its two branches. $\endgroup$ – Rahul Jun 30 '14 at 22:16
  • $\begingroup$ @Rahul Good point. I guess neither does it qualify as a region (open connected set) in $\mathbb{R}^2$. $\endgroup$ – Eric Auld Jun 30 '14 at 22:46
  • $\begingroup$ Well, I mean you can compactify to making it another version of a circle on a sphere. $\endgroup$ – Adam Hughes Jul 1 '14 at 0:20
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Both ellipse and hyperbola are conics, and as such the solution to a quadratic equation. Along any line, there are up to two points where the quadric intersects the line. These two points can either be distinct real points, in which case the curve of the quadric switches side at each such point. Or it can be a pair of complex points which are conjugate to one another in those cases where the line passes by the quadric. Or as a limit between these two cases, the two points of intersection can coincide, in which case the quadric will stay on one side of the line. That last one is exactly your tangential situation. So there you see that you have only one point of intersection for tangents. The core is the quadratic nature of the equation.

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  • $\begingroup$ Exactly what I was looking for!! Thank you so much. I knew the answer was as simple as that. $\endgroup$ – Eric Auld Jun 30 '14 at 21:40
  • $\begingroup$ For the hyperbola, how would you prove that a tangent to one branch does not intersect the other branch? $\endgroup$ – marty cohen Jun 30 '14 at 21:52
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    $\begingroup$ @martycohen Let a line be $\{(t\xi+a, t\eta+b): t\in \mathbb{R}\}$ for given $(a,b)$ and $(\xi,\eta)$ (point/slope). Given $x^2-y^2=1$, for example, plug in for $x$ and $y$ and you have a degree 2 polynomial in $t$, which can have at most two roots (counting multiplicity). I claim that a tangent line is a double root. You can check this because at an even root there is no "crossing", i.e. in a neighborhood of the root the polynomial does not change sign. $\endgroup$ – Eric Auld Jun 30 '14 at 21:55

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