# Show that the ellipse and the hyperbola are convex

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?

• That's basically the definition of convex graphs as suprema of a family of linear functions. – Adam Hughes Jun 30 '14 at 21:26
• 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. – Eric Auld Jun 30 '14 at 21:39
• 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. – Rahul Jun 30 '14 at 22:16
• @Rahul Good point. I guess neither does it qualify as a region (open connected set) in $\mathbb{R}^2$. – Eric Auld Jun 30 '14 at 22:46
• Well, I mean you can compactify to making it another version of a circle on a sphere. – Adam Hughes Jul 1 '14 at 0:20

• @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. – Eric Auld Jun 30 '14 at 21:55