How would one approach proving the following statements:

(a) A plane can be covered with the interiors of finite number of hyperbolas

(b) A plane cannot be covered with the interiors of finite number of parabolas


  1. interior is the part of the plane that contains a focus (foci) of the aforementioned shapes.

  2. It is clear that without losing generality one can take $y=ax^2$ and $xy=a$ for all meaningful values of the parameter $a$ (e.g. $a\ne0$) and all possible rotations of these curves around the origin.

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2 Answers2


For (a), the origin-centered rotations of $xy=1$ by $0^\circ$, $45^\circ$, $90^\circ$, and $135^\circ$ leave a bounded octagonal hole that's easily covered.

For (b), consider a finite collection of parabolas, and choose a line $\ell$ parallel to none of their axes. (Since there are finitely-many axes, we can make such a choice.) The intersection of this $\ell$ with the interior any parabola, if not empty, is a finitely-long (possibly zero-length) segment. An infinitely-long line certainly contains a point not covered by a finite collection of those.

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Presumably you are allowed to have the interiors overlap. For a, you need to find such a finite collection. Think about $xy=1$ shifted up and right by some large amount, say $10$ units plus a copy shifted down and to the left by $10$ units. The center gets covered nicely. We still haven't got the whole plane, but ...

For b, think about trying to cover a very large circle. To be specific, we will think about how much is covered by the interior of $y=x^2$. If you take the circle $x^2+y^2=R^2$, you only cover from roughly $(-R,R^2)$ to $(R,R^2)$ (I know these aren't on the circle, but they almost are for large $R$ ). This represents an angle of about $\frac 2R$ radians, which can be made arbitrarily small by taking $R$ large enough. It works even if the parabola doesn't go through the origin, you just need to work on $R$. So given a finite set of parabolas, take $R$ large enough that... There are a lot of holes to fill, but the idea is there.

Ross Millikan
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