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Without knowing the equation for an ellipse, but rather just with the geometric definition i.e. the ellipse is the locus of points with $PF_1 + PF_2 = k > F_1F_2$.

I tried to use a classic construction: $\omega = \odot(F_1,k)$, so for $Q \in \omega$, the point $P$ of meeting of the perpendicular bisector between $Q$ and $F_2$, with $QF_1$ is in the ellipse.

By the converse: given $P$ in the ellipse, the point $Q = PF_1 \cap \odot(P,PF_2)$, not in between $P$ and $F_1$, is in $\omega$.

The problem is that this last operation transforms lines in weird curves, while I was hoping it would transform lines in another lines and thus, a line would meet a circle in more than two points.

Notice that it is not obvious from this definition that an ellipse can be transformed in a circle by dilation on some axis.

hellofriends
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  • You might as well prove it for closed (strictly) convex curves in general. – dxiv Jul 31 '21 at 06:01
  • eh not really an algebraic question – hellofriends Jul 31 '21 at 06:04
  • Doesn't have to be algebraic. You can prove that the interior of the ellipse is a convex set geometrically. – dxiv Jul 31 '21 at 06:13
  • Yes, I believe so. – hellofriends Jul 31 '21 at 06:16
  • The proof the interior is convex is kinda simple: Let $AB$ be in the interior of $\mathcal E$, then in the $\triangle F_1AB$, if we take a point $P$ on side $AB$, we must have $\min {F_1A,F_1B}< F_1P < \max {F_1A,F_1B}$ and thus $\min {F_2A,F_2B} < F_2P < \max{F_2A,F_2B}$. From this I think we can conclude that $P$ is in the interior as well. – hellofriends Jul 31 '21 at 21:46
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    That only proves that $,F_1P+F_2P \lt \max(F_1A, F_1B) + \max(F_2A, F_2B),$ but that's not enough to prove $,F_1P+F_2P \lt \max(F_1A+F_2A, F_1B+F_2B),$ which is what's needed. However, you can prove it using only the triangle inequality, see here for example. – dxiv Jul 31 '21 at 22:56
  • @dxiv I'll try to translate it to geometry – hellofriends Jul 31 '21 at 23:03

3 Answers3

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Here is a simple argument that does not directly use the sum of distances from the focal points:

We can begin with the assumption that this property holds for a circle, that is, a line can't meet a circle in more than two points. Then we note that if the plane of the circle is viewed from not top but a tilted direction, then the circle is seen as an ellipse. In fact for any given ellipse, one can find a point of view at which the circle is seen like that given ellipse. Also, such change in point of view does not essentially change lines. Therefore, our assumption of the property holding for circles can be translated to the statement you wanted to demonstrate.

Saeed
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    "Notice that it is not obvious from this definition that an ellipse can be transformed in a circle by dilation on some axis." – hellofriends Aug 01 '21 at 00:20
  • @hellofriends Here is a beautiful proof showing the equivalence of the locus property of ellipse (constant sum of distances from foci) and the conic section definition, which demonstrates how, by rotation of a plane, a circle becomes an ellipse: https://mathworld.wolfram.com/DandelinSpheres.html – Saeed Aug 01 '21 at 01:05
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There is a simple convexity argument. $d(x,y)=\sqrt{(x-x_1)^2+(y-y_1)^2}$ is a convex function, so the same holds for $f(x,y)=\sqrt{(x-x_1)^2+(y-y_1)^2}+\sqrt{(x-x_2)^2+(y-y_2)^2}$. Let $(x_1,y_1)$ and $(x_2,y_2)$ be the foci. Assume that a line meets the ellipse at three distinct points. Without loss of generality we may assume that the line has equation $y=0$, then consider $g(x)=f(x,0)$. $g(x)$ is convex, so it cannot attain the same value three times. We have got a contradiction.

Jack D'Aurizio
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In the style of classical geometry:

Let $\ell$ be the line, $F_1, F_2$ be the two foci, and $G_2$ be the reflection of $F_2$ through $\ell$. Then for any point $P$ on the line, $PF_2 = PG_2$, so the intersection of $\ell$ with the ellipse consists of all points $P$ where $F_1P + PG_2 = k$.

Let $Q$ be intersection of $F_1 G_2$ with $\ell$: $F_1 P + PG_2$ is minimized when $P=Q$, by the triangle inequality. Suppose that $Q, R, S$ appear on line $\ell$ in that order. Then $R$ is in the interior of $\triangle F_1 G_2 S$, so ray $G_2R$ must intersect $F_1S$ at some point $X$:

enter image description here

By a series of triangle inequalities, $F_1S + SG_2 > F_1 R + R G_2$, because

\begin{align} F_1S + SG_2 &= F_1X + XS + SG_2 \\ &> F_1X + XG_2 \\ &= F_1X + XR + RG_2 \\ &> F_1R + RG_2. \end{align} In particular, we cannot have $F_1S + SG_2 = k$ and $F_1R + RG_2 = k$ at the same time. Therefore points of intersection between the line and the ellipse cannot be on the same side of $Q$ - so there can be at most two of them.

I suppose I assumed that the line $\ell$ does not pass between $F_1$ and $F_2$: if it does, then we do the same thing, except without the reflection step first.

Misha Lavrov
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