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Given a straight line $D$, a point $F\notin D$ and a positive real number $e$, a conic is a subset of ${\cal P}_2$ defined as: $$ \mathcal{C}(e,F,D) = \{M\in {\cal P}_2,\, d(F,M)=e\,d(M,D) \} $$ where ${\cal P}_2$ is the Euclidean plane and $d$ is the Euclidean distance.

It is well known that when $e$ tends to $0$, ${\cal C}(e,F,D)$ tends to a circle and I would like to see this from this definition without taking any algebraic computations. (we can show this fact from algebraic computations computing the distances in some coordinates system but I would like to avoid this approach.)

I see that $e,F,D$ are not independent in the sense that if $e$ tends to $0$ then $M$ has to tend to $F$ and we do not obtain a circle so there should be some connections between $e$, $D$, $F$. But then it means that the previous definition is not "well-posed". Moreover a conic being defined by a polynomial equation of degree less than 2 (so 5 coefficients seen as parameters), there should not have connection between $e$ (1 parameter), $F$ (2 parameters), and $D$ (2 parameters)... What do I miss here ?

A second point assume there is a connection between $e,F,D$ and I have the feeling that when $e$ tends to $0$ then $D$ as to be seen as a point at infinity. This framework should be related to projective geometry and I would like some developments in this direction so that I can deduce the circle as a limit. An idea is that since we have a point to infinity, the object has to be invariant by any rotation. Moreover it is a convex set so it is a circle... I need to put some maths on this idea and I think projective geometry and the study of a group of transformations acting on the set should lead to find the symmetries.

Smilia
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    $D$ does not have 3 degrees of freedom, only $2$. –  Apr 02 '19 at 12:50
  • In order to get a circle of a set radius, you want $d(F, M)$ to be constant, so instead of letting $M$ get closer to $F$, you move $D$ further away. Increasing $d(M,D)$ while $d(F,M)$ is constant forces $e \to 0$. – Paul Sinclair Apr 02 '19 at 17:01

3 Answers3

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If you are willing to do this in three dimensions then you can use a Dandelin sphere. Given $e < 1,$ $F,$ and $D$ you can construct the line $\ell_a$ perpendicular to $D$ through $F$ and find the two distinct points $M_1, M_2$ on $\ell_a$ such that $d(F,M_1)=ed(M_1,D)$ and $d(F,M_2)=ed(M_2,D)$.

Let $\pi_1$ be the plane in which $D$ and $F$ lie. Place a sphere tangent to the plane $\pi_1$ at $F.$ Choose the radius of the sphere so that in the plane through $M_1,$ $M_2,$ and the center of the sphere you can find a point $P$ from which lines through $P$ and $M_1$ or $M_2$ are tangent to the sphere. (There is one radius for which the tangents would be parallel; choose any other.) Use $P$ as the apex of a right circular cone tangent to the sphere. Let $\mathcal C'$ be the intersection of this cone and the plane in which $F$ and $D$ lie.

There a beautiful proof that shows that the curve $\mathcal C'$ satisfies the conditions by which $\mathcal C$ was defined; you can see it in this answer. The argument in that answer was meant to derive the eccentricity formula from the cone, but you should be able to show uniqueness of this solution, which means it also derives the cone from the formula.

In particular, in the proof we find that the sphere is tangent to the cone along a circle, and the plane through that circle intersects the plane $\pi_1$ along the line $D.$

Now if you rotate the cone about the center of the sphere, the plane through the tangent circle of the cone and the sphere moves and so does its intersection with the plane $\pi_1.$ If you rotate the cone so that the angle between the planes is decreased, the intersection is further from $F$ and the cone intersects $\pi_1$ in a new conic section with a smaller eccentricity. If we continue to call the line of intersection $D$ and the eccentricity $e,$ in the limit as the apex of the cone approaches the line through the center of the sphere perpendicular to $\pi_1,$ the line $D$ goes toward infinity and the eccentricity $e$ goes to zero. Moreover, in the limiting case (where the planes are parallel) the conic section is a circle.

David K
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Keep $D$ fixed and reduce $e$ so that $M$ gets closer and closer to $F$. At the same time, $d(M,D)$ becomes more and more constant and the locus tends to a circle of radius $e\,d(F,D)$.

If you want the radius to be $R$, dilate the whole figure by $\dfrac{R}{e\,d(F,D)}$. For eccentricity $0$, you get a perfect circle with a line at infinity.

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For this definition $e\to 0$ actually implies the curve tending to a circle. The 'problem' might be, it is a vanishing circle – it's an ellipse with semiaxes' ratio tending to $1$ whilst both semiaxes tend to $0$, because $d(M,D)\to d(F,D)$ is bounded, nonzero, hence $d(M,F) = e\,d(M,D)\to 0$.

CiaPan
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  • ok for this circle seen as a point.

    But can't we look the case where $D$ is at infinity ?

     – Smilia Apr 02 '19 at 12:38
  • Actually, if you fix a focus point $F$ and the conic's apex $A$, then reducing the eccentricity towards zero will cause the curve to tend to a circle with centre $F$ and radius $FA$ while the directrix $D$ flows away to infinity. Which is equivalent to zooming-in the previous set-up, in which $D$ is fixed but $A$ approaches $F$. – CiaPan Apr 02 '19 at 12:42