# The distance between the center of a circle touching the a conic cut at two points, and between the midpoint of the two tangent points

A few days ago, while using GeoGebra, I was able to generalize a well-known property of the parabola: The distance between the center of a circle tangential to a parabola at two points and between the midpoints of the two tangential points is equal to the distance between the focus and the guide

And now the new formulation is valid for all conic sections, but I couldn't prove it, please, any help would be welcome, this new form:

The distance between the center of a circle touching a conic section at two points and the midpoint of the two tangent points is equal to the length of the radius of the circle kissing its apex multiplied by the ratio between the distance between the midpoints of the two tangent points from the center of the cut and the radius of the principal diameter of the conic section.

in another meaning:

Let $$P$$ be a conic with center $$S$$, principal vertex $$A$$, kissing circle with vertex at $$A$$ is $$R$$, center $$O$$, and let $$C$$ be a circle tangent to $$P$$ at two points whose midpoint is $$M$$ and center of $$C$$ is $$N$$, then $$MN=OA×\frac{SM}{SA}$$

The formulation works for all conic sections, where the parabola is a special case of an ellipse whose center is a point at infinity or a hyperbola whose center is a point at infinity, and the ratio becomes equal to $$1$$ Here are some graphics for easier understanding:

If $$P$$ is a point on the ellipse of foci $$F$$, $$F'$$ and centre $$S$$, let $$PN$$ be the normal at $$P$$ (bisector of $$\angle FPF'$$) and $$M$$ the projection of $$P$$ on the major axis $$AA'$$. Segment $$MN$$ is called the subnormal at $$P$$. Your result is well known and a geometrical proof can be found, for instance, in Drew's book on conic sections.
Let $$H$$, $$H'$$ be the projections of $$P$$ on the directrices, which cut line $$AA'$$ at $$X$$ and $$X'$$. From the angle bisector theorem we have:
\begin{align} F'N:FN &=F'P:FP \\ &=H'P:HP\\ &=X'M:XM, \end{align} hence $$(F'N-FN):(F'N+FN)=(X'M-XM):(X'M+XM),$$ that is: $$2SN:F'F=2SM:X'X.$$ From this it follows: \begin{align} SN:SM &=F'F:X'X \\ &=SF:SX\\ &=SF^2:SA^2, \end{align} where in the last equality I used the relation $$SX=SA/e=SA^2/SF$$. Finally: $$(SM-SN):SM=(SA^2-SF^2):SA^2$$ that is $$MN:SM={SA^2-SF^2\over SA}:SA,$$ where $$(SA^2-SF^2)/SA$$ is the radius $$OA$$ of the osculating circle at $$A$$.
A more common way to rewrite this relation is: $$MN=(1-e^2)SM,$$ where $$e=SF/SA$$ is the eccentricity.