TITLE: an important property of an ellipse
QUESTION [10 upvotes]: Good morning everybody.
I would like to know the proof of the following observation on the ellipse.

A circle is drawn with the right latus rectum as diameter. Another circle is drawn with its center on the major axis such that it is internally tangent to both the circle given above, and the auxiliary circle (the circumcircle of the ellipse). This circle is said to be right associate circle of ellipse. (The left associate circle is defined similarly.) All the circles whose diameters are right focal chords are tangent to the right associate circle, and similarly, all the circles whose diameters are left focal chords are tangent to the left associate circle. 

Also, can we see something similar in the other conic sections?

REPLY [4 votes]: How might one discover the Associate Circle? It's the envelope of the family of Focal Chord Circles.
Finding an envelope of a family of curves --say, parameterized by $\theta$-- is conceptually straightforward (though often computationally thorny): eliminate $\theta$ from the system
$$\begin{eqnarray*}
F(\theta) &=& 0 \\
\frac{\partial}{\partial\theta} F(\theta) &=& 0
\end{eqnarray*}$$
where $F(\theta)=0$ is the common equation for the curve family. What remains is a formula for the curve --the "envelope"-- that is tangent to each curve in the family.
For the FCCs, we start by recalling the center and radius from my previous answer:
$$\begin{eqnarray*}
C = \frac{p e \cos\theta}{1 - e^2\cos^2\theta}(\cos\theta,\sin\theta) =: (c \cos\theta, c\sin\theta) \hspace{0.5in} r = \frac{p}{1-e^2\cos^2\theta}
\end{eqnarray*}$$ 
The standard equation for the circle is
$$\begin{eqnarray*}
(x - c \cos\theta )^2 + ( y - c \sin\theta )^2 = r^2
\end{eqnarray*}$$
so that we can write
$$\begin{eqnarray*}
F(\theta) &=& (x-c \cos\theta)^2+(y-c\sin\theta)-r^2 \\
&=& x^2 + y^2 - 2 x c \cos\theta - 2 y c\sin\theta + c^2 - r^2 \\
&=& x^2 + y^2 - 2 x \frac{p e \cos^2\theta}{1-e^2\cos^2\theta} - 2 y \frac{p e \cos\theta\sin\theta}{1-e^2\cos^2\theta} - \frac{p^2}{1-e^2\cos^2\theta}
\end{eqnarray*}$$
I'll leave it to the reader to compute $\frac{\partial}{\partial \theta}F$ and work through the elimination of $\theta$. In my attempt, I arrived at a large polynomial, one of whose factors yields the equation
$$\begin{eqnarray*}
p^2 + e p x - x^2 ( 1 + e ) - y^2 ( 1 + e ) = 0
\end{eqnarray*}$$
so that,
$$\begin{eqnarray*}
\left( x - \frac{p e}{2(1+e)}\right)^2 + y^2 = \left( 
\frac{p(2+e)}{2(1+e)} \right)^2
\end{eqnarray*}$$
This circle has center and radius that agree with $K$ and $s$ in my previous answer; it's the Associate Circle!
But ... hold on ... a second factor of my polynomial is a kind of "conjugate" of the above:
$$\begin{eqnarray*}
\left( x - \frac{p e}{2(1-e)}\right)^2 + y^2 = \left( 
\frac{p(2-e)}{2(1-e)} \right)^2
\end{eqnarray*}$$
For $0 \le e < 1$, this circle is tangent to the "other ends" of the Auxiliary Circle and the ellipse, as well as tangent to the "outside" of the Focal Chord Circles:

(Hmmmm ... I seem to get J.M.'s "jittery gif" effect after upload. What's up with that? The original renders perfectly on my Mac.)
For the parabolic case $e=1$, the factor's equation becomes $p^2+px = 0$ so that $x = -p$. Here, the "circle" is a vertical line tangent to the Auxiliary Circle. We could've suspected the existence of this line from the parabolic animation in J.M.'s answer: observe how the red circles all hug the left edge of the plot.
For the hyperbolic case $e>1$, we get an appropriate variation:

So, in fact, we have two Associate Circles, what we might call "inner" (@srujana's original) and "outer".
(A third factor of my polynomial represents a point-circle at $(-pe/2,0)$. The final factor is a strange degree-six polynomial with an elaborate plot. These would seem to be extraneous "solutions" to the system, and may not have appeared if I had gone a different route in my $\theta$-elimination process.)
Of course, despite having revealed an Associate twin, this methodology is no more insightful a solution than with the coordinate approach. Indeed, it appears less insightful, since the final equations just happen to factor out of a polynomial mess. I wonder what's really going on ...