an important property of an ellipse

Solution 1:

I present here a rather elaborate coordinate geometry proof. (I'd sure like to see a simpler way of going about this!)

Take the ellipse with eccentricity $\varepsilon\in(0,1)$ and semilatus rectum $p$ to have the polar equation

$$r=\frac{p}{1-\varepsilon\cos\,\theta}$$

such that one of the ellipse's foci is at the origin, and the circle whose diameter is the latus rectum has the equation $r=p$. From this configuration, we find that the associate circle corresponding to the focus at the origin has its center at $\left(\dfrac{p}{2}\left(1-\dfrac1{1+\varepsilon}\right),0\right)$ and a radius of $\dfrac{p}{2}\left(1+\dfrac1{1+\varepsilon}\right)$.

Let a focal chord corresponding to the focus at the origin be at an angle $\varphi$ from the horizontal axis. We find that the line $\theta=\varphi$ intersects the ellipse at the angle values $\theta=\varphi$ and $\theta=\varphi+\pi$, corresponding to the points

$$\left(\frac{\pm p\cos\,\varphi}{1\mp\varepsilon\cos\varphi},\frac{\pm p\sin\,\varphi}{1\mp\varepsilon\cos\,\varphi}\right)$$

(I thank Blue for noting this simplification; the previous version of this answer took a more circuitous route.)

The circle whose diameter is the segment joining these two points has its center at $\left(\dfrac{\varepsilon p\,\cos^2\varphi}{1-(\varepsilon\cos\,\varphi)^2},\dfrac{\varepsilon p\cos\,\varphi\sin\,\varphi}{1-(\varepsilon\cos\,\varphi)^2}\right)$ and a radius of $\dfrac{p}{1-(\varepsilon\cos\,\varphi)^2}$. To verify that this circle is tangent to the associate circle, we find the radical line of these two circles (which coincides with the common tangent line of two tangent circles); we find the equation of the radical line to be

$$p\left(\frac{2\varepsilon}{\sec^2\varphi-\varepsilon^2}+\frac1{1+\varepsilon}-1\right)x+\frac{2 \varepsilon p \tan\,\varphi}{\sec^2\varphi-\varepsilon^2}y+\frac{\varepsilon p^2 (\varepsilon+\sec^2\varphi)}{(1+\varepsilon)(\sec^2\varphi-\varepsilon^2)}=0$$

The point of tangency is then found to be

$$\left(\frac{p(\tan^2\varphi-\varepsilon-1)}{(\varepsilon+1)^2+\tan^2\varphi},-\frac{p(\varepsilon+2)\tan\,\varphi}{(\varepsilon+1)^2+\tan^2\varphi}\right)$$

I'll leave the verification that the line perpendicular to the radical line at the point of tangency passes through the center of the associate circle to you.

Here is a Mathematica demonstration:

With[{p = 1, e = 1/Sqrt[2], n = 31},
     Animate[PolarPlot[{p/(1 - e*Cos[t]), p}, {t, -Pi, Pi}, 
       Epilog -> {{Green,
                   Circle[{(p/2)*(1 - 1/(1 + e)), 0},
                          (p/2)*(1 + 1/(1 + e))]}, 
                  {Red,
                   Circle[{(e*p*Cos[th]^2)/(1 - (e*Cos[th])^2),
                           (e*p*Cos[th]*Sin[th])/(1 - (e*Cos[th])^2)},
                           p/(1 - (e*Cos[th])^2)],
                   Line[{{p*Cos[th]/(1 - e*Cos[th]),
                          p*Sin[th]/(1 - e*Cos[th])}, 
                         {-p*Cos[th]/(1 + e*Cos[th]),
                          -p*Sin[th]/(1 + e*Cos[th])}}]}},
       Frame -> True], {th, 0, 2 Pi, 2 Pi/(n - 1)}]]

Here's a parabolic version:

Some tweaking in the code given above is needed to handle the hyperbolic case e > 1; I'll leave this as an exercise to the reader.

Solution 2:

Here's a slightly different approach, which also uses coordinates, and which fails to enlighten as to why the phenomenon works.

Let $C = C(\theta)$ and $r = r(\theta)$ be the center and radius of the Focal Chord Circle (FCC) for which the focal chord makes angle $\theta$ with the $x$ axis. Let $K$ and $s$ be the center and radius of the Auxiliary Circle (AC).

Say that ray $CK$ meets the FCC at $T$; then, $|CT| = r$. If also $|KT| = s$, then $T$ must be a point of tangency for FCC and AC, so let's verify that $|KT| = s$ ... or, equivalently, that $|CK| = r-s$.

With the right focus at the origin, and major axis aligned with the $x$-axis, the endpoints of the focal chord have these Cartesian coordinates:

$$\begin{eqnarray*} \frac{\pm p}{1\mp e\cos\theta} (\cos\theta,\sin\theta) \end{eqnarray*}$$

where $p$ is the length of the semi-latus rectum and $e$ is the eccentricity of the conic. The point $C$ is the midpoint of the chord, having coordinates equal to the average of the endpoints' coordinates; $r$ is the half the distance between those endpoints:

$$\begin{eqnarray*} C = \frac{p e \cos\theta}{1-e^2\cos^2\theta}(\cos\theta,\sin\theta) \hspace{0.5in} r = \frac{p}{1-e^2\cos^2\theta} \end{eqnarray*}$$

The $x$-axis meets the Left Associate Circle at the point $(\frac{-p}{1+e},0)$ (where the LAC meets the conic) and at the point $(p,0)$ (where the LAC meets the Auxiliary Circle). Thus, the LAC has center and radius given by ...

$$\begin{eqnarray*} K = \left(\frac{p e}{2(1+e)}, 0\right) \hspace{0.5in} s = \frac{p(2+e)}{2(1+e)} \end{eqnarray*}$$

Now, simply compute ...

$$\begin{eqnarray*} |CK| &=& \frac{p e ( 1 + 2 e \cos^2\theta + e^2 \cos^2\theta )}{2(1+e)(1-e^2 \cos^2\theta)} \\\\ &=& \frac{p ( 2(1+e) - ( 2 + e )( 1 - e^2 \cos^2\theta ) )}{2(1+e)(1-e^2 \cos^2\theta)} \\\\ &=& \frac{p}{(1-e^2 \cos^2\theta)} - \frac{p ( 2 + e )}{2(1+e)} \\\\ &=& r - s \hspace{0.25in} QED \end{eqnarray*}$$

The argument would seem to apply to all conics, with this caveat: for $e > 1$, we restrict the domain of $\theta$ so that $|\cos\theta|\le 1/e$, which keeps the endpoints of the focal chord on the same branch of the hyperbola.

Solution 3:

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 ...