# an important property of an ellipse

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?

-
Hello, Please answer my question.... – srujana Oct 27 '11 at 10:00
Yoda not far. Patience. Soon you will be with him. – joriki Oct 27 '11 at 10:36
Related... – user13838 Oct 27 '11 at 10:37

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.

-
JM is it just at my end or does your animation have a problem? I see the lower part jump around with results like this – t.b. Oct 27 '11 at 10:45
It seems to be running smoothly here, @t.b. I'll wait for somebody to report if s/he has the same problem as you, and then I'll investigate... – J. M. Oct 27 '11 at 10:47
For what it's worth, the GIF shows up fine on my end (in both Firefox and Chrome on Linux). – Rahul Oct 27 '11 at 13:02
@J.M. for me it looks perfect now. – t.b. Oct 27 '11 at 13:29
Back to the math ... Since you start with the polar equation for the ellipse with focus at origin, there's no need to take a side-track through $\tan m = \theta$ to find the points where the focal chord intersects the ellipse. For a given $\theta$, the polar coordinates of the points are simply $(\theta, r(\theta))$ and $(\pi+\theta, r(\pi+\theta))$; hence, in Cartesian coordinates, $\frac{\pm p}{1\mp\epsilon \cos\theta}\left(\cos\theta,\sin\theta\right)$. – Blue Oct 29 '11 at 11:54

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.

-

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 ...

-
Neat! $\phantom{}$ – J. M. Nov 20 '11 at 9:34