# What is the path equation that is created with the middle point of a fixed length line segment that touching both ends to an ellipse.

Ellipse equation is $(\frac{x}{a})^2+(\frac{y}{b})^2=1$ and the length of line segment is $2k$, if we move the line segment all around of the ellipse while touching both ends to the ellipse. What is the path equation of middle point of the line segment that moved on ellipse?

My sense says that the path equation is an ellipse but need to proof it and need to find what the middle point path equation is.

Note: If a=b then the ellipse turns to a circle and the middle point path equation can be shown easily that ${x}^2+{y}^2=a^2-{k^2}$

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Parameterize the ellipse as $(a \cos \bullet, b \sin \bullet)$. Take points $P := ( a \cos p, b \sin p)$ and $Q := (a \cos q, b \sin q)$ on the ellipse, with midpoint $M := (P+Q)/2$.

If $|PQ| = 2k$, then

$$a^2 (\cos p - \cos q)^2 + b^2 (\sin p - \sin q)^2 = 4k^2$$

The coordinates of $M$ are $$\begin{eqnarray*}x &=& \frac{a}{2}(\cos p + \cos q)\\ y &=& \frac{b}{2}(\sin p + \sin q) \end{eqnarray*}$$

Eliminating $\cos q$ and $\sin q$ (say, via the method of resultants) from these coordinate expressions based on the $k^2$ equation gives (barring transcription errors) the following implicit formulas for $x$ and $y$:

$$\begin{eqnarray*} 0 = x^4 ( a^2 - b^2 )^2 &-& 2 x^3 a ( a^2 - b^2 )( 2 a^2 - b^2 ) \\ &+& x^2 a^2 \left( a^2 b^2 - 2 ( a^2 - b^2 ) k^2 + (a^2-b^2)(6a^2-b^2)\cos^2 p \right) \\ &-& 2 x a^3 \cos p \left( a^2 b^2 + 2 a^2 ( a^2 - b^2 ) \cos^2 p - k^2 ( 2 a^2 - b^2 )\right) \\ &+& a^4 \left( a^2 \cos^2 p - k^2 \right)\left( b^2 - k^2 + ( a^2 - b^2 )\cos^2 p \right) \\ 0 = y^4 ( a^2 - b^2 )^2 &-& 2 y^3 b ( a^2 - b^2 )( a^2 - 2b^2 ) \\ &+& y^2 b^2 \left( a^2 b^2 + 2 ( a^2 - b^2 ) k^2 + (a^2-b^2)(a^2-6b^2)\sin^2 p \right) \\ &+& 2 y b^3 \sin p \left( a^2 b^2 - 2 b^2 ( a^2 - b^2)\sin^2 p + k^2 ( a^2 - 2 b^2 )\right) \\ &+& b^4 \left( b^2 \sin^2 p - k^2 \right)\left( a^2 - k^2 - ( a^2 - b^2 )\sin^2 p \right) \end{eqnarray*}$$

Eliminating the trigonometric quantities from these equations yields a monster polynomial in $x$ and $y$. Assuming that the 12-thousand-term factor is extraneous, we get this implicit formula for the midpoint curve:

$$b^6 x^4 + a^6 y^4 + x^2 y^2 a^2 b^2 (a^2 + b^2 ) - a^2 b^4 x^2( b^2 - k^2 ) - a^4 b^2 y^2( a^2 - k^2) = 0$$

This certainly does not appear to be the template for an ellipse.

Here's the polar form, with $x = r \cos\theta$ and $y = r\sin\theta$, after factoring-out (and ignoring) an $r^2$ that gives an extraneous origin-point:

$$r^2 = \frac{a^2 b^2 \left( a^2( a^2 - k^2 )\sin^2\theta + b^2( b^2 - k^2 )\cos^2\theta \right)}{\left(a^2\sin^2\theta+b^2\cos^2\theta\right)\left(a^4\sin^2\theta+b^4\cos^2\theta\right)}=\frac{a^2b^2}{a^2\sin^2\theta+b^2\cos^2\theta}-\frac{a^2b^2k^2}{a^4\sin^2\theta+b^4\cos^2\theta}$$

(In the second equality, note that the first term is the formula for the original ellipse, whereas the second term could be considered the formula for an ellipse with semi-major axis $ak/b$ and semi-minor axis $bk/a$; the major-minor ratio of this auxiliary ellipse is the square of that of the original.)

We can see here that the curve is defined when $r^2 \geq 0$, so that we must have $$k^2 \leq \frac{a^4 \sin^2\theta + b^4 \cos^2\theta}{a^2\sin^2\theta+b^2\cos^2\theta}$$

Observe that the special case $a = b \neq 0$ (with $k\leq a$) reduces to a circle: $$x^2 + y^2 = a^2 - k^2 \hspace{1in} r^2 = a^2 - k^2$$

... while the special case $k=0$ reduces to the original ellipse: $$b^2 x^2 + a^2 y^2 = a^2 b^2 \hspace{1in} r^2 = \frac{a^2 b^2}{a^2 \sin^2\theta + b^2 \cos^2\theta}$$

Here's an image of various curves for the ellipse with $a = 5$ and $b = 2$, where $k$ takes values from $0$ (matching the ellipse) to $5$ (just the origin) in increments of $0.5$. The fattest "figure-8" touching the origin is for $k = 2 = b$.

Note. (I should make an animation to illustrate this.) Imagine $P$ is at $(5,0)$ and $Q$ is somewhere counter-clockwise around the ellipse, making a chord of length $2k = 2b$. As $P$ moves counter-clockwise itself, it pushes $Q$ further left, until $Q$ reaches $(-5,0)$, with $P$ somewhat to the left of the minor axis. Eventually, we expect $Q$ to wind up to the right of $P$, but this can't happen in a continuous way with $P$ traveling ever-counter-clockwise: the $PQ$ chord would need to pass through being vertical, but the only vertical chord of length $2b$ is the minor axis, whereas $P$ has already passed that axis. On the other hand, if we think of $Q$ as dragging $P$ behind it, then as $Q$ rounds the left bend, we find that it ends up pushing $P$ back a bit until the chord aligns with the minor axis in a continuous way (putting $M$ at the center, and pinching off a lobe of our figure-8). This is all well and good, but as $Q$ continues counter-clockwise, it tugs $P$ to the right; as $Q$ rounds the right bend, we find that it's pushing $P$ in the manner that $P$ was pushing $Q$ in the first scenario. Thus, we can't continuously plot the midpoint curve in this case by moving either endpoint of the chord in single direction around the ellipse. It would appear that the same can be said for any $k$ that causes its curve to bend inward at the left and right ends (as is already happening with $k=1$ in the figure); I wonder what the threshold on $k$ is for this phenomenon. (It should be easy enough to figure out: find the smallest $k$ such that the curve has a vertical tangent line at a point with $y\neq 0$.)

Edit. Let's tackle the bend threshold.

Differentiating the curve equation gives this differential relation: $$\begin{eqnarray*} 0 &=& b^2 x \left( 2 b^4 x^2 + a^2 ( a^2 + b^2 ) y^2 - a^2 b^2 ( b^2 - k^2 ) \right) \; \mathrm{d}x \\ &+& a^2 y \left( 2 a^4 y^2 + b^2 ( a^2 + b^2 ) x^2 - a^2 b^2 ( a^2 - k^2 ) \right) \; \mathrm{d}y \end{eqnarray*}$$ We'll have a vertical tangent line when $\mathrm{d}x = 0$. We're interested in the cases where $y \neq 0$, so we consider $2 a^4 y^2 + b^2 ( a^2 + b^2 ) x^2 - a^2 b^2 ( a^2 - k^2 ) = 0$. Using this to eliminate $x$ from the curve equation gives the following possibilities for $y$: $$(1)\qquad y^2 = -\frac{b^2(a+k)(ak+b^2)}{a(a^2-b^2)} \hspace{1in} (2) \qquad y^2=\frac{b^2(a-k)(ak-b^2)}{a(a^2-b^2)}$$ The possibilities in (1) are no possibilities at all, since (with $a > b$) the right-hand side is always negative; in (2), clearly requiring $a>k$, we find that the right-hand side becomes non-negative ---hence, that we get our additional vertical tangents--- once $k$ becomes larger than $b^2/a$, the length of the ellipse's semi-latus rectum! (In the figure, that's $2^2/5 = 0.8$, consistent with the visible bend for the $k=1$ curve and lack of bend in the $k=0.5$ curve.)

(By the way, when $k=b^2/a$, the "auxiliary ellipse" ---mentioned after the polar representation of the curve equation--- has semi-major axis $b$.)

So, what we've learned here is that a latus-rectum-length (or shorter, but not longer) chord can be pushed continuously in one direction around the circumference of an ellipse. (Of course, the latus-rectum-length chord will go vertical when it aligns with an actual latus rectum; that is, at each of the ellipse's foci.) Neat!

I suspect that we'd get similar results for hyperbolas and parabolas. I also suspect that this is a known property of conics, which means it might have a cleaner derivation than the one I've given here.

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Thank you very much for detailed answer. Really great. Also It can be wonderful to see an animation that makes clear view of the path for everybody. When I start to solve the problem as first approach, I took very small length of line segment in draw and it looked like as ellipse to me but now I can see that if we take length of line as 2b and while a>b we can see easily that the path passes from orgin. Again thanks for clear steps and great explanations. Do you know what the name of such paths is? –  Mathlover Mar 8 '12 at 10:14
@Mathlover Thanks for the thanks. I may not get around to the animation, but I may revisit the "threshold on $k$" issue mentioned at the revised end of my answer; it's the place where the paths stop looking deceptively ellipse-like and start exhibiting inward bends. By the way, I don't know that these paths have a particular name. –  Blue Mar 10 '12 at 10:49
@DayLateDon: +1. But how did you eliminate the cos and sin terms from the two equations? It must have been an arduous task! –  user17762 Mar 11 '12 at 4:16
@Sivaram: I just threw Mathematica's Resultant[] at the problem a few times. (I've gotten lazy. :) Specifically ... Writing my answer's first three equations in "$=0$" form, let $P_k$, $P_x$, and $P_y$ be the "polynomials" on the non-"0" sides of those equations. Likewise, let $P_p$ be the "polynomial" "$-1+\cos^2 p+\sin^2 p$" and $P_q$ be "$-1+\cos^2 q+\sin^2 q$". Then, use Resultant[] and $P_q$ and $P_p$ to eliminate sines from everything; then I use Resultant[] to eliminate the cosines. (Alternatively, I could express sines and cosines as complex exponentials, and eliminate those.) –  Blue Mar 11 '12 at 5:00
@DayLateDon: Nice. Thanks! –  user17762 Mar 11 '12 at 5:02