A Reader's Solution: The Locus of Midpoints of Fixed-Length Chords of an Ellipse
About half a year ago, I used the "circle-transformation method" to solve the problem of the locus of midpoints of fixed-length chords within an ellipse, and derived the equation of the locus. A few days ago, I received an email from a reader going by the name "理想" (Ideal), who proposed his own solution to this problem and arrived at an equation of a different form, which left him puzzled about whether the two were equivalent. After checking, I found that his locus equation is essentially equivalent to mine, though his always passes through the origin, which is a minor shortcoming. Still, his equation somehow looks tidier. This is genuinely surprising, because judging from his simplification process, there's a certain flavor of "simplifying into something more complicated" along the way, yet the final answer turns out to be quite neat — truly interesting.
With the reader's permission, I'm posting his derivation here to share with everyone! A PDF document is attached below, and you're welcome to download and read it. I hope to see more traces left behind by readers here on Scientific Spaces.
A Solution to the Locus of Midpoints of Fixed-Length Chords of an Ellipse
Author: 理想 (Ideal)
This article introduces a method for computing the locus of midpoints of fixed-length chords of an ellipse. Let the semi-major and semi-minor axes of the ellipse be $2a$ and $2b$ respectively, and let the chord length be $2r$. As the two endpoints of the chord slide along the ellipse, the locus traced out by the chord's midpoint is:
$$(\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1)(\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{r^2}{a^2b^2}) + \frac{r^2}{a^2b^2} = 0$$
It is not an ellipse, but rather a higher-order curve. more
Let the semi-major and semi-minor axes of the ellipse be $2a$ and $2b$ respectively, and let the chord length be $2r$. Let the two endpoints of the chord be $A(x_1, y_1)$ and $B(x_2, y_2)$, and let the chord's midpoint be $P(x, y)$. We have the following relations:
$$\begin{eqnarray*}x = \frac{x_1 + x_2}{2}\\ y = \frac{y_1 + y_2}{2}\end{eqnarray*}$$
Condition 1: Points $A(x_1, y_1)$ and $B(x_2, y_2)$ lie on the ellipse, satisfying the ellipse equation:
$$\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1\tag{1}$$
$$\frac{x_2^2}{a^2} + \frac{y_2^2}{b^2} = 1\tag{2}$$
Condition 2: the chord length is $|AB| = 2r$:
$$(x_1 - x_2)^2 + (y_1 - y_2)^2 = 4r^2\tag{3}$$
First key equation: adding the ellipse equations for points A and B: (1) + (2):
$$\begin{aligned}\frac{x_1^2 + x_2^2}{a^2} + \frac{y_1^2 + y_2^2}{b^2} = 2 \\ \frac{(x_1 + x_2)^2 + (x_1 - x_2)^2}{a^2} + \frac{(y_1 + y_2)^2 + (y_1 - y_2)^2}{b^2} = 4\end{aligned}\tag{4}$$
Let $x_1 - x_2 = 2w$, $y_1 - y_2 = 2h$, and combine with $x_1 + x_2 = 2x$, $y_1 + y_2 = 2y$; substituting into (4):
$$\begin{aligned}\frac{4x^2 + 4w^2}{a^2} + \frac{4y^2 + 4h^2}{b^2} = 4 \\ \frac{x^2 + w^2}{a^2} + \frac{y^2 + h^2}{b^2} = 1\end{aligned}\tag{5}$$
This gives the first key equation (5).
Second key equation: (1) − (2):
$$\begin{aligned}\frac{x_1^2 - x_2^2}{a^2} + \frac{y_1^2 - y_2^2}{b^2} = 0 \\ \frac{(x_1 + x_2)(x_1 - x_2)}{a^2} + \frac{(y_1 + y_2)(y_1 - y_2)}{b^2} = 0 \\ \frac{2x\cdot 2w}{a^2} + \frac{2y\cdot 2h}{b^2} = 0\end{aligned}$$
Rearranging terms and squaring both sides (to eliminate the sign):
$$\frac{x^2}{a^2}\cdot\frac{w^2}{a^2} = \frac{y^2}{b^2}\cdot\frac{h^2}{b^2}\tag{6}$$ This gives the second key equation (6).
Third key equation: let $2w = x_1 - x_2$, $2h = y_1 - y_2$, and substitute into equation (4):
$$w^2 + h^2 = r^2\tag{7}$$ This gives the third key equation (7).
Putting these together, the three key equations obtained are as follows:
$$\begin{eqnarray*}\frac{x^2 + w^2}{a^2} + \frac{y^2 + h^2}{b^2} = 1......(8) \\ \frac{x^2}{a^2}\cdot\frac{w^2}{a^2} =\frac{y^2}{b^2}\cdot\frac{h^2}{b^2} ......(9)\\ w^2 + h^2 =r^2......(10) \end{eqnarray*}$$
As long as we eliminate the terms $w^2$ and $h^2$ from these, we obtain a curve equation involving only $x^2$ and $y^2$. Here is one way to do it: let
$$\begin{eqnarray*}p = \frac{x}{a}......(11) \\q = \frac{y}{b}......(12) \\m = \frac{w}{a}......(13) \\n = \frac{h}{b}......(14)\end{eqnarray*}$$
The system of equations becomes
$$\begin{eqnarray*}p^2 + q^2 + m^2 + n^2 = 1......(15) \\p^2m^2 = q^2n^2......(16) \\a^2m^2 + b^2n^2 = r^2......(17) \end{eqnarray*}$$
From $(17) \times p^2 + (17) \times q^2$ we get:
$$a^2m^2p^2 + b^2n^2p^2 + a^2m^2q^2 + b^2n^2q^2 = p^2r^2 + q^2r^2\tag{18}$$ From (16), $p^2m^2 = q^2n^2$; substituting into the above equation (18),
$$\begin{aligned}a^2n^2q^2 + b^2n^2p^2 + a^2m^2q^2 + b^2m^2p^2 = p^2r^2 + q^2r^2 \\ (a^2q^2 + b^2p^2)(m^2 + n^2) = (p^2 + q^2)r^2\end{aligned}\tag{19}$$ $(15) \times (a^2q^2 + b^2p^2)$, we get:
$$\begin{aligned}(a^2q^2 + b^2p^2)(p^2 + q^2 + m^2 + n^2) = (a^2q^2 + b^2p^2) \\ (a^2q^2 + b^2p^2)(p^2 + q^2) + (a^2q^2 + b^2p^2)(m^2 + n^2) = (a^2q^2 + b^2p^2)\end{aligned}$$
Combining this with equation (19) to eliminate the $(m^2 + n^2)$ term gives:
$$\begin{aligned}(a^2q^2 + b^2p^2)(p^2 + q^2) + (p^2 + q^2)r^2 = (a^2q^2 + b^2p^2) \\ (p^2 + q^2)(a^2q^2 + b^2p^2 + r^2) = (a^2q^2 + b^2p^2 + r^2) - r^2 \\ (p^2 + q^2 - 1)(a^2q^2 + b^2p^2 + r^2) + r^2 = 0 \\ (p^2 + q^2 - 1)(\frac{p^2}{a^2} + \frac{q^2}{b^2} + \frac{r^2}{a^2b^2}) + \frac{r^2}{a^2b^2} = 0\end{aligned}$$
Substituting into
$$\begin{aligned}p = \frac{x}{a} \\ q = \frac{y}{b}\end{aligned}$$
we obtain the final result:
$$(\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1)(\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{r^2}{a^2b^2}) + \frac{r^2}{a^2b^2} = 0$$
Conclusion: The locus of midpoints of fixed-length chords of an ellipse is actually not that hard to solve — it's just that the resulting curve is not itself an ellipse, and even after solving for the functional equation, it's not easy to visualize the shape of the curve.
PDF Download: A Solution to the Locus of Midpoints of Fixed-Length Chords of an Ellipse.pdf
Translated automatically with claude-sonnet-5; all equations are reproduced verbatim from the source. Copyright remains with the original author.