About six months ago, I used the “circle transformation method” to solve the problem of the midpoint locus of a constant-length chord in an ellipse and derived the trajectory equation. A few days ago, I received an email from a netizen named “Li Xiang,” who proposed his own solution to this problem and obtained a trajectory equation in a different form, expressing doubt about the equivalence of the two. After verification, his trajectory equation and mine are basically equivalent. However, the trajectory equation he derived always includes the origin, which is a minor drawback. Nevertheless, his trajectory equation looks somewhat more elegant. This is indeed surprising, because looking at his simplification process, it feels like “simplifying by complicating,” yet he arrived at a quite concise answer, which is truly interesting.
With the netizen’s consent, I am posting his process here to share with everyone! A PDF document is attached at the end for download. I hope to see more traces left by readers in Scientific Space.
A Solution for the Locus of the Midpoint of a Constant-Length Chord in an Ellipse
Author: Li Xiang
This article introduces a method for calculating the locus of the midpoint of a constant-length chord in an ellipse. Let the semi-major and semi-minor axes of the ellipse be a and b, respectively, and the chord length be 2r. As the two ends of the chord slide along the ellipse, the locus formed by the midpoint of the chord is: \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1\right)\left(\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{r^2}{a^2b^2}\right) + \frac{r^2}{a^2b^2} = 0 It is not an ellipse, but a higher-order curve.
Let the semi-major and semi-minor axes of the ellipse be a and b, and the chord length be 2r. Let the two ends of the chord be A(x_1, y_1) and B(x_2, y_2), and the midpoint of the chord be P(x, y). We have the following relationships: \begin{aligned} x &= \frac{x_1 + x_2}{2} \\ y &= \frac{y_1 + y_2}{2} \end{aligned}
Condition 1: Points A(x_1, y_1) and B(x_2, y_2) are 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: Chord length |AB| = 2r: (x_1 - x_2)^2 + (y_1 - y_2)^2 = 4r^2 \tag{3}
The first key equation: (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 and y_1 - y_2 = 2h. Combining this with x_1 + x_2 = 2x and y_1 + y_2 = 2y, and 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).
The 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 and squaring both sides (to eliminate the negative 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).
The third key equation: Let 2w = x_1 - x_2 and 2h = y_1 - y_2, and substitute into (3): w^2 + h^2 = r^2 \tag{7} This gives the third key equation (7).
In summary, the three key equations obtained are as follows: \begin{aligned} \frac{x^2 + w^2}{a^2} + \frac{y^2 + h^2}{b^2} &= 1 \tag{8} \\ \frac{x^2}{a^2} \cdot \frac{w^2}{a^2} &= \frac{y^2}{b^2} \cdot \frac{h^2}{b^2} \tag{9} \\ w^2 + h^2 &= r^2 \tag{10} \end{aligned}
By eliminating the w^2 and h^2 terms, we can obtain the curve equation containing only x^2 and y^2. Here is one solution: Let \begin{aligned} p &= \frac{x}{a} \tag{11} \\ q &= \frac{y}{b} \tag{12} \\ m &= \frac{w}{a} \tag{13} \\ n &= \frac{h}{b} \tag{14} \end{aligned} The system of equations becomes: \begin{aligned} p^2 + q^2 + m^2 + n^2 &= 1 \tag{15} \\ p^2m^2 &= q^2n^2 \tag{16} \\ a^2m^2 + b^2n^2 &= r^2 \tag{17} \end{aligned}
(17) \times p^2 + (17) \times q^2 gives: 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 this into (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) gives: \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 with equation (19) to eliminate the (m^2 + n^2) term: \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)\left(\frac{p^2}{a^2} + \frac{q^2}{b^2} + \frac{r^2}{a^2b^2}\right) + \frac{r^2}{a^2b^2} &= 0 \end{aligned} Substituting p = \frac{x}{a} and q = \frac{y}{b} back into the equation: \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1\right)\left(\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{r^2}{a^2b^2}\right) + \frac{r^2}{a^2b^2} = 0
Conclusion: The locus of the midpoint of a constant-length chord in an ellipse is actually not difficult to solve, but it is not an elliptical curve. Even if the functional equation is solved, it is not easy to see the shape of the curve.
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