I saw this answer for a question about finding an ellipse from four points in general position; its centre and three points on its circumference, and wrote an edit with the following changes:
- Define $C_{12}$ as twice its value in the original version; this adds two $/2$'s in its use in the initial matrix representation, but removes $21$ multiplications by $2$ elsewhere
- restate the formulas for $C_{11},C_{22},C_{12}$ coefficients as a vector equality, to avoid duplicating division by $|M|$, and rewrite the numerators; they were arranged haphazardly previously (like the raw output of a CAS) rather than as a sum over cyclic index changes. From$$\begin{array}{ccl} C_{11}&=&\frac{y_2^22x_3y_3-2x_2y_2y_3^2+2x_1y_1y_3^2-y_1^22x_3y_3+y_1^22x_2y_2-2x_1y_1y_2^2}{|M|}\\C_{22}&=&\frac{2x_2y_2x_3^2-x_2^22x_3y_3+x_1^22x_3y_3-2x_1y_1x_3^2+2x_1y_1x_2^2-x_1^22x_2y_2}{|M|}\\C_{12}&=&\frac{x_2^2y_3^2-y_2^2x_3^2+y_1^2x_3^2-x_1^2y_3^2+x_1^2y_2^2-y_1^2x_2^2}{|M|} \end{array}$$to$$\begin{pmatrix}C_{11}\\C_{22}\\C_{12}\end{pmatrix}=\frac{\begin{pmatrix}x_1y_1(y_3^2-y_2^2)&+&x_2y_2(y_1^2-y_3^2)&+&x_3y_3(y_2^2-y_1^2)\\x_1y_1(x_2^2-x_3^2)&+&x_2y_2(x_3^2-x_1^2)&+&x_3y_3(x_1^2-x_2^2)\\x_1^2(y_2^2-y_3^2)&+&x_2^2(y_3^2-y_1^2)&+&x_3^2(y_1^2-y_2^2)\end{pmatrix}}{|M|}$$
- Factorise $|M|$, from $$x_1^2(y_2^22x_3y_3-2x_2y_2y_3^2)+y_1^2(2x_2y_2x_3^2-x_2^22x_3y_3)+2x_1y_1(x_2^2y_3^2-y_2^2x_3^2)$$ to $$(x_1y_2-x_2y_1)(x_2y_3-x_3y_2)(x_3y_1-x_1y_3)$$
- Change ellipse characterisation in terms of polar parameters, from $$\left(\frac{(b^2-a^2)\cos^2\theta}{a^2b^2}+\frac{1}{b^2}\right)x^2+\left(\frac{(a^2-b^2)\cos^2\theta}{a^2b^2}+\frac{1}{a^2}\right)y^2+\frac{2(a^2-b^2)\sin\theta\cos\theta}{a^2b^2}xy-1=0$$to$$\begin{aligned}0&=\left(\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\right)x^2+\left(\frac{\cos^2\theta}{b^2}+\frac{\sin^2\theta}{a^2}\right)y^2+\left(\frac1{b^2}-\frac1{a^2}\right)\sin(2\theta)xy-1\\&=\left(\left(\frac1{a^2}-\frac1{b^2}\right)\cos^2\theta+\frac{1}{b^2}\right)x^2+\left(\left(\frac1{b^2}-\frac1{a^2}\right)\cos^2\theta+\frac{1}{a^2}\right)y^2+\left(\frac1{b^2}-\frac1{a^2}\right)\sin(2\theta)xy-1\end{aligned}$$I kept the original (minorly changed), because it's computationally more efficient if less straightforward to parse, though doing so felt somewhat verbose.
However, it was rejected by users amWhy and Another User, both using the default message,
The edit does not improve the quality of the post. Changes to the content are unnecessary or make the post more confusing.
which isn't specific as to which changes were objectionable. Each of these seems as though it reduces confusion, and I would hope they'd at least tell me how to salvage it from whatever didn't. Could anyone less invested explain in more detail?
I noticed a later answer used an equivalent matrix definition to the original version of this, is it a standard convention?
Also, unrelatedly, four other answers seem as though they should be deleted.
- one answer on 2013-03-23 effectively restates the problem,
- another on 2013-03-23 notes the aforementioned condition of 'in general position,' which could be merged into the question,
- a third on 2015-11-08 gives a diagram (where two of the three specified circumference points are antipodal, so implied by each other; an infinitude satisfy this) and link to a prewritten answer to an entirely different question,
- a fourth on 2015-11-09 (by the author of the third) states only two circumference points $B,C$ are necessary to find an ellipse (as $\left\{\begin{pmatrix}B\sin t\\C\cos t\end{pmatrix}: t\in[0,2\pi)\right\}$), which again fails to uniquely characterse it.