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In indical notation, we can write equation of conic as:

$$ s_{ij} = Ax_i x_j + B(x_i y_j + x_j y_i) + C y_i y_j + F(x_i + x_j) + G(y_i + y_j) + H=0$$

Where $i$ and $j$ stands in for $(x_i, y_i)$ and $(x_j,y_j)$. Now my question is there, an algorithim for finding a $k$ such that,

$$ s_{ij} = s_{kk}$$

Like plugging in a pair of points is equal to a plugging a single point into the equation?

Refer this

With blue's comment, I now realized, the correct question to ask is if there is a "natural point" which we may take.

tryst with freedom
  • 11,538
  • Take $y_k$ to be (almost) anything you like, and solve the resulting quadratic for $x_k$. The "almost" accounts for wanting the discriminant to be non-negative, but the point is that $(x_k,y_k)$ is typically not unique. A reasonable question to ask is: Is there a "natural" choice for $(x_k,y_k)$? One candidate is the pole of the line through the given pair of points, but I don't know if this works. – Blue Jan 10 '21 at 21:52
  • That's something new to learn :P – tryst with freedom Jan 10 '21 at 21:54
  • Of course, the locus of $(x_k,y_k)$ is simply another conic, just with a different $H$ coefficient. Perhaps a "natural" point is one on this conic whose tangent is parallel to the line between the given points. – Blue Jan 11 '21 at 00:27

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