Why does this method of finding the equation of the tangent line to a second-degree curve work?

Question

Given any second degree curve with the general equation $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ and a point $(x', y')$ on it, if I transform the curve equation as follows \begin{align} (1) &&x^2 &\to& xx'\\ (2) &&y^2&\to& yy'\\ (3) &&2xy &\to& xy' + x'y\\ (4) &&2x &\to& x + x'\\ (5) &&2y &\to& y + y'\\ \end{align} I get the equation of the tangent to the curve at the point $(x', y')$ or if the point does not lie on the curve, I get the equation of the chord of contact (which is more general). My question is why does it work, like is there something deeper going on and are there similar transformations for higher degree curves?

Answer 1

This is linked to the duality between lines and points in the projective plane $\mathbf P_2$. You can take look at Pole and Polar on Wikipedia.

I'll add that a number of properties in projective geometry can be proved by a so-called ‘transformation by reciprocal polars’, considering the projective coordinates $[x:y:t]$ of a point are the projective coordinates of a projective line, and turning a problem about points into a problem about lines and vice-versa. For instance, to prove three points are aligned, you'll show the corresponding dual lines of these points have a common point.