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I am studying projective geometry and am stuck on understanding the following:

For a parametric curve

$$ x = x(t), y = y(t) $$

the dual curve is given by

$$ X=\frac{-y′}{xy′-yx'}, Y=\frac{x′}{xy′−yx′} $$

This is explained on Wikipedia, in Chapter 1, Section 2 of Discriminants, Resultants, and Multidimensional Determinants by Gelfand, Kapranov, and Zelevinsky (referred by @Jan-Magnus-Økland; see this question), and in this document.

I feel like something so simple and fundamental is escaping me...how was this derived? Is Cramer's Rule involved?

adam.hendry
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  • There's a subtlety here: take $x^2+y^2-1=0$ and a parametrization, say, $x=\frac{t^2-1}{t^2+1},y=\frac{2t}{t^2+1}.$ Then using the above formulas you get $X=-\frac{t^2-1}{t^2+1},Y=-\frac{2t}{t^2+1}$ and line corresponding becomes $xX+yY-1=0,$ while via the matrix formulation the circle is self-dual and for $X^2+Y^2-1=0$ a corresponding line is $xX+yY+1=0.$ – Jan-Magnus Økland Mar 12 '21 at 06:27
  • @Jan-MagnusØkland Are the two lines $xX+yY-1=0$ and $xX+yY+1=0$ equivalent because we could multiply one or the other by $-1$ or $1$ since the coordinates are homogeneous? – adam.hendry Mar 13 '21 at 18:52
  • It's true the line comes from xX+yY+zZ=0 homogeneous, maybe z=1 and Z=-1? Still it's baffling that the two methods don't agree. – Jan-Magnus Økland Mar 13 '21 at 19:21
  • Sorry for the confusion which stems from wikipedia, For the circle (and ellipses centered at the origin) both $xX+yY-1=0$ and $xX+yY+1=0$ are tangents, but for general conics $xX+yY+1=0$ is the correct tangent. – Jan-Magnus Økland Mar 14 '21 at 09:36
  • @Jan-MagnusØkland Not at all. Thank you for looking into it! If the methods weren't consistent/didn't agree, I certainly wouldn't want to use one or the other. Your appreciation for mathematics is readily apparent and very valuable :). – adam.hendry Mar 14 '21 at 16:29

1 Answers1

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Let

$$ Xx + Yy + 1 = 0 \tag{1} $$

be the equation of a line in line coordinates. A point $(x)$ and line $[X]$ (in the notation of Coxeter) are said to be incident in projective geometry if their inner product is zero:

$$ \{xX\} = 0. $$

The tangent to our parametric equation at point $t_0$ is given by $<x'(t_0),y'(t_0)>$. Hence, the point $(x')$ is incident on our line if

$$ \{x'X\} = 0. $$

That is,

$$ Xx' + Yy' = 0 \tag{2} $$

Thus, we have two equations in two unknowns ($(1)$ and $(2)$), which via Cramer's rule yield

$$ X=\frac{-y'}{xy'-yx'}, Y=\frac{x'}{xy'-yx'}. $$

Q.E.D.

adam.hendry
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