Dual of a parametric curve in affine chart

Question

Consider a curve $C \subset P^2$ with coordinates $(x_0:x_1:x_2)$, and let's look at this curve in affine chart $x_0 = 1$. Suppose in this chart the curve is parametric $$ (1:x(t):y(t)) $$ I want to find dual curve $C^*$ and how to prove from this that the double dual curve $C^{**}$ is equal to $C$? I found the formula of $C^*$ in wikipidea $$ \left( 1 : \dfrac{-y}{xy' - yx'} : \dfrac{x}{xy'-yx'} \right) $$ but it looks very strange to me and i didn't understand the proof in wikipidea at all, so i tryid to calculate it by myself. Isn't the tangent line to parametric curve $(1:x(t):y(t))$ given by $(1:x'(t):y'(t))$? So the dual curve should have coordinates $(1:x'(t):y'(t))$

Answer 1

Too long for a comment

From wikipedia:

Consider a parametrically defined curve $(x,y)=(x(t),y(t)),$ in projective coordinates $(x,y,z)=(x(t):y(t):1).$ Its projective tangent line is a linear plane spanned by the point of tangency and the tangent vector, with linear equation coefficients given by the cross product: $$(P:Q:R)=(x,y,1)\times (x',y',0)=(-y':x':xy'-yx').$$

Making $p=\frac{P}{R},q=\frac{Q}{R}.$

From Gelfand, Kapranov & Zelevinsky (p 17):

A local parametric equation of $X$ has the form $x = x(t), y = y(t),$ where $t$ is a local coordinate on $X,$ and $x(t), y(t)$ are analytic functions. By definition,The dual curve $X^{\vee}$ has the parametrization $p = p(t), q = q (t),$ where $p(t)x +q (t) y+ 1 = 0$ is the affine equation of the tangent line to $X$ at $(x(t), y(t)).$ Hence, the parametric representation of $X^{\vee}$ has the form $$p(t)= \dfrac{-y}{xy' - yx'} , q(t)= \dfrac{x}{xy'-yx'}$$ This formula readily implies biduality for plane curves. Indeed, applying it once again, we find that the parametric representation of $X^{\vee\vee}$ is $x = u(t), y = v(t),$ where $$u(t)= \dfrac{-q}{pq' - qp'} , v(t)= \dfrac{p}{pq'-qp'}$$ Substituting here the values of $p(t)$ and $q(t)$ from above, we find that $u(t) = x(t), v(t) = y(t),$ hence $X^{\vee\vee}=X$.

Maybe you can say what is unclear, so we can target answers to you?

Edit

$(x:y:1)$ is the point of tangency and $(x':y':0)$ is the tangent there given by differentiation, note that $1'=0.$ Here we use the screen $z=1$

in 3-space, remembering that points in the projective plane correspond to lines through the origin in 3-space, and a line in the projective plane corresponds to a 2-plane in 3-space. Now one way to get this plane through the origin spanned by two vectors you take their cross product. The components of the normal vector belong to the dual projective plane, of lines in the plane. Remember we have the universal line $au+bv+cw=0,$ which can be read both ways as a line in projective space and as a plane in 3-space. Here $a,b,c$ coordinates of the dual plane and $u,v,w$ coordinates of the plane with screens $c=1$ and $w=1$ making the universal line $au+bv+1=0,$ and we have let $a=p(t),b=q(t).$ Comparing the two $p(t)u+q(t)v+w=0$ and $-y'u+x'v+(xy'-yx')w=0$ (remember we set $w=1$) we get the result.