Is the "equidistant curve" to an algebraic curve algebraic?

Question

Let $ L \subseteq \mathbb{R}^2 $ be a smooth real algebraic curve. Let's fix some parameter $ \delta \in \mathbb{R} $ and for every point $ (x,y) \in L $ define $$ L_{\delta}(x,y) = (x,y) + \delta n(x,y), $$ where $ n(x,y) $ is a normal vector to $L$ at $(x,y)$. The equidistant curve is then $$ L_\delta = \{L_\delta (x,y) : (x,y) \in L\}. $$

The idea is very simple: we take a normal vector with length $ \delta $ and "roll" it along the curve. Note that this is not the same as the set of points at the same distance $\delta$ from $L$.

I am trying to find the equation of $L$. But I do not even understand if it is algebraic. For example, let's consider the parabola $ L$ given by $y - x^2 = 0 $. It is easy to find a parametrization of $ L_\delta $: $$ x = t - \delta \frac{2t}{\sqrt{4t^2+1}}, $$ $$ y = t^2 + \delta \frac{1}{\sqrt{4t^2+1}}. $$ But how do I find the equation oh $x,y$ from this one? I looked up some literature about computational algebraic geometry. As I understood, there are only theorems for the case when the parametrization is rational in the variable $t$, which is not the case here. $ \mathbb{R} $ is not algebraicaly closed, maybe this is also an issue.

For me the question is interesting when $L$ is an algebraic surface in $ \mathbb{R}^n $, but the case of curves on the plane looks so simple that I am sure somebody has already solved it. Any help would be appreciated!

Answer 1

Yes, $L_\delta$ is algebraic. You can find its equations by elimination theory as follows: Let $L$ be defined by the polynomial equation $F(x,y) = 0$. Now consider the polynomial equations $$ F(x,y)=(u{-}x)-aF_x(x,y)= (v{-}y)-aF_y(x,y)= a^2(F_x(x,y)^2+F_y(x,y)^2)-\delta^2=0. $$ for (x,y,a,u,v). This is 4 equations for 5 unknowns. You can now use elimination theory to find a polynomial equation $G(u,v)=0$ in the ideal of the above system of equations that does not involve $x$, $y$, or $a$. This will be the algebraic equation of the parallel curve you want.

In your example of $y-x^2$, you find, for example, that the curve has degree $6$ and has the equation $$ \begin{aligned} 0=&16\,{u}^{6}+16\,{u}^{4}{v}^{2}-40\,{u}^{4}v + \left( -48\,{b}^{2}+1 \right) {u}^{4}\\ &\quad-32\,{u}^{2}{v}^{3}+ \left( -32\,{b}^{2}+32 \right) {u }^{2}{v}^{2}+ \left( 8\,{b}^{2}-2 \right) {u}^{2}v+ \left( 48\,{b}^{4} -20\,{b}^{2} \right) {u}^{2}\\ &\quad+16\,{v}^{4}+ \left( -32\,{b}^{2}-8 \right) {v}^{3}+ \left( 16\,{b}^{4}-8\,{b}^{2}+1 \right) {v}^{2}+ \left( 32\,{b}^{4}+8\,{b}^{2} \right) v\\ &\quad-16\,{b}^{6}-8\,{b}^{4}-{b}^{2 } \end{aligned} $$ where, to save typing, I have writteen $b$ instead of $\delta$.

Remark: If you apply this method to Willie Wong's 'counterexample', you get the equation $$ 0 = \bigl((u-v)^2-2b^2\bigr)\bigl((u+v)^2-2b^2\bigr), $$ i.e., the equation of the 4 lines $u\pm v = \pm b\sqrt2$, as expected.