Let $$A = \left\{ (x, y) \in \mathbb{R}^2 : y \geq x^2 +c, \, c \in \mathbb{R} \right\}$$ Find the polar set for $A$.
My attempts: $(u, v) \in \mathbb{R}^2; l(x, y) := ux+vy, \partial A: t \rightarrow (x(t), y(t)) = ( ct, \sqrt{c}t).$ I'm trying to find max of $f(t) = l(x(t), y(t)) = uct+v\sqrt{c}t$. But it is $\infty$. Help me please!
The boundary of the polar set $A^\circ=\{b\in\mathbb{R}^2: b^\top a\leq 1 \forall a\in A\}$ belongs to the conic section given by the equation $x^2 -4 c y^2 + 4 y=0$. This will give you an ellipse for $c<0$, a parabola for $c=0$, and a hyperbola for $c>0$.
For fixed non-zero $a\in A$, the set $\{b\in\mathbb{R}^2: b^\top a\leq 1\}$ is a half-plane that contains $A^\circ$. For convexity reasons, one always has $X^\circ=\mathrm{conv}(X)^\circ$ if $X$ is an arbitrary set and $\mathrm{conv}$ refers to taking the convex hull.
$$A^\circ =\bigcap_{a\in A}\{b\in\mathbb{R}^2: b^\top a\leq 1\}=\bigcap_{a\in \mathrm{ext}(A)}\{b\in\mathbb{R}^2: b^\top a\leq 1\}$$
where $\mathrm{ext}(A)$ is the set of extreme points of $A$, i.e., the smallest set $X$ for which $\mathrm{conv}(X)=A$. In our case $\mathrm{ext}(A)=\{a=(a_1,a_2)\in\mathbb{R}^2:a_2=a_1^2+c\}$ is the entire boundary of the parabola $A$.
For $a\in\mathrm{ext}(A)$, the half-space $\{b\in\mathbb{R}^2: b^\top a\leq 1\}$ supports $A^\circ$, i.e., $A^\circ \subset \{b\in\mathbb{R}^2: b^\top a\leq 1\}$ but $A^\circ\not\subset\{b\in\mathbb{R}^2: b^\top a< 1\}$. In other words, the bounding straight line $\{b\in\mathbb{R}^2: b^\top a= 1\}$ of the half-plane $\{b\in\mathbb{R}^2: b^\top a\leq 1\}$ will be a tangent of (the boundary curve of) $A^\circ$.
Computing a parametrized curve from its tangents means that you compute the envelope of a one-parameter set of straight lines. Here, the family of straight lines is given by the implicit equation $F(t,x,y)=0$, where $F:\mathbb{R}^3\to\mathbb{R}$, $F(t,x,y)=tx+(t^2+c)y-1$. (Take $b=(x,y)$ and $a=(t,t^2+c)\in\mathrm{ext}(A)$, then $F(t,x,y)=0$ means $b^\top a=1$. For computing the envelope, we need to also have $0=\frac{\partial}{\partial t}F(t,x,y)=2ty+x$.
Solving $0=2ty+x$ for $x$ gives $x=-2ty$. Plug this into $tx+(t^2+c)y-1=0$ and solving for $y$ yields $y=\frac{1}{c-t^2}$. Plugging this back into $x=-2ty$, we obtain the parametrization $(x(t),y(t))=(\frac{-2t}{c-t^2},\frac{1}{c-t^2})$.
To get a parameter-free equation, we start with $x=-2ty$. Square both sides and get $x^2=4t^2y^2$. From our parameterization, we know $t^2=c-\frac{1}{y}$. We obtain $x^2=4(c-\frac{1}{y})y^2=4cy^2-4y$.
Remark 1: In case $c>0$, the conic section is a hyperbola but the polar set is just the convex hull of the branch below the $x$-axis. (Recall that the polar set is defined as an intersection of half-planes, not just by its tangents.)
Remark 2: In case $c\geq 0$, we cannot have $t=0$ in the parameterization. This is where the point where the tangent touches $A^\circ$ jumps from one arm of the parabola to the other ("at infinity", $c=0$), or from one branch of the hyperbola to the other ($c<0$).
