Determining the polar of the following set: We were given this problem, but I'm confused as to how approach it;
$$A = {x=(x_1,x_2) \in \Bbb{R}^2: (x_1-1)^2+x_2^2 \leq 1}$$
All our professor has given us so far is the definition of a polar of a set. I'm familiar with separation theorems and extreme points as well. We worked a similar example in class, but instead of computing it directly, we used the fact that an ellipsoid is the image of the Euclidean ball under an affine transformation - which I don't see how I would apply here. 
 A: Define $$A^{o}=\{y \ | \ \langle x,y \rangle \le 1 \text{ for all } x \in A\}$$
Then $$A^{\star}=A^{o}\cap (-A^{o})$$
If $A$ is symmetric with respect to $0$ then $A^{o}=-A^{o}$. 
We will determine first $A^{o}$. If $(u,v)$ is on the boundary of $A^{o}$ then 
the function $(x,y) \mapsto u x + v y$ has a maximum $1$ on the region $A$. Let $(x_0, y_0)$ a point of maximum. Then the plane in the $(x,y)$ domain 
$u x + v y = 1$ is a supporting plane for $A$. If the boundary of $A$ is defined by $f(x,y)=$ const then the gradient of $f$ at $(x_0, y_0)$ is proportional to $(u,v)$. Therfore we have the system
$$u x_0 + v y_0 =1\\
u \frac{\partial f(x_0, y_0)}{\partial y} - v\frac{\partial f(x_0, y_0)}{\partial x}=0\\
f(x_0, y_0) = 0$$
Eliminate  $x_0$, $y_0$ from the system above to get an equation in $u$, $v$. This works in general if $f(x,y)$ is a an algebraic function. 
(If $f$ is transcendental then in principle one could approximate $A$ from below and above with semialgebraic sets to get approximations for $A^{o}$).
For instance, in our case the system is 
$$u x + v y -1=0\\
u y - v(x-1) = 0\\
(x-1)^2 + y^2-1=0$$
We get $2 u + v^2  -1 =0$. Therefore, 
$$A^{o}= \{(x,y)\ | \ x\le \frac{1-y^2}{2} \}$$
and so 
$$-A^{o}=\{(x,y)\ | \ x\ge \frac{y^2-1}{2} \}$$
and finally
$$A^{\star}=\{(x,y)\ |  -\frac{1-y^2}{2}\le x \le \frac{1-y^2}{2} \}$$
See the plot  with Wolfram Alpha. 

Note also that if in the above system we use $(ux+vy)^2-1=0$ we get directly the equation for the boundary of $A^{\star}$. 
A: Let $(u,v)\in{\mathbb R}^2$ be given, and define $\ell(x,y):=ux+vy$. The point $(u,v)$ belongs to the polar set $A^\circ$ of the compact disc $A:=\bigl\{(x,y)\,\bigm|\,(x-1)^2+y^2\leq1\bigr\}$ iff
$$\max\bigl\{\ell(x,y)\,\bigm|\,(x,y)\in A\bigr\}\leq 1\ .\tag{1}$$
The linear function $\ell$ has no critical points. It therefore takes its maximum on the boundary
$$\partial A:\quad t\mapsto\bigl(x(t),y(t)\bigr):=(1+\cos t,\sin t)\qquad(0\leq t\leq 2\pi)\ .$$
The pullback
$$f(t):=\ell\bigl(x(t),y(t)\bigr)=u+u\cos t+v\sin t$$
has maximum value $u+\sqrt{u^2+v^2}$, which is then the maximum considered in $(1)$. It follows that
$$(u,v)\in A^\circ\quad\Longleftrightarrow\quad u+\sqrt{u^2+v^2}\leq 1\ .$$
The condition on the RHS enforces $u\leq1$ and is then equivalent with $u^2+v^2\leq1-2u+u^2$,
or
$$u\leq{1\over2}(1-v^2)\ .\tag{2}$$
It follows that $A^\circ$ is the closed parabolic region described in $(2)$.
