Find the polar set of $\{(x,y)\in\mathbb R^2: x^2+y^4\le1\}$ Let $A \subset \mathbb{R}^{n}$ be non-empty. The set $$ A^{\circ} = \lbrace c \in \mathbb{R}^{n} |\; \langle c,x \rangle \leq 1 \quad \forall x \in A  \rbrace $$
is called the polar of $A$. 
I'm trying to find the polar of $ \lbrace (x,y) \in \mathbb{R}^{2} |\; x^{2} + y^{4} \leq 1 \rbrace$. 
I tried to take the function $f(x,y)=ax+by-1 \leq0$ and do a lagrange optimisation thing to try and find points $(a,b)$ subject to this constraint $g(x,y)=x^{2}+y^{4}-1$, but now I'm just confused and have no idea how to do this.
I don't know any advanced properties of polars, I only know a few basic ones and the bipolar theorem. 
Any help would be appreciated. 
 A: Computing the norm induced by $A$
The set $A$ is compact and convex. Moreover, the origin $(0,0)$ lies in the interior of $A$. Therefore, $A$ is the unit ball of a unique norm $\|\cdot\|_A$ on $\mathbb{R}^2$. This norm is can be written as the Minkowski functional of $A$:
$$
\|(x,y)\|_A=\inf\{\lambda >0:(x,y)\in\lambda A\}.
$$
For every fixed $(x,y)\in\mathbb{R}^2\setminus\{(0,0)\}$, there exists a unique point $(x^\prime,y^\prime)$ on the ray starting at $(0,0)$ and passing through $(x,y)$, which lies on the boundary of $A$. (Existence is a consequence of $(0,0)\in\operatorname{int}(A)$ and the compactness of $A$, uniqueness follows from convexity of $A$.) Obviously, there exists a number $\lambda>0$ such that $(x^\prime,y^\prime)=\lambda(x,y)$. This point satisfies the equation $x^2+y^4-1=0$ which defines the boundary of $A$. That is $(\lambda x)^2+(\lambda y)^4-1=0$, which results in
$$
\lambda =\sqrt{-\frac{x^2}{2y^4}+\sqrt{\frac{x^4}{4y^8}+\frac{1}{y^4}}}
$$
for $y\neq 0$, and $\lambda =1/|x|$ otherwise. The value of $\|(x,y)\|_A$ is the inverse of that:
$$
\|(x,y)\|_A=\left(\sqrt{-\frac{x^2}{2y^4}+\sqrt{\frac{4x^4}{y^8}+\frac{1}{y^4}}}\right)^{-1}.
$$
for $y\neq 0$ and $\|(x,y)\|_A=|x|$ otherwise.
Subgradients of convex functions
If $f:\mathbb{R}^d\to \mathbb{R}$ is a convex function. By definition, $\xi$ is a subgradient of $f$ at $x$ if $f(y)-f(x)\geq \langle \xi,y-x\rangle$ for all $y\in\mathbb{R}^d$. In particular, this inequality is true if $f(y)\leq f(x)$, that is, $0\geq f(y)-f(x)\geq \langle\xi, y-x\rangle$ or $\langle \xi,y\rangle \geq\langle\xi,x\rangle$ if $f(y)\leq f(x)$. Therefore, subgradients of convex functions are outer normal vectors of supporting hyperplanes of sublevel sets. Note that the subdifferential of any norm $\|\cdot\|$ is constant on the rays emanating from $(0,0)$ (without the origin itself). More precisely, if $\|\cdot\|_\ast$ is the dual norm of $\|\cdot\|$, one has 
$$
\partial \|\cdot\| (x)=\{\xi\in \mathbb{R}^d: \langle\xi,x\rangle=\|x\|, \|\xi\|_\ast=1\}.
$$
(keywords: norming functionals, duality mapping) This is an exposed face of the dual unit ball. In our case, the primal norm is $\|\cdot\|_A$ with unit ball $A$, so the dual unit ball will be the polar set $A^\circ$. It's magic! Since the union of all exposed faces of a convex body is its boundary, we just have to evaluate the subdifferential of $\|\cdot\|_A$ at "enough" points. 
Convex differentiable functions
If a convex function is differentiable at a point, its convex subdifferential coincides with the set containing only the usual gradient. Here the boundary of $A$ is smooth (supporting hyperplanes = supporting lines are unique at every boundary point), the norm $\|\cdot\|_A$ is differentiable on $\mathbb{R}^2\setminus\{(0,0)\}$.
Combining our knowledge
Exploiting the axial symmetries of $A$, it suffices to compute the part of the boundary of $A^\circ$ which lies in the quadrant $(0,\infty)\times (0,\infty)$. (Note that $A^\circ$ is also symmetric with respect to the coordinate axes. The intersection points of $\operatorname{bd}(A^\circ)$ with the axes can be treated separately but they turn out to be $(\pm 1,0)$ and $(0,\pm 1)$, as one would expect.)
Fix $(x,y)\in(0,\infty)\times (0,\infty)$. Then 
$$
\|(x,y)\|_A=\frac{\sqrt{2}y^2}{\sqrt{-x^2+\sqrt{x^4+4y^4}}}
$$
whose gradient at $(x,y)$ is
$$
\nabla \|\cdot\|_A(x,y)=\left(\frac{\sqrt{2}xy^2}{\sqrt{x^4+4y^4}\sqrt{-x^2+\sqrt{x^4+4y^4}}},\frac{2\sqrt{2}x^4y+4y^5-2\sqrt{2}x^2y\sqrt{x^4+4y^4}}{\sqrt{x^4+4y^4}\left(-x^2+\sqrt{x^4+4y^4}\right)^{3/2}}\right).
$$
If $t\mapsto (x(t),y(t))$, $t\in I$, is a curve that meets every direction in $(0,\infty)\times (0,\infty)$ exactly once, $\nabla \|\cdot\|_A(x(t),y(t))$ is a parametrization for the desired part of the boundary of $A^\circ$. Let us take
$$
x(t)=\sqrt{2}\sqrt{\cos(t)}, \qquad y(t)=\sqrt{\sin(t)}, \qquad I=(0,\pi/2).
$$
Then
\begin{align*}
&\nabla \|\cdot\|_A(x(t),y(t))\\
&=\left(\frac{2\sqrt{\cos(t)}\sin(t)}{2\sqrt{-2\cos(t)+2}},\frac{8\sqrt{2}\cos^2(t)\sqrt{\sin(t)}+4\sqrt{2}(\sin(t))^{5/2}-8\sqrt{2}\cos(t)\sqrt{\sin(t)}}{2(-2\cos(t)+2)^{3/2}}\right)\\
&=\left(\frac{\sqrt{\cos(t)}\sin(t)}{\sqrt{2}\sqrt{-\cos(t)+1}},\frac{2\cos^2(t)\sqrt{\sin(t)}+(\sin(t))^{5/2}-2\cos(t)\sqrt{\sin(t)}}{(-\cos(t)+1)^{3/2}}\right)\\
&=\left(\frac{\sqrt{\cos(t)}\sin(t)}{\sqrt{2}\sqrt{-\cos(t)+1}},\sqrt{\sin(t)}\frac{2\cos^2(t)+\sin^2(t)-2\cos(t)}{(-\cos(t)+1)^{3/2}}\right)\\
&=\left(\frac{\sqrt{\cos(t)}\sin(t)}{\sqrt{2}\sqrt{-\cos(t)+1}},\sqrt{\sin(t)}\frac{\cos^2(t)+1-2\cos(t)}{(-\cos(t)+1)^{3/2}}\right)\\
&=\left(\frac{\sqrt{\cos(t)}\sin(t)}{\sqrt{2}\sqrt{-\cos(t)+1}},\sqrt{\sin(t)}\frac{(-\cos(t)+1)^2}{\left(-\cos(t)+1\right)^{3/2}}\right)\\
&=\left(\frac{\sqrt{\cos(t)}\sin(t)}{\sqrt{2}\sqrt{-\cos(t)+1}},\sqrt{\sin(t)}\sqrt{-\cos(t)+1}\right).
\end{align*}
