Mixing Numerical Range and inner product Let $\mathbf{A}$ and $\mathbf{b}$ be a symmetric $N\times N$ real matrix and $N\times 1$ real vector respectively. Then consider the set of points in $\mathbb{R}^2$ defined as
\begin{align}
\mathbb{S}=\{\left(\mathbf{u}^T\mathbf{A}\mathbf{u},\mathbf{b}^T\mathbf{u}\right)\in\mathbb{R}^2,\mathbf{u}^T\mathbf{u}=1\}
\end{align}
That $\mathbb{S}$ is bounded is obvious. I was wondering if it had any other properties. In particular I am looking at convexity, similar to numerical ranges. If not, what are the conditions on $\mathbf{A}$ and $\mathbf{b}$ for the convexity and so on. 
Motivation:
Consider the optimization problem 
\begin{align}
&\min_{\mathbf{u}\in \mathbb{R}^N}\mathbf{u}^T\mathbf{A}\mathbf{u}+\mathbf{b}^T\mathbf{u}  \\\
\mbox{subject to }\\\
&\mathbf{u}^T\mathbf{u}=1
\end{align}
Then using the definition of $\mathbb{S}$ as earlier, I can write the above problem as
\begin{align}
\min_{(x,y)\in\mathbb{S}}x+y
\end{align}
Thus if $\mathbb{S}$ is convex, the above problem will be convex. 
 A: I don't know if your question admits a general answer, but for sure $\mathbb{S}$ is not always convex. Indeed, let us consider the following example:
$$\mathbf{A} = \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix}1\\ 0\end{bmatrix}.$$
Then we have $\mathbb{S}=\{(u_1^2+2u_2^2,u_1): \mathbf{u}^T \mathbf{u} = 1\}$. With few simple manipulations, we can parametrize $\mathbb{S}$ in the following way: 
$$\mathbb{S} = \{(2-\cos^2(t),cos(t)): t \in [0,2\pi)\},$$ 
and this is a continuous section of the parabola $\{x = 2-y^2\}$, that is not convex in $\mathbb{R}^2$.
A: We can solve directly your problem. We assume that $b$ is not an eigenvector of $A$. Let $f:u\rightarrow u^TAu+b^Tu,\;g:u\rightarrow u^Tu-1$.
There is a real $\lambda$ and a unitary $u=(u_i)_i$ s.t. $Df_u+\lambda Dg_u=0$, that is $(A+\lambda I)u=-b/2, \;u^Tu=1$. We may assume that $A=diag(\lambda_i)_i$ where the $\lambda_i$ are real. One obtains, for every $i$, $u_i=\dfrac{-b_i}{2(\lambda+\lambda_i)}$ with $\sum_i\dfrac{b_i^2}{(\lambda+\lambda_i)^2}=4$. Thus $\lambda$ is a root of a polynomial $P$ of degree $2N$. The sequel is easy. 
EDIT: The degree of the previous polynomial $P$ is $2N$ (edited). Unfortunately, $P$ seems to have no special properties ; indeed numerical experiments, when $N=3,4$, give for $P$ a Galois group equal to $S_6,S_8$.
