Let $\mathbf C$ be a positive-definite $k\times k$ matrix. For all vectors $\mathbf u\in \mathbb R^k$ of length $\|\mathbf u\|=1$, consider vectors $\mathbf {uu}^\top\mathbf{Cu}$; they form a surface in $\mathbb R^k$. What is this surface? In particular, what is it in case of $k=2$?
Here is an example for $\mathbf C = \left(\begin{array}{cc}4&2\\2&2\end{array}\right)$:
I do realize that the "main axes" (the longest and the shortest cuts) of this curve are given by the eigenvectors of $\mathbf C$ scaled by the respective eigenvalues. But the whole shape looks weird.
If you write your vectors in the basis of eigenvectors $f_1,\ldots,f_k$ of $C$, then if $u=\sum t_jf_j$ we have $$ u^TCu=\sum t_j^2\lambda_j, $$ where $\lambda_1,\ldots,\lambda_k$ are the eigenvalues of $C$ (counting multiplicities). So, in each direction, you are stretching the unit circle by $\sum t_j^2\lambda_j$ a convex combination of the eigenvalues. In particular, when $u$ is the $j^{\rm th}$ eigenvector, the value is precisely $\lambda_j$.
More explicitly, $$ \{uu^TCu:\ \|u\|=1\}=\left\{\left(\sum_{j=1}^kt_j^2\lambda_j\right)\,u:\ u=\sum t_jf_j,\ \sum t_j^2=1\right\}. $$ When $k=2$, $$ \{uu^TCu:\ \|u\|=1\}=\{(\lambda_1\cos^2\theta+\lambda_2\sin^2\theta)\,u:\ \theta\in[0,2\pi],\ u=(\cos\theta,\sin\theta)\} $$