Is an ellipsoid defined as a quadric always symmetric? In Wikipedia, one can read that points $\mathbf{x}\in{\rm I\!R}^3$ of an ellipsoid of center $\mathbf{v}\in {\rm I\!R}^3$ are solution of the equation
$$
(\mathbf{x}-\mathbf{v})^T A (\mathbf{x}-\mathbf{v}) = 1
$$
where $A$ is a positive definite matrix.
Is the matrix $A$ also symmetric ? I read here that any matrix $Q$ of a quadratic form $x^T Q x$ can be made symmetric using $\frac{1}{2}(Q+Q^T)$ instead.
What happen if the matrix $A$ is not positive definite ? If one of its eigenvalues is null or negative, what is the shape described by such equation? I feel like a cone or a cylinder can be represented. 
Does it means that there exist a matrix $V$ such that $A=V^T V$ ? I read this statement here. In this case, what are the properties of $V$?
 A: Usually when someone says "positive definite", they are using the definition of "positive definite" which implies "symmetric". I.e. positive definite matrices are usually symmetric by definition.
If one of the eigenvalues is null, you get a degenerate conic section. If one of the eigenvalues is negative, you get a conic section with negative curvature, analogous to a hyperbola, i.e. hyperboloids in $\mathbb{R}^3$. https://en.wikipedia.org/wiki/Quadric#Euclidean_space
Yes, it is true that every positive definite matrix has a square root, i.e. a matrix $V$ such that $V^\top V = A$, one way to find $V$ is to do a Cholesky factorization, another way is to take the eigenvalue decomposition of $A = U D U^\top$. All of the entries of $D$ are positive, so one can define the entrywise square root, $\sqrt{D}$, so then $A = (U \sqrt{D}) (\sqrt{D} U^\top) = (U^\top \sqrt{D})^\top (U^\top \sqrt{D})$, so define $V = \sqrt{D} U^\top$. (Note also that the singular value decomposition and eigenvalue decomposition coincide for a positive definite matrix, so one could also get the matrix square root using the SVD.)
A: Any quadric surface, not just an ellipsoid, can be written 
$$\begin{bmatrix}x&y&z&1\end{bmatrix}\begin{bmatrix}a&h&g&s\\h&b&f&p\\g&f&c&q\\s&p&q&r\end{bmatrix}\begin{bmatrix}x\\y\\z\\1\end{bmatrix}=[0].$$
