Why is a positive definite matrix needed in the ellipsoid matrix representation? An ellipsoid centered at the origin is defined by the solutions $\mathbf{x}$ to the equation $\mathbf{x}^TM\mathbf{x} = 1$, where M is a positive definite matrix.
How can I see why M needs to be positive definite, based on the equation of an ellipse $Ax^2 + Bxy + Cy^2 = 1$ where $B-4AC < 0$? It looks like the idea is to make $B-4AC < 0$ equate to the requirement that all eigenvalues of $M$ are positive for a 2x2 matrix, but I can't seem to make it work.
Also, what other shapes can we represent with $\mathbf{x}^TM\mathbf{x} = 1$ when $M$ is not positive definite?
 A: Consider the factorization of the matrix $A$ which is always possible since $A$ is a real symmetric matrix, and the :
$A=P\Lambda P^{-1} =P\Lambda P^T$, 
where the matrix $P$ is orthogonal because $A$ is symmetric, and the diagonal matrix $\Lambda$ contains the eigenvalues of $A$.
Then the ellipsoid can be rewritten as:
$$x^TP\Lambda P^Tx = (P^Tx)^T\Lambda(P^Tx) = y^T\Lambda y = 1$$
where, of course, $y$ is derived from the linear map 
 $\varphi:\mathbb{R}^n \rightarrow \mathbb{R}^n$, defined as $y = \varphi(x) = P^Tx$.
Now, $P$ represents a pure rotation (since it is orthogonal, it preserves the norm of a vector) so that this new expression describes an ellipsoid having its axes aligned with the canonical base axes $e_1, ..., e_n$.
Since the matrix is diagonal, the quadratic form gives:
$$y^T\Lambda y =\lambda_1y_1^2+...+\lambda_ny_n^2 = 1$$
Of course, you could recognise the canonical form of an ellipsoid:
$$\frac{y_1^2}{c_1^2} +...+ \frac{y_n^2}{c_n^2}=1$$ where $c_i$ is the length of the $i$-th semiaxis
But, if the matrix is not positive definite, then the eigenvalues could assume also zero or even negative values.
In the first case, $c_i^2$ cannot exist since it should be obtained as $\frac{1}{\lambda_i}$, but dividing by zero is not allowed. In a suggestive way of thinking, it is like the ellipsoid were infinitely extended along that direction $e_i$, then it cannot be an ellipsoid.
In the second case, the $c_i$ become complex numbers, because they are the root of a negative number, and if all are negative, the ellipsoid is said 'imaginary' for trivial reasons.
