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I have this question about linear geometry. For the equation $$2x^2+3y^2+az^2-2xy+4z-5=0$$ I have to find for which values of $a$ do I get:

  1. a parabolic cylinder,
  2. a pair of parallel planes.

I've a lead: for $a>0$, $a<0$ and $a=0$, bring each case to its canonical form and then I am stuck.

Thanks for the help. Benny.

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I have recently dealt with a similar question.

A quadric may be defined by an implicit matrix equation as follows: $ \mathbf{X}^T \mathbf{A} \mathbf{X} = 0$ where $\mathbf{X} = \begin{bmatrix} x & y & z & 1 \end{bmatrix}^T $ is the position vector in homogeneous coordinates and $\mathbf{A} $ is a symmetric matrix in the form of: \begin{equation} \mathbf{A} = \begin{bmatrix} a & b & c & d \\ b & e & f & g \\ c & f & h & i \\ d & g & i & j \end{bmatrix} \end{equation}

Explicitly, this expands to: \begin{equation} 0 = a{x}^{2} + +e{y}^{2} + h{z}^{2} + +2bxy + 2cxz + 2fyz + +2dx + 2gy + 2iz + j \end{equation}

Now, some information we need later:

The principal submatrix of $\mathbf{A}$ is denoted $\mathbf{A}_u$ and consists of the upper-left $3\times3$ of matrix, e.g.

$ \mathbf{A}_u = \begin{bmatrix} a & b & c \\ b & e & f \\ c & f & h \end{bmatrix} $

Now, we have enough information to classify the surface.

A quadric surface may be identified by examining the Eigenvalues of the defining matrix and the principle submatrix. The so-called inertia of a matrix is defined as a pair of integers consisting of number of strictly positive eigenvalues $\sigma_+$ and the number of strictly negative eigenvalues $\sigma_-$. Since a quadric may be multiplied by a non-zero real scalar without altering the shape of the curve, $\sigma_+$ and $\sigma_-$ may be arbitrarily switched and hence inertia may be redefined as the pair

$\big(\max(\sigma_+, \sigma_-), \min(\sigma_+, \sigma_-) \big)$

The inertia of the quadric matrix is denoted$(S_1, S_2)$ and inertia of the principle submatrix is denoted as $(S_{u_1}, S_{u_2})$.

The two inertia pairs $(S_1, S_2)$ and $(S_{u_1}, S_{u_2})$ uniquely define the shape of the quadric with the classification found in Dupont (Table 1 pp173).

So, for a pair of planes, we need inertia pairs [(1,1), (1,0)] and for a parabolic cylinder, we need [(2, 1), (1, 0)].

From here, it is a matter of sorting out the values for $a$ that make the signs of the eigenvalues match the required inertia pairs and your problem is solved.

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