How can the level curves of a quadratic form be a pair of lines?

Question

$x_1^2+4x_1x_2+4x_2^2=1\Rightarrow A\begin{pmatrix}1&2\\2&4\end{pmatrix}\Rightarrow f_A(A)=\begin{vmatrix}1-\lambda&2\\2&4-\lambda\end{vmatrix}\Rightarrow\begin{Bmatrix}\lambda_1=0\\\lambda_2=5\end{Bmatrix}\\ \Rightarrow q\,(\vec{x})=5c_2^2 $

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Why is this a pair of lines, and not a parabola - how does $5c_2^2=1$ yield anything other than a quadratic function?

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EDIT:

After some tinkering, I'm convinced that the level curves look like this:

What's unclear is how I would graph these w.r.t both $x_1,x_2$ as well as the principal axes $c_1, c_2$ as we do with ellipses and hyperbolae. Where would these two lines be on this graph? Neither of them are functions of $c_1, c_2$. Is there some obvious change of coordinates that I'm not seeing here?

Answer 1

$$x_1^2 + 4x_1x_2+4x_2^2 = (x_1+2x_2)^2$$ Therefore the function is constant on lines of constant $x_1+2x_2$, which are straight lines.

(Note also that the determinant of the quadratic form's matrix is equal to zero, so you have at least one eigenvector with eigenvalue zero, which is bascally the same thing.)

Answer 2

Think of a quadratic form as a "norm" on the vector space. The level curves of the quadratic form are the "spheres" of the norm.

In this case, since $\lambda_1=0$, the "norm" of every vector is $5$ times its component in the $\lambda_2$ direction.

To your question about $5c_2^2=1$, I should hope that this does not give a parabola! This equation involves only one variable, not two.