Conic sections - quadratic curve

[Nick R]

Here is what I know:

1) All quadratic curves of 2 variables correspond to a conic section.

$$ax^2 + 2bxy +cy^2 + 2dx + 2fy + g = 0$$

$$a, b, c$$ are not all $$0$$

2) The definitions of parabola (in terms of a directrix and focus), ellipse (in terms of 2 foci), hyperbola (in terms of directrix and focus).

3) The determinate of a 2x2 matrix is the area of the parallelogram formed by the 2 row vectors.

Question:

The above quadratic equation can be found to be either an ellipse, parabola or hyperbola depending on the value of the determinate

$$\left| \begin{array}{ccc} \ a & b \\ b & c\end{array} \right|$$

I haven't seen any sort of derivation, or even a hint, as how to arrive at the significance of this determinate.

Can anyone point me to one?

 

[HallsofIvy]

That's an example of a symmetric bilinear form. The matrix multiplication
$$\left[\begin{array}{cc}x & y\end{array}\right]\left[\begin{array}{cc}a & b \\ b & c \end{array}\right]\left[\begin{array}{c} x \\ y \end{array}\right]$$
gives ax²+ 2bxy+ c.

It can be shown, in linear algebra, that any symmetric matrix, A, can be "diagonalized"- that is, that there exist an orthogonal matrix P such that PAP^-1= PAP^T= D where D is a matrix having only 0s off the main diagonal. It can further be shown that the numbers on the diagonal are the eigenvalues of A and that the rows of P are eigenvectors corresponding to those eigenvalues. If, in the above equation, X^TAX, we replace A by P^TDP, we have X^T(P^TDP)X= (PX)^TD(PX). If we let PX= Y= <x', y'> and the diagonal elements of D are $\lambda_1$ and $\lambda_2$, then that last multiplication is
$$\left[\begin{array}{cc} x' & y' \end{array}\right]\left[\begin{array}{cc}\lambda_1 & 0 \\ 0 & \lambda_2 \end{array}\right]\left[\begin{array}{c} x' \\ y' \end{array}\right]$$
= $\lambda_1 x'^2+ \lambda_2 y'^2$
And that last is
1) an ellipse if $\lambda_1$ and $\lambda_2$ are both the same sign.

2) a hyperbola if $\lambda_1$ and $\lambda_2$ are of different signs.

3) a parabola if one of $\lambda_1$ and $\lambda_2$ is 0.

Since, as I said before, $\lambda_1$ and $\lambda_2$ are the eigenvalues of the orignal matrix, the conic section is a parabola if and only if that matrix has a 0 eigenvalue.

 

[Nick R]

So if $$P = \left[\begin{array}{cc}\ e_x_1 & e_y_1 \\ e_x_2 & e_y_2 \end{array}\right]$$


$$Y^TDY = ax^2 + 2bxy + cy^2 = \lambda_1 (xe_1_x + y_e_2_x)^2 + \lambda_2 (xe_1_y + ye_2_y)^2$$

The square terms are always positive, so basically this means that

if both eigenvalues are positive/negative, the quadratic term of the conic section is always positive/negative (ellipse).

The eigenvalues are of different signs, the quadratic term of the conic section may be positive or negative depending on (x,y)... (hyperbola)

If one eigenvalue is 0, the quadratic term is always (the sign of the remaining eigenvalue) (parabola).

Is there something obvious I'm missing here that would make the signficance of these eigenvalues of clear in the context of the shape drawn corresponding to the quadratic term?

Thanks for the response too, very useful.

 

[HallsofIvy]

I believe I had already addressed that. Changing to the x', y' coordinate system, where <x', y'>= PX. The equation becomes $\lambda_1 x'^2+ \lambda_2 y'^2$ in that coordinate system so that whether it is an ellipse, hyperbola, or parabola depends on what $\lambda_1$ and [/itex]\lambda_2[/itex] are. Of course, an ellipse is an ellipse, a hyperbola is a hyperbola, and a parabola is a parabola, no matter what coordinate system you write the equation in!

 

[Nick R]

Wow I didn't get the significance of that at first.

So P (which is composed of the eigenvectors of A), is a linear transformation to some other coordinate system in which the "cross term" in the curve is eliminated...

So its sort of a rotation (with "squishing" if the eigenvectors are not orthogonal).

Using a graphing tool I can clearly see that the shapes of parabola, ellipse, hyperbola follow from those properties of the eigenvalues.

I know that A applied to an eigenvector results in a multiple of the eigenvector. So I suspect what is going on here is that some sort of symmetry (in reference to the graph of ax^2 + 2bxy + cy^2) is guaranteed about the eigenvectors.

Is that correct? This is great thanks.