1. ## Conic to Quadratic Form (How To?)

I am reviewing an old Linear Algebra textbook, and am going through a chapter about going from a Quadratic Form to a Conic Section.

For example, starting with the following Quadratic Form:

$\displaystyle F(x,y) = a x^{2} + b x y + c y^{2}$

Let $\displaystyle \vec{v} = \begin{bmatrix}x\\ y\end{bmatrix}$

Then find a symmetric Matrix, A, such that $\displaystyle F(x,y) = A \vec{v}\cdot \vec{v}$

Then find an orthogonal matrix, Q, such that $\displaystyle Q^{T}A Q = D$, where D is a diagonal matrix.

Also, let $\displaystyle \vec{v'} = Q' \vec{v}$

At this point, we have enough information to determine if the original Quadratic Form represented a parabola, ellipse, or hyperbola.

However, I would like to go in the reverse direction: from a Conic to a Quadratic Form.
For example, say I have an ellipse of eccentricity, e.
How can I work backwards to create the Q, D, and A matrices, compute their eigenvalues and eigenvectors, etc.

Can anybody suggest an easy-to-understand reference for working through these steps?

2. ## Re: Conic to Quadratic Form (How To?)

I worked through this problem, so I am posting my work (maybe somebody else would find this informative.)

- - -

I am interested in seeing a relationship for standard ellipses as the eccentricity varies, so there is no rotation and the major axis, a, is 1, and aligned along the x-axis.

So, given the standard form of an ellipse

$\displaystyle \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and knowing that $\displaystyle b = a(1 - \textit{e}^{2})$

the equation can be rearranged into the following form:

$\displaystyle (1 - \textit{e}^{2})^2 x^{2} + y^{2} = (1 - \textit{e}^{2})^2$

This means that the diagonal matrix, D, is

$\displaystyle D = \begin{bmatrix}(1 - \textit{e}^{2})^2 &0\\ 0 &1\end{bmatrix}$

Because there is no rotation (the a- and b-axes are aligned with the x- and y-axes), Q, which can be considered a rotation matrix, is

$\displaystyle Q = \begin{bmatrix}\cos(\theta) &-\sin(\theta) \\ \sin(\theta) &\cos(\theta) \end{bmatrix} = \begin{bmatrix}1 &0 \\ \0 &1 \end{bmatrix}$

det(Q) is 1, so it does, indeed, serve as a rotation matrix.

The A matrix can now be computed as follows:

$\displaystyle A = QDQ^{T}$

which works out to

$\displaystyle A = \begin{bmatrix}(1 - \textit{e}^{2})^2 &0\\ 0 &1\end{bmatrix}$

which is the same as the D matrix.

$\displaystyle det(A) = (1 - \textit{e}^{2})^2$

The equation in its original quadratic form is the same as the one found above:

$\displaystyle F(x,y) = (1 - \textit{e}^{2})^2 x^{2} + y^{2} = (1 - \textit{e}^{2})^2$

3. ## Re: Conic to Quadratic Form (How To?)

I get a different result from using the substitution $\displaystyle b = a(1-e^2)$:

\displaystyle \begin{align*}\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} & = 1 \\ \dfrac{x^2}{a^2} + \dfrac{y^2}{a^2(1-e^2)^2} & = 1 \\ (1-e^2)^2x^2 + y^2 & = a^2(1-e^2)^2\end{align*}

Edit: Also, from what I recall about eccentricity (which I haven't looked at in a long time, so forgive me if I am mistaken), $\displaystyle e = \sqrt{1-\dfrac{b^2}{a^2}}$ so $\displaystyle b=a\sqrt{1-e^2}$, which would give:

\displaystyle \begin{align*}\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} & = 1 \\ \dfrac{x^2}{a^2} + \dfrac{y^2}{a^2(1-e^2)} & = 1 \\ (1-e^2)x^2 + y^2 & = a^2(1-e^2)\end{align*}

4. ## Re: Conic to Quadratic Form (How To?)

Oops! You are correct! Thanks.

$\displaystyle p=a (1-e^2)$, another variable relevant to ellipses.

$\displaystyle b=a\sqrt{1-e^2}$

Goes to show why it is always good to have more than one pair of eyes look things over.

5. ## Re: Conic to Quadratic Form (How To?)

Originally Posted by DavidB
Oops! You are correct! Thanks.

$\displaystyle p=a (1-e^2)$, another variable relevant to ellipses.

$\displaystyle b=a\sqrt{1-e^2}$

Goes to show why it is always good to have more than one pair of eyes look things over.
Ah, ok. Then when using that substitution, I get $\displaystyle (1-e^2)^2x^2 + y^2 = p^2$, which is slightly different from what you got. Perhaps $\displaystyle Q^\prime = aQ$ would work for your matrix. It is a rotation and stretch matrix. But, that would give you $\displaystyle A^\prime = Q^\prime D (Q^\prime)^T$ with $\displaystyle \det(A^\prime) = p^2$.