First, I would like to introduce some background for the problem. I am currently dealing with a formula in crystallography which can be written as, $$\frac{1}{d^2}=\frac{1}{\sin^2\beta}\left(\frac{h^2}{a^2}+\frac{l^2}{c^2}-\frac{2hl\cos\beta}{ac}\right).$$
Basically I have a bunch of $(h,l,d)$ experimental data and I want to use the data points I have to fit for parameters $a$ and $c$ ($\beta$ is known). I prefer linear fitting which is reliable to me and easy to work with.
I did some simple algebra to make the formula neat. Taking $\tilde{a}=\frac{1}{a}$, $\tilde{c}=\frac{1}{c}$, $\tilde{h}=\frac{hd}{\sin\beta}$ and $\tilde{l}=\frac{hl}{\sin\beta}$, and it is easy to get, $$\tilde{a}^2\tilde{h}^2+\tilde{c}^2\tilde{l}^2-2\tilde{a}\tilde{c}\tilde{h}\tilde{l}\cos\beta=1.$$
I believe this is an equation describing an tilted ellipse curve in the $(\tilde{h},\tilde{l})$ plane. My question is, can I further separate the variables in the ellipse equation? Can it be rewritten in the form $f(\tilde{h},\tilde{l})=A(\tilde{a},\tilde{c})g(\tilde{h},\tilde{l})+B(\tilde{a},\tilde{c})$, where $f, g$ are both functions only of $\tilde{h},\tilde{l}$, and $A,B$ are both functions only of $\tilde{a},\tilde{c}$? I have tried hard but I can not make it. Could anyone please tell me if it is doable?
You have $$ b^2 x^2 - 2ab \cos \beta \; \; \; xy + a^2 y^2 = a^2 b^2 \; . $$
You want to orthogonally diagonalize the symmetric matrix $$ \left( \begin{array}{cc} b^2 & - ab \cos \beta \\ -ab \cos \beta & a^2 \end{array} \right) $$ with trace $a^2 + b^2,$ determinant $a^2 b^2 \sin^2 \beta,$ eigenvalues $$ \lambda = \frac{a^2 + b^2 \pm \sqrt{a^4 + (2 - 4 \sin^2 \beta) a^2 b^2 + b^4}}{2} $$ You then find the eigenvectors and make them unit length.