Deriving the equation of an ellipse from another related equation

Question

Consider the equation for $x,y, \phi \in \mathbb{R}$ $$ \left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 - 2 \cos (\phi) \frac{xy}{ab} = \sin^2 (\phi) $$ It is supposed to be an elementary exercise to show that the vector $(x,y)$ traces out an ellipse, but I am not quite seeing it. If I attempt to solve for some constant $A,B,C,D,R \in \mathbb{R}$ such that $$ \frac{(x - A)^2}{B} + \frac{(y - C)^2}{D} = R^2$$ would match the above equation, then I always end up with the non-linear term $xy$ causing me problems. On the other hand with $$ \frac{(x - Ay)^2}{B} + \frac{(y - Cx)^2}{D} = R^2$$ We have $$ x^2 \left( \frac{1}{B} + \frac{C^2}{D} \right) - 2 xy \left( \frac{ A}{B} + \frac{C}{D}\right) + y^2 \left( \frac{A^2}{B} + \frac{1}{D}\right) = R^2$$ Now we would have $$ \frac{1}{a^2} = \frac{1}{B} + \frac{C^2}{D}$$ $$ \frac{1}{b^2} = \frac{1}{D} + \frac{A^2}{B}$$ $$ \frac{\cos (\phi)}{ab} = \frac{A}{B} + \frac{C}{D}$$ $$ \sin^2 \phi = R^2$$ However now the equation of the ellipse is in the wrong "base". Any ideas?

Answer 1

HINT:

The $xy$ term is not nonlinear, it has the effect of tilting the principal axes.

Like e.g., in: [ $ x^2 - xy + y^2 = 1 $] .

$ \phi = \pi/2 $ is ellipse standard form.