<s>[INST] Parametric form of an ellipse given by $ax^2 + by^2 + cxy = d$ If $c = 0$, the parametric form is obviously $x = \sqrt{\frac{d}{a}} \cos(t), y = \sqrt{\frac{d}{b}} \sin(t)$.
When $c \neq 0$ the sine and cosine should be phase shifted from each other.  How do I find the angular shift and from there how do I adjust the factors multiplying the sine and cosine?
 [/INST] You would complete squares: $\left(a x + \frac{1}{2} c y\right)^2 + \left(a b - \frac{c^2}{4} \right) y^2 = a d$.
From there: $a x + \frac{c}{2} y = \sqrt{a d} \sin(t)$ and $\sqrt{a b - \frac{c^2}{4}} y = \sqrt{a d} \cos(t)$, assuming $c^2 < 4 a b$, and $a d > 0$. 
Solving for $x$ and $y$ and denoting $\mathcal{D} = 4 a b - c^2$
$$
   x(t) = \sqrt{\frac{d}{a}} \left( \sin(t) - \frac{c}{\sqrt{\mathcal{D}} } \cos(t) \right) \qquad
   y(t) = \frac{2 \sqrt{a d}}{\sqrt{\mathcal{D}}} \cos(t)
$$ 
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