<s>[INST] how to parameterize the ellipse $x^2 + xy + 3y^2 = 1$ with $\sin \theta$ and $\cos \theta$ I am trying draw the ellipse $x^2 + xy + 3y^2 = 1$ so I can draw it.  Starting from the matrix:
$$ \left[ \begin{array}{cc} 1 & \frac{1}{2} \\ \frac{1}{2} & 3  \end{array}\right]$$
I computed the eigenvalues $2 \pm \frac{1}{2}\sqrt{5}$ and the eigenvectors (not normalized):
$$\left[ \begin{array}{c} x\\  y  \end{array}\right] 
= \left[ \begin{array}{c} 1\\  2\pm \sqrt{5}  \end{array}\right] $$
So then I tried writing down some combination of the data I generated:
$$ \left[ \begin{array}{c} x(\theta)\\  y(\theta)  \end{array}\right] 
= \cos \theta \left[ \begin{array}{c} 1\\  2+ \sqrt{5}  \end{array}\right] + \sin \theta \left[ \begin{array}{c} 1\\  2- \sqrt{5}  \end{array}\right]
$$
However, I have a hard time checking the ellipse equation holds true for all $\theta$:
$$x(\theta)^2 + x(\theta)y(\theta) + 3y(\theta)^2 = 1$$
What are the correct functions $x(\theta), y(\theta)$ ?

Following the  comments, rescaling the eigenvectors and multiplying the eigenvalues:
$$ \left[ \begin{array}{c} x(\theta)\\  y(\theta)  \end{array}\right] 
= \frac{2 + \frac{1}{2}\sqrt{5}}{\sqrt{10 + 4 \sqrt{5}}} \left[ \begin{array}{c} 1\\  2+ \sqrt{5}  \end{array}\right]\cos \theta + 
\frac{2 - \frac{1}{2}\sqrt{5}}{\sqrt{10 - 4 \sqrt{5}}}\left[ \begin{array}{c} 1\\  2- \sqrt{5}  \end{array}\right]\sin \theta
$$
Is it clear that the ellipse equation is satisfied?  I am not sure how to check this.
 [/INST] If you need a parametrization, it is best to just complete the square, rather then exploiting the full power of the spectral theorem.
$$ x^2+xy+3y^2 = 1 $$
is equivalent to:
$$ \left(2x+y\right)^2 + 11 y^2 = 4$$
hence $2x+y=2\cos\theta,y=\frac{2}{\sqrt{11}}\sin\theta$ is a valid parametrization, that leads to:
$$ x = \cos\theta-\frac{1}{\sqrt{11}}\,\sin\theta,\quad y=\frac{2}{\sqrt{11}}\,\sin\theta.$$
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