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.
 A: 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:
$$ x = \cos\theta-\frac{1}{\sqrt{11}}\,\sin\theta,\quad y=\frac{2}{\sqrt{11}}\,\sin\theta.$$
A: You have $x^2 +xy + 3y^2=(x+1/2 y)^2+11/4 y^2=1$.
You can then take:
$$\begin{array}{lll}
\sin \theta & = & x+1/2 y\\
\cos \theta &= &\sqrt{11}/2 y
\end{array}$$
Which is equivalent to:
$$\begin{array}{lll}
y &= 2/\sqrt{11} \cos \theta\\
x &=\sin \theta - 1/\sqrt{11} \cos \theta
\end{array}$$
to get a parametrization like the one you're looking for. 
A: let us define $\theta$ by $$\cos\theta = \frac1{\sqrt{10+4\sqrt 5}}, \sin \theta=\frac{2+\sqrt 5}{\sqrt{10+4\sqrt 5}}.$$ then you can verify that 
$$\pmatrix{1&1/2\\1/2&3}\pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta} = \pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta}\pmatrix{2+\sqrt5/2&0\\0&2-\sqrt5/2}$$ that is 
$$\pmatrix{1&1/2\\1/2&3}= 
\pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta}
\pmatrix{2+\sqrt5/2&0\\0&2-\sqrt5/2}
\pmatrix{\cos \theta&\sin \theta\\-\sin \theta&\cos \theta}$$ that is $$A = U^\top DU, U^\top U = I \text{ where } U = \pmatrix{\cos \theta&\sin \theta\\-\sin \theta&\cos \theta} . $$
you can now define the new coordinate transformation by  $\xi, \eta$ by the relation 
$$\pmatrix{\xi\\\eta} = U\pmatrix{x\\y}, \pmatrix{x\\y} = U^\top \pmatrix{\xi\\\eta}.$$ 
with this we get 
$$\begin{align}x^2 + xy + 3y^2 &=\pmatrix{x&y} A\pmatrix{x\\y}\\ &=
\pmatrix{x&y} U^\top D U \pmatrix{x\\y}\\
&= \pmatrix{\xi & \eta}D\pmatrix{\xi\\\eta} \\
&= (2 + \sqrt5/2)\xi^2 + (2 - \sqrt5/2)\eta^2\end{align}$$
