How does one parameterize $x^2 + xy + y^2 = \frac{1}{2}$? 
Parameterize the curve $C$ that intersects the surface
  $x^2+y^2+z^2=1$ and the plane $x+y+z=0$.

I have this replacing equations:
$$ x^2+y^2+(-x-y)^2=1$$
and clearing have the following:
$$ x^2+xy+y^2=1/2$$
which it is the equation of an ellipse but I find it difficult parameterization values
Any advice will be of much help, thanks in advance

 A: Hint:
Rewrite the equation as a sum of squares of linear forms:
$$x^2+xy+y^2=\Bigl(x+\frac12y\Bigr)^2 +\frac34 y^2=\frac12,$$
whence the equation in standard form\fracSetting $\;X=x+\frac12 y$, $Y=\frac{\sqrt{3}}2y$, you get the equation$$X^2+Y^2=\frac12.$$
Can you parametrise now?
A: Better yet, do you know how to make a coordinate change in $\Bbb R^3$, with new coordinates $(x',y',z')$, that will make the plane $x+y+z=0$ become the plane $z'=0$? Then you will have a unit circle in that plane, which you certainly know how to parametrize.
Hint: Choose an orthonormal basis $e_1,e_2,e_3$ for $\Bbb R^3$ with $e_3 = \dfrac1{\sqrt3}(1,1,1)$. 
A: Since $2 x^2 + 2 x y + 2 y^2$ is a quadratic form, we write
$$\begin{bmatrix} x\\ y\end{bmatrix}^T \begin{bmatrix} 2 & 1\\ 1 & 2\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix} = 1$$
We now compute the eigendecomposition of the matrix above
$$\begin{array}{rl} \begin{bmatrix} 2 & 1\\ 1 & 2\end{bmatrix} &= \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{bmatrix} \begin{bmatrix} 3 & 0\\ 0 & 1\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{bmatrix}\\\\ &= 3 \begin{bmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{bmatrix}^T + \begin{bmatrix} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}\end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}\end{bmatrix}^T\end{array}$$
Thus, $2 x^2 + 2 x y + 2 y^2 = 1$ can be written as
$$3 \left(\frac{x+y}{\sqrt{2}}\right)^2 + \left(\frac{x-y}{\sqrt{2}}\right)^2 = 1$$
It should now be easy to parameterize the ellipse.
A: Let 
\begin{equation}
x=\sqrt{\frac{2}{3}}\sin(t)
\end{equation}
\begin{equation}
y=\sqrt{\frac{2}{3}}\sin\left(t+\frac{2\pi}{3}\right)
\end{equation}
\begin{equation}
z=\sqrt{\frac{2}{3}}\sin\left(t-\frac{2\pi}{3}\right)
\end{equation}
Then $x+y+z=0$ and $x^2+y^2+z^2=1$
