Easier way to parametrize an ellipse. I want to calculate the parametrization of the curve $$E=\{(x,y,z):z=x^2+y^2,x+y+z=1\}$$
To do so I did a change of variables taking as basis the vectors $\frac{1}{\sqrt{2}}(-1,1,0),\frac{1}{\sqrt{6}}(-1,-1,2),\frac{1}{\sqrt{3}}(1,1,1)$ which are the normal vector of the plane, and two other perpendicular vecors of the plane. With this basis the plane $x+y+z=1$ becomes $z'=\sqrt{3}/3$, and $$\begin{cases}x=-x'/\sqrt{2}-y'/\sqrt{6}+z'/\sqrt{3}\\y=x'/\sqrt{2}-y'/\sqrt{6}+z'/\sqrt{3} \\z=2y'/\sqrt{6}+z'/\sqrt{3}\end{cases}$$
Now using this change of variables into $E$ I have that $$\frac{(x')^2}{\frac{1}{(3^{(1/4)})^2}}+\frac{(y'-\sqrt{2/3})^2}{(3^{(1/4)})^2}=1$$ and therefore the parametrization of the ellipse is $$x'=3^{1/4}\cos\theta,\;y'=\sqrt{2/3}+3^{1/4}\sin\theta,\; z'=\sqrt{3}/3,\; 0\leq \theta \leq2\pi$$ 
As I checked with wolfram alpha, this is correct. However it requires a lot of calculations to just parametrize this curve, is there any faster way to do it?
 A: You can parametrize in the original coordinates. Solving the plane for $z$ you have:
$$x+y+z=1 \iff z = 1-x-y \tag{$\star$}$$
so substituting into the paraboloid and rewriting gives:
$$z=x^2+y^2 \implies 1-x-y=x^2+y^2 \iff \left(x+\tfrac{1}{2}\right)^2+\left(y+\tfrac{1}{2}\right)^2=\tfrac{3}{2}$$
This is easily parametrized as $x+\tfrac{1}{2}=\sqrt{\tfrac{3}{2}}\cos t$ and $y+\tfrac{1}{2}=\sqrt{\tfrac{3}{2}}\sin t$ and $z$ follows from $(\star)$:
$$z = 1-x-y=1-\left(\sqrt{\tfrac{3}{2}}\cos t-\tfrac{1}{2}\right)-\left(\sqrt{\tfrac{3}{2}}\sin t-\tfrac{1}{2}\right)=2-\sqrt{\tfrac{3}{2}}\cos t-\sqrt{\tfrac{3}{2}}\sin t$$
to get:
$$\gamma : [0,2\pi]\to\mathbb{R}^3: t \mapsto
\left( \sqrt{\tfrac{3}{2}}\cos t-\tfrac{1}{2} ,
 \sqrt{\tfrac{3}{2}}\sin t-\tfrac{1}{2} ,
 2-\sqrt{\tfrac{3}{2}}\cos t-\sqrt{\tfrac{3}{2}}\sin t \right)$$
A: $$z=x^2+y^2$$ hints the use of polar coordinates and the equation of the plane becomes $$\rho^2+\rho(\cos\theta+\sin\theta)-1=0.$$
Solve for $\rho$ in terms of $\theta$ and you are done.
