Consider the paraboloid $z=x^2+y^2$. The plane $2x-4y+z-6=0$ cuts the paraboloid, its intersection being a curve. Find "the natural" parameterization of this curve.
I have set each equation equal to each other by solving for $z$, completed the square to reach $(x-1)^2+(y+2)^2=11$. This is where I am stuck. Any help would be appreciated!
Plugging $z=x^2+y^2$ into $2x-4y+z-6=0$ we get \begin{align*} 2x-4y+x^2+y^2-6&=0\\ x^2+2x+\color{red}{1}+y^2-4y+\color{red}{4}&=6+\color{red}{1+4}\\ (x+1)^2+(y-2)^2&=11 \end{align*}
Then, a parameterization for the curve is $$\begin{cases}x&=\sqrt{11}\cos t-1 \\ y&=\sqrt{11}\sin t+2\\z&=2\sqrt{11}(2\sin t -\cos t)+16\end{cases}\qquad 0\le t\le 2\pi$$