I have to find the intersection between these two surfaces in $\mathbb{R}^{3}$: this cylinder $$x^{2}+y^{2}-8x-8y+28=0$$ and this plane: $$x-y=0$$ and then find a parametric curve $\gamma(t)=(x(t),y(t),z(t))$ of the intersection between them.
Well, from the plane, I got: $$x=y$$ and in the cylinder, I got the parabola $$2x^{2}-16x+28=0$$ My question is: how do I parametrize it? I tried something like: $$\gamma(t)=(2t^{2}-16t+28,t,???)$$ But I really don't know how to do it.
The intersection is not a parabola. After simplifying you get
$$ x^2 - 8x + 14 = 0 $$
which can be written as
$$ (x - 4)^2 = 2 $$
and whose solutions are
$$ x = \pm \sqrt{2} + 4 $$
So the parametric form of the solution is just
$$ \gamma(t) = (\pm \sqrt{2} + 4, \pm \sqrt{2} + 4, t) $$
I marked those in thick black lines in the figure below