Answer
$(0,0, 0)$ and $(1,0,1)$
Work Step by Step
General form of parametric equation for $r(t)= ti +(2t-t^2) k$ is:
$x=t; y=0, z=2t-t^2$
We are given that the paraboloid equation is, $z=x^2+y^2$
This gives: $2t-t^2=(t)^2+(0)^2 \\2t-2t^2=0\\ 2t(1-t) =0 \\t=0,1$
Set the value of $t$ in order to get the point of intersection.
This implies that we have
$x=0; y=0$ and $z=2(0)-(0)^2=0\\ x=1; y=0$ and $z=2(1)-(1)^2=1$
So, the the point of intersection are: $(0,0, 0)$ and $(1,0,1)$
