\(\mathbf r(t)=x(t)\,\mathbf i+y(t)\,\mathbf j+z(t)\,\mathbf k=t\,\mathbf i+(4t-t^2)\,\mathbf k\)
\(\implies \begin{cases}x(t)=t\\y(t)=0\\z(t)=4t-t^2\end{cases}\)
So the paraboloid is given by
\(z(t)=x(t)^2y(t)^2\iff 4t-t^2=0\)
which holds for \(t^2-4t=t(t-4)=0\), i.e. when \(t=0\) and \(t=4\).
To make sure this is correct, plug in \(t=0\) and \(t=4\) to find out the value of \(\mathbf r(t)\), and see if these points are on the paraboloid.
\(\mathbf r(0)=0\,\mathbf i+0\,\mathbf j+0\,\mathbf k\implies (0,0,0)\)
\(0=0^2\times0^2\)
\(\mathbf r(4)=4\,\mathbf i+0\,\mathbf j+(4\times4-4^2)\,\mathbf k\implies (4,0,0)\)
\(0=4^2\times0^2\)
Both of these hold, so the answer is correct.
