Answer
True
Work Step by Step
Given: $r(t)=(0,t^{2},4t)$
Here, $x=0,y=t^{2},z=4t$
As $z=4t$ gives $t=\frac{z}{4}$
and $y=t^{2}$ becomes
$y=(\frac{z}{4})^{2}$
$y=\frac{z^{2}}{16}$
which represents a standard form of the equation of a parabola.
Hence, the statement is true.