Answer
$\left\{\begin{array}{l}
x=t^{2}+1,\\
y=t
\end{array}\right.\qquad t\leq 0$
(sample answer)
Work Step by Step
$x-1=y^{2},\ \quad y\leq 0$
This represents the lower half of the graph.
Defining a parameter $t=y$, we have:
$\left\{\begin{array}{l}
x=t^{2}+1,\\
y=t
\end{array}\right.\qquad t\leq 0$
as a possible parametrization.