Answer
$\left\{\begin{array}{l}
x=t^{2}+1,\\
y=t
\end{array}\right.\qquad t\leq 0$
Work Step by Step
Add a restriction on y, to obtain the lower portion of the graph:
$x-1=y^{2},\ \quad y\leq 0$
$x=y^{2}+1,\ \quad y\leq 0$
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.
