Answer
A parametrization of the intersection is
${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,4{{\cos }^2}t} \right)$, ${\ \ }$ for ${\ \ }$ $0 \le t \le 2\pi $.
Work Step by Step
Suppose there exists some parameter $t$ such that $x = \cos t$ and $y = \sin t$, for $0 \le t \le 2\pi $.
Since ${\cos ^2}t + {\sin ^2}t = 1$, so
${x^2} + {y^2} = {\cos ^2}t + {\sin ^2}t = 1$
Thus, the $x$- and $y$-coordinates of points on the surface ${x^2} + {y^2} = 1$ can be represented by the parametrization $\left( {x,y} \right) = \left( {\cos t,\sin t} \right)$.
Hence, the intersection of the surfaces ${x^2} + {y^2} = 1$ and $z = 4{x^2}$ can be parametrized by
${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,4{{\cos }^2}t} \right)$, ${\ \ }$ for ${\ \ }$ $0 \le t \le 2\pi $.
