Our aim is to find $a$ for which there exist $t$ and $s$ such that $r_1(t)=r_2(s)$. This yields us

$$at^2 =s\\ 2t=s\\ (t+1)^2=s^2$$

Substitute $s=2t$ in $(t+1)^2=s^2$ gives us

$$\begin{aligned}t^2+2t+1=4t^2&\implies 3t^2-2t-1=0\\&\implies t=\frac{2\pm \sqrt{4+12}}{6}=\frac{2\pm 4}{6}=1,-\frac 13\end{aligned}$$

Now $at^2 =s$ gives us

$$\begin{aligned}&\text{For } t=1, \quad s=2\implies a=2\\ &\text{For } t=-\frac 13, \quad s=-\frac 23\implies \frac a9=-\frac 23\implies a=-6\end{aligned}$$

Thus the the correct option is $a=2$, that is **Option A**.