JFET biasing problem
You "feel like there's something missing" in the task... and you are right, there are missing two pieces of information to determine the required bias (\$ V_{GS} \$).

From the given information you can write:

$$ I_D = \frac{V_{DD}}{2 \cdot (R_S + R_D)} = \frac{k \cdot V_{DD} - V_{GS}}{R_S} = I_{DSS} \cdot \left(1 - \frac{V_{GS}}{V_P}\right)^2 $$ where \$k\$ is the \$ R_1, R_2 \$ divider ratio \$ \frac{R_2}{R_1+R_2} \$.

So you have to know either:  


\$ R_S, k \$ and you can calculate the required \$ V_{GS} \$, (then \$ I_D \$ and \$ R_D  \$), or  
\$ R_S, R_D \$ and you can calculate \$ I_D \$, then the required \$ V_{GS} \$ (and finally \$k\$), or  
\$ R_D, k \$ and again calculate the rest


Then, to solve the \$ R_1, R_2 \$ divider out of the \$k\$ value, you have to know (choose) value of one of the resistors (or their sum), of course.