11. If one side of the square $A B C D$ lies on the line $y=2 x-17$, and the other two vertices are on the parabola $y=x^{2}$, then the minimum value of the area of the square is $\qquad$ .
Let $A\left(t, t^{2}\right)$, line $A B: y-t^{2}=2(x-t) \Rightarrow y=2 x+t^{2}-2 t$.
$$
\begin{array}{l}
\text { Combining }\left\{\begin{array}{l}
y=x^{2}, \\
y=2 x+t^{2}-2 t
\end{array} \Rightarrow x^{2}-2 x=t^{2}-2 t \Rightarrow(x-1)^{2}=(t-1)^{2}\right. \\
\Rightarrow B\left(2-t,(2-t)^{2}\right) \Rightarrow|A B|=2 \sqrt{5}|1-t| . \\
\text { Thus } \frac{\left|t^{2}-2 t+17\right|}{\sqrt{5}}=2 \sqrt{5}|1-t| \Rightarrow t^{2}-2 t+17=10|1-t| \\
\Rightarrow\left\{\begin{array} { l } 
{ t \leqslant 1 , } \\
{ t ^ { 2 } + 8 t + 7 = 0 }
\end{array} \text { or } \left\{\begin{array}{l}
t>1, \\
t^{2}-12 t+27=0
\end{array} \Rightarrow t=-1,-7 \text { or } 3,9 .\right.\right.
\end{array}
$$

Therefore, there are only two such squares, with side lengths $|A B|=4 \sqrt{5}$ and $|A B|=16 \sqrt{5}$. Thus, the minimum area of the square is $(4 \sqrt{5})^{2}=80$.