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 $\qquad$
11.80.

Let the side $AB$ of a square lie on the line $y=2x-17$, and the coordinates of the two vertices on the parabola be $C(x_1, y_1)$ and $D(x_2, y_2)$. Then the equation of the line $l$ on which $CD$ lies is $y=2x+b$. By solving the system of equations formed by the line $l$ and the parabola, we get $x^2=2x+b$. The solutions are
$$
x_{1,2}=1 \pm \sqrt{b+1}.
$$

Let the side length of the square be $a$, then
$$
\begin{array}{l}
a^2=(x_1-x_2)^2+(y_1-y_2)^2=5(x_1-x_2)^2 \\
=20(b+1).
\end{array}
$$

Take any point $(6,-5)$ on the line $y=2x-17$, the distance from this point to the line $y=2x+b$ is $a$, then
$$
a=\frac{|17+b|}{\sqrt{5}}.
$$

By solving equations (1) and (2) together, we get $b_1=3, b_2=63$.
Thus, $a^2=80$ or $a^2=1280$.