10. (20 points) As shown in Figure 1, there is a square with three vertices $A, B, C$ on the parabola $y=x^{2}$. Find the minimum value of the area of such a square.
10. Let $A\left(x_{1}, x_{1}^{2}\right), B\left(x_{2}, x_{2}^{2}\right), C\left(x_{3}, x_{3}^{2}\right)$.

Assume $x_{1} < x_{2} < x_{3}$ and $k > 0$.
From $A B \perp B C \Rightarrow -\frac{1}{k} = \frac{x_{2}^{2} - x_{1}^{2}}{x_{2} - x_{1}}$
$$
\Rightarrow x_{1} + x_{2} = -\frac{1}{k}.
$$

Since $|A B| = |B C|$, we have
$$
\left(x_{1} - x_{2}\right)^{2} \left(1 + \frac{1}{k^{2}}\right) = \left(x_{2} - x_{3}\right)^{2} \left(1 + k^{2}\right).
$$

Substituting equations (1) and (2) into the above equation, we get
$$
\begin{array}{l}
\left(2 x_{2} + \frac{1}{k}\right) = k \left(k - 2 x_{2}\right) \\
\Rightarrow x_{2} = \frac{k^{3} - 1}{2 k (1 + k)}.
\end{array}
$$

Thus, the area of the square is
$$
\begin{array}{l}
S = |B C|^{2} = \left(1 + k^{2}\right) \left(x_{2} - x_{3}\right)^{2} \\
= \frac{\left(k^{2} + 1\right)^{3}}{k^{2} (k + 1)^{2}} = \frac{\left(k^{2} + 1\right)^{2} \left(k^{2} + 1\right)}{k^{2} (k + 1)^{2}} \\
\geqslant \frac{4 k^{2} \times \frac{1}{2} (k + 1)^{2}}{k^{2} (k + 1)^{2}} = 2.
\end{array}
$$

When $k = 1$, $S_{\text{min}} = 2$.