Example 8 Given that there are three vertices $A, B, C$ of a square on the parabola $y=x^{2}$, find the minimum value of the area of such a square.

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As shown in Figure $11-5$, without loss of generality, assume that two of the vertices are on the right side of the $y$-axis (including the $y$-axis), and let the coordinates of points $A, B, C$ be $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$, respectively, with the slope of $B C$ being $k(>0)$. Then we have $y_{3}-y_{2}=k\left(x_{3}-x_{2}\right), y_{1}-$ $y_{0}=-\frac{1}{k}\left(x_{1}-x_{2}\right)$. Since points $A, B, C$ lie on the parabola, we have $y_{1}=x_{1}^{2}, y_{2}=x_{2}^{2}, y_{3}=x_{3}^{2}$. Substituting these into the above two equations yields $x_{3}=k-x_{1}, x_{1}=-\frac{1}{k}-x_{2}$. (1) Since $|A B|=|B C|$, i.e., $\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}=$ $\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}}$,
we have $\sqrt{1+\frac{1}{k^{2}}}\left(x_{2}-x_{1}\right)=\sqrt{1+k^{2}}\left(x_{3}-x_{2}\right)$,
thus $x_{2}-x_{1}=k\left(x_{3}-x_{2}\right)$.
From (1), we get: $\frac{1}{k}+2 x_{2}=k\left(k-2 x_{2}\right)$. Therefore, $k^{2}-\frac{1}{k}=(2 k+2) x_{2} \geqslant 0$,
solving this gives $k \geqslant 1$, and $x_{2}=\frac{k^{3}-1}{2 k(k+1)}$, so the side length of the square is
$$
\begin{array}{l}
\sqrt{k^{2}+1}\left(x_{3}-x_{2}\right)=\sqrt{k^{2}+1}\left(k-2 x_{2}\right)=\sqrt{k^{2}+1}\left[k-\frac{k^{3}-1}{k(k+1)}\right]=\frac{k^{2}+1}{k} \cdot \frac{\sqrt{k^{2}+1}}{k+1} \\
\geqslant \frac{2 k}{k} \cdot \frac{\sqrt{k^{2}+1}}{k+1} \geqslant \frac{2 k}{k} \cdot \frac{\sqrt{\frac{1}{2}(k+1)^{2}}}{k+1}=\sqrt{2},
\end{array}
$$

The equality holds if and only if $k=1$, i.e., when point $B$ is at the origin. Therefore, the minimum area of the square is 2.