Example 14 (1998 Shanghai High School Mathematics Competition) As shown in Figure 5-13, it is known that on the parabola $y=$ $x^{2}$, there are three vertices $A, B, C$ of a square. Find the minimum value of the area of such a square.
Solution: It is evident that we need to express the side length of the square and then find its minimum value.
Without loss of generality, assume that two of the three vertices are on the right side of the $y$-axis (including the $y$-axis), as shown in Figure 5-13, and let the coordinates of points $A$, $B$, and $C$ be $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$, respectively, with the slope of $BC$ being $k (k > 0)$.
Then, $y_{3} - y_{2} = k(x_{3} - x_{2})$,
$y_{1} - y_{2} = -\frac{1}{k}(x_{1} - x_{2})$.
Since points $A$, $B$, and $C$ lie on the parabola,
we have $y_{1} = x_{1}^2$, $y_{2} = x_{2}^2$, $y_{3} = x_{3}^2$. Substituting these into equations (1) and (2),
we get $x_{3} = k - x_{2}$, $x_{1} = -\frac{1}{k} - x_{2}$.
Since $|AB| = |BC|$,

i.e., $\sqrt{(x_{1} - x_{2})^2 + (y_{1} - y_{2})^2} = \sqrt{(x_{3} - x_{2})^2 + (y_{3} - y_{2})^2}$.
Thus, $\sqrt{1 + \frac{1}{k^2}}(x_{2} - x_{1}) = \sqrt{1 + k^2}(x_{3} - x_{2})$
$$
\begin{array}{l}
\Rightarrow x_{2} - x_{1} = k(x_{3} - x_{2}) \\
\Rightarrow \frac{1}{k} + 2x_{2} = k(k - 2x_{2}) \\
\Rightarrow k^2 - \frac{1}{k} = (2k + 2)x_{2} \geq 0 \Rightarrow k^3 \geq 1 \Rightarrow k \geq 1 .
\end{array}
$$

and $x_{2} = \frac{k^3 - 1}{2k(k + 1)}$. The side length of the square is
$$
\begin{aligned}
\sqrt{k^2 + 1}(x_{3} - x_{2}) & = \sqrt{k^2 + 1}(k - 2x_{2}) = \sqrt{k^2 + 1}\left[k - \frac{k^3 - 1}{k(k + 1)}\right] = \frac{k^2 + 1}{k} \cdot \frac{\sqrt{k^2 + 1}}{k + 1} \\
& \geq \frac{2k}{k} \cdot \frac{\sqrt{\frac{1}{2}(k + 1)^2}}{k + 1} = \sqrt{2} .
\end{aligned}
$$

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