Example 5. Through the origin, any line intersects the parabola $y^{2}=x^{2}-x+3$ at points $A$ and $B$. Find the locus of the midpoint of $AB$.

Keep the original text's line breaks and format, and output the translation result directly.
Let the equation of the line passing through the origin be
$$
\mathrm{y}=\mathrm{kx} . \quad(\mathrm{k} \text { is a parameter) }
$$

Then the coordinates of points A and B are the solutions to the system of equations
$$
\left\{\begin{array}{l}
y=k x, \\
y=x^{2}-x+3
\end{array}\right.
$$

Eliminating \( y \) from (1) and (2) gives
$$
\mathrm{x}^{2}-(\mathrm{k}+1) \mathrm{x}+3=0 \text {. }
$$

Eliminating \( x \) from (1) and (2) gives
$$
\mathrm{y}^{2}-(\mathrm{k}+1) \mathrm{k} y+3 \mathrm{k}=0 .
$$

If the coordinates of points A and B are \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\), respectively, then \(x_{1}\) and \(x_{2}\) are the roots of equation (3), and \(y_{1}\) and \(y_{2}\) are the roots of equation (4). Let the coordinates of the midpoint of \(A B\) be \((x, y)\). Then, by the midpoint formula and Vieta's formulas, we have:
$$
\begin{array}{l}
x=\frac{1}{2}\left(x_{1}+x_{2}\right)=\frac{k+1}{2}, \\
y=\frac{1}{2}\left(y_{1}+y_{2}\right)=\frac{k+1}{2} \cdot k .
\end{array}
$$

Therefore, \(\mathrm{y}=\frac{\mathrm{k}+1}{2} \cdot\left(2 \cdot \frac{\mathrm{k}+1}{2}-1\right)\)
$$
=\mathrm{x}(2 \mathrm{x}-1),
$$
$$
y=2 x^{2}-x
$$