18. (15 points) As shown in Figure 4, draw two perpendicular chords $OA$ and $OB$ from the origin $O$ to the parabola $y^{2}=2 p x(p>0)$, and then draw the angle bisector of $\angle AOB$ intersecting $AB$ at point $C$. Find the equation of the locus of point $C$.
18. Let \( A\left(2 p t_{1}^{2}, 2 p t_{1}\right) 、 B\left(2 p t_{2}^{2}, 2 p t_{2}\right) \), then
\[
k_{O A}=\frac{2 p t_{1}}{2 p t_{1}^{2}}=\frac{1}{t_{1}}, k_{O B}=\frac{2 p t_{2}}{2 p t_{2}^{2}}=\frac{1}{t_{2}} \text {. }
\]

Since \( O A \perp O B \), we have \( k_{O A} \cdot k_{O B}=-1 \), i.e., \(\frac{1}{t_{1}} \cdot \frac{1}{t_{2}}=-1\)
Thus, \( t_{2}=-\frac{1}{t_{1}} \).
Let \( C(x, y) \), then \( k_{O C}=\frac{y}{x} \).
Since \( O C \) bisects \( \angle A O B \), we have \( \angle A O C=45^{\circ} \). Therefore,
\[
\frac{k_{O A}-k_{O C}}{1+k_{O A} k_{O C}}=\tan 45^{\circ}=1 \text {, }
\]

i.e., \(\frac{\frac{1}{t_{1}}-\frac{y}{x}}{1+\frac{1}{t_{1}} \cdot \frac{y}{x}}=1\). Solving this, we get \( t_{1}=\frac{x-y}{x+y} \).
Since \( A 、 C 、 B \) are collinear, when \( t_{1}+t_{2} \neq 0 \), we have \(\frac{2 p t_{1}-y}{2 p t_{1}^{2}-x}=\frac{2 p t_{1}-2 p t_{2}}{2 p t_{1}^{2}-2 p t_{2}^{2}}\),

i.e., \(\frac{2 p t_{1}-y}{2 p t_{1}^{2}-x}=\frac{1}{t_{1}+t_{2}}=\frac{t_{1}}{t_{1}^{2}-1}\).
Simplifying, we get \( y\left(1-t_{1}^{2}\right)-2 p t_{1}+x t_{1}=0 \).
Substituting \( t_{1}=\frac{x-y}{x+y} \) and simplifying, we obtain
\[
x^{3}+3 x y^{2}-2 p\left(x^{2}-y^{2}\right)=0 \text {. }
\]

When \( t_{1}+t_{2}=0 \), i.e., \( t_{1}-\frac{1}{t_{1}}=0 \), we get
\[
t_{1}^{2}=1, t_{2}^{2}=\left(-t_{1}\right)^{2}=1 \text {. }
\]

In this case, the point \( C(2 p, 0) \) satisfies equation (1).
Therefore, equation (1) is the trajectory equation of point \( C \).