8. Given $O$ as the origin,

curve $C_{1}: x^{2}-y^{2}=1$ intersects with curve $C_{2}: y^{2}=2 p x$ at points $M$ and $N$. If the circumcircle of $\triangle O M N$ passes through point $P\left(\frac{7}{2}, 0\right)$, then the equation of curve $C_{2}$ is $\qquad$ .
8. $y^{2}=\frac{3}{2} x$.

As shown in Figure 2, let points $M\left(x_{0}, y_{0}\right), N\left(x_{0},-y_{0}\right)$, and $T$ be the intersection of $M N$ with the $x$-axis. Then $T\left(x_{0}, 0\right)$.

Since points $O, M, P, N$ are concyclic, by the intersecting chords theorem, we have
$$
\begin{array}{l}
O T \cdot T P=M T \cdot T N \\
\Rightarrow x_{0}\left(\frac{7}{2}-x_{0}\right)=y_{0}^{2}=2 p x_{0}\left(x_{0}>0\right) .
\end{array}
$$

Solving this, we get $x_{0}=\frac{7}{2}-2 p$,
$$
y_{0}^{2}=2 p x_{0}=7 p-4 p^{2} \text {. }
$$

Substituting into the equation of curve $C_{1}$, we get
$$
\begin{array}{l}
\left(\frac{7}{2}-2 p\right)^{2}-\left(7 p-4 p^{2}\right)=1 \\
\Rightarrow 32 p^{2}-84 p+45=0 .
\end{array}
$$

Solving this, we get $p=\frac{15}{8}$ (discard) or $p=\frac{3}{4}$.
Thus, the equation of curve $C_{2}$ is $y^{2}=\frac{3}{2} x$.