8. Given point $O$ as the origin, curve $C_{1}: x^{2}-y^{2}=1$ and curve $C_{2}: y^{2}=2 p x$ intersect 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 $\qquad$ ?
Reference answer: As shown in the figure, let the coordinates of points $M$ and $N$ be $\left(x_{0}, y_{0}\right)$ and $\left(x_{0},-y_{0}\right)$, respectively, and let $T$ be the intersection of $M N$ with the $x$-axis, so the coordinates of $T$ are $\left(x_{0}, 0\right)$. Since points $O, M, P, N$ are concyclic, by the intersecting chords theorem, we have $O T \cdot T P = M T \cdot T N$, which is $x_{0} \cdot\left(\frac{7}{2}-x_{0}\right) = y_{0}^{2} = 2 p x_{0}$, where $x_{0} > 0$.

Solving this, we get $x_{0} = \frac{7}{2} - 2 p$, $y_{0}^{2} = 2 p x_{0} = 7 p - 4 p^{2}$.
Substituting into the equation of curve $C_{1}$, we get $\left(\frac{7}{2} - 2 p\right)^{2} - \left(7 p - 4 p^{2}\right) = 1$, which simplifies to $32 p^{2} - 84 p + 45 = 0$.
Solving this, we get $p = \frac{15}{8}$ (discard) or $p = \frac{3}{4}$. Therefore, the equation of curve $C_{2}$ is $y^{2} = \frac{3}{2} x$.