15. As shown in the figure, $O$ is the origin, point $F$ is the focus of the parabola $C_{1}: x^{2}=2 p y$ $(p>0)$, and the tangent line at point $P$ on the parabola $C_{1}$ is tangent to the circle $C_{2}: x^{2}+y^{2}=1$ at point $Q$.
(1) When the equation of line $P Q$ is $x-y-\sqrt{2}=0$, find the equation of the parabola $C_{1}$;
(2) When the positive number $p$ varies, let $S_{1}, S_{2}$ be the areas of $\triangle F P Q$ and $\triangle F O Q$, respectively. Find the minimum value of $\frac{S_{1}}{S_{2}}$.
(1) $y^{\prime}=\frac{x}{p}$, the tangent line equation at point $P\left(2 p t, 2 p t^{2}\right)$ on the parabola $C_{1}$ is $y-2 p t^{2}=2 t(x-2 p t)$ $\Rightarrow y=2 t x-2 p t^{2}$, thus $\frac{2 p t^{2}}{\sqrt{4 t^{2}+1}}=1 \Rightarrow 4 p^{2} t^{4}-4 t^{2}-1=0 \Rightarrow t^{2}=\frac{1+\sqrt{p^{2}+1}}{2 p^{2}}$. When $2 t=1 \Rightarrow t^{2}=\frac{1+\sqrt{p^{2}+1}}{2 p^{2}}=\frac{1}{4}$, $2 \sqrt{p^{2}+1}=p^{2}-2 \Rightarrow p^{2}=8 \Rightarrow p=2 \sqrt{2}$. Therefore, the equation of $C_{1}$ is $x^{2}=4 \sqrt{2} y$.
(2) Solving the system $\left\{\begin{array}{l}y=2 t x-2 p t^{2}, \\ x^{2}+y^{2}=1\end{array} \Rightarrow x_{Q}=\frac{4 p t^{3}}{4 t^{2}+1}\right.$. Extending $P Q$ to intersect the $y$-axis at point $M\left(0,-2 p t^{2}\right)$, then $S_{1}=\frac{1}{2}\left(\frac{p}{2}+2 p t^{2}\right)\left|2 p t-\frac{4 p t^{3}}{4 t^{2}+1}\right|=\frac{1}{2} p^{2}|t|\left(2 t^{2}+1\right)$, $S_{2}=\frac{1}{2} \cdot \frac{p}{2} \cdot \frac{4 p\left|t^{3}\right|}{4 t^{2}+1}=\frac{p^{2}\left|t^{3}\right|}{4 t^{2}+1} \Rightarrow \frac{S_{1}}{S_{2}}=\frac{\left(2 t^{2}+1\right)\left(4 t^{2}+1\right)}{2 t^{2}}=4 t^{2}+\frac{1}{2 t^{2}}+3 \geqslant 2 \sqrt{2}+3$.
Equality holds when $t^{2}=\frac{\sqrt{2}}{4} \Rightarrow p=\sqrt{2 \sqrt{2}+2}$, so the minimum value of $\frac{S_{1}}{S_{2}}$ is $2 \sqrt{2}+3$.