Problem 1310

Given that $\left(\sqrt{2}-\frac{3}{4}\right)-(\sqrt{2}-1)=\frac{1}{4}$, the equation of the parabola $C_{1}$ is $x=4 \cdot \frac{1}{4}(y-1)^{2}+\sqrt{2}-1$, which simplifies to $y^{2}-2 y-x+\sqrt{2}=0$.

Suppose the tangents to $C_{1}$ and $C_{2}$ at their intersection point $A\left(x_{0}, y_{0}\right)$ are perpendicular. The tangent to $C_{1}$ at point $A$ is given by $y_{0} y-\left(y+y_{0}\right)-\frac{1}{2}\left(x+x_{0}\right)+\sqrt{2}=0$, with slope $k_{1}=\frac{1}{2 y_{0}-2}$. Similarly, the slope of the tangent to $C_{2}$ at point $A$ is $k_{2}=-\frac{1}{2 y_{0}-a}$. Since $k_{1} \cdot k_{2}=-1$, we have $4 y_{0}^{2}-2(a+2) y_{0}+2 a-1=0$,
(1) Also, since $\left(x_{0}, y_{0}\right)$ lies on both $C_{1}$ and $C_{2}$, we have
$$y_{0}^{2}-2 y_{0}-x_{0}+\sqrt{2}=0, \quad y_{0}^{2}-a y_{0}+x_{0}+2 b=0 .$$

Adding these two equations, we get $2 y_{0}^{2}-(2+a) y_{0}+2 b+\sqrt{2}=0$,
(2) Subtracting (2) multiplied by 2 from (1), we eliminate $y_{0}$ and get: $-a+2 b+\sqrt{2}+\frac{1}{2}=0$. Therefore, the curve $y^{2}-a y+x+2 b=0$ passes through the fixed point $\left(\sqrt{2}-\frac{1}{2}, 1\right)$.