$15(1,0)$ is a fixed point, $P(0, b)$ is a moving point on the $y$-axis, and point $M(a, 0)$ satisfies $\overrightarrow{P M} \cdot \overrightarrow{P F}=0$. If point $N$ satisfies $2 \overrightarrow{P N}+\overrightarrow{N M}=\overrightarrow{0}$.
(1) Find the equation of the trajectory curve $C$ of point $N$;
(2) Find the trajectory of the intersection points of any two mutually perpendicular tangents to curve $C$.
15 (1) It is easy to know that $\overrightarrow{P M}=(a,-b), \overrightarrow{P F}=(1,-b)$, and from $\overrightarrow{P M} \cdot \overrightarrow{P F}=0$, we get $a+b^{2}=0$. Let the coordinates of point $N$ be $(x, y)$, then $\overrightarrow{P N}=(x, y-b), \overrightarrow{N M}=(a-x, -y$. Thus, we have
$$
2(x, y-b)+(a-x,-y)=\overrightarrow{0} .
$$

This gives $a+x=0, y-2 b=0$. Therefore, $x=-a=b^{2}=\frac{y^{2}}{4}$, and the equation of curve $C$ is $y^{2}=4 x$.
(2) When $y>0$, we have $y=2 \sqrt{x}, y^{\prime}=\frac{1}{\sqrt{x}}=\frac{2}{y}$; when $y<0, y=-2 \sqrt{x}, y^{\prime}=$ $-\frac{1}{\sqrt{x}}=\frac{2}{y}$. In summary, the slope of the tangent line at any point $(x, y)$ on curve $C$ except the origin $(0,0)$ is $\frac{2}{y}$. Let $l_{1}$ and $l_{2}$ be two perpendicular tangent lines of curve $C$, with points of tangency $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, respectively. Then their equations are
$$
\begin{array}{l}
l_{1}: y=\frac{2}{y_{1}}\left(x-x_{1}\right)+y_{1} \\
l_{2}: y=\frac{2}{y_{2}}\left(x-x_{2}\right)+y_{2} .
\end{array}
$$

The x-coordinate of the intersection point satisfies
$$
\begin{aligned}
& 2 x\left(\frac{1}{y_{1}}-\frac{1}{y_{2}}\right)+2\left(\frac{x_{2}}{y_{2}}-\frac{x_{1}}{y_{1}}\right)+\left(y_{1}-y_{2}\right) \\
= & 2 x\left(\frac{y_{2}-y_{1}}{y_{1} y_{2}}\right)+2\left(\frac{y_{2}}{4}-\frac{y_{1}}{4}\right)-\left(y_{2}-y_{1}\right) \\
= & \left(y_{2}-y_{1}\right)\left(\frac{2 x}{y_{1} y_{2}}+\frac{1}{2}-1\right) \\
= & 0 .
\end{aligned}
$$

Since $l_{1} \perp l_{2}$, we know $y_{1} y_{2}=-4$, and because $y_{2} \neq y_{1}$, we have $-\frac{x}{2}+\frac{1}{2}-1=0$, which gives $x=$ -1. Therefore, the line $x=-1$ is the required trajectory.

Note: This problem should also prove that any point on the line $x=-1$ is the intersection of two perpendicular tangent lines of curve $C$, but this is not required here, and obtaining $x=-1$ is sufficient for full marks.