15. (12 points) $F(1,0)$ 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 $2 \overrightarrow{P N} + \overrightarrow{N M} = 0$, find:
(1) The equation of the trajectory curve $C$ of point $N$;
(2) The trajectory of the intersection points of any two mutually perpendicular tangents to curve $C$.

(1) Notice that $\overrightarrow{P M}=(a,-b), \overrightarrow{P F}=(1,-b)$.
From $\overrightarrow{P M} \cdot \overrightarrow{P F}=0$, we get $a+b^{2}=0$.
Let point $N(x, y)$. Then
$\overrightarrow{P N}=(x, y-b), \overrightarrow{N M}=(a-x,-y)$.
Thus, $2(x, y-b)+(a-x,-y)=0$, which means $a+x=0, y-2 b=0$.
Therefore, $x=-a=b^{2}=\frac{y^{2}}{4}$.
Hence, the equation of curve $C$ is $y^{2}=4 x$.
(2) For curve $C$:
When $y>0$, we have
$y=2 \sqrt{x}, y'=\frac{1}{\sqrt{x}}=\frac{2}{y}$;
When $y<0$, we have
$$
y=-2 \sqrt{x}, y'=-\frac{1}{\sqrt{x}}=\frac{2}{y} \text {. }
$$

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 $x$ satisfies
$$
\begin{array}{l}
2 x\left(\frac{1}{y_{1}}-\frac{1}{y_{2}}\right)+2\left(\frac{x_{2}}{y_{2}}-\frac{x_{1}}{y_{1}}\right)+\left(y_{1}-y_{2}\right) \\
=2 x\left(\frac{y_{2}-y_{1}}{y_{1} y_{2}}\right)+2\left(\frac{y_{2}}{4}-\frac{y_{1}}{4}\right)-\left(y_{2}-y_{1}\right) \\
=\left(y_{2}-y_{1}\right)\left(\frac{2 x}{y_{1} y_{2}}+\frac{1}{2}-1\right)=0 .
\end{array}
$$

Since $l_{1} \perp l_{2}$, we know $y_{1} y_{2}=-4$.
Also, $y_{2} \neq y_{1}$, so
$$
-\frac{x}{2}+\frac{1}{2}-1=0 \Rightarrow x=-1 \text {. }
$$

Therefore, the line $x=-1$ is the required trajectory.