15 Let vectors $\vec{i}$ and $\vec{j}$ be the unit vectors in the positive directions of the $x$-axis and $y$-axis, respectively, in a rectangular coordinate plane. If $\vec{a}=(x+2) \vec{i}+y \vec{j}$, $\vec{b}=(x-2) \vec{i}+y \vec{j}$, and $|\vec{a}|-|\vec{b}|=2$.
(1) Find the equation of the locus of point $P(x, y)$ that satisfies the above conditions;
(2) Let $A(-1,0)$ and $F(2,0)$. Is there a constant $\lambda(\lambda>0)$ such that $\angle P F A=\lambda \angle P A F$ always holds? Prove your conclusion.
15（1）From the condition $|\vec{a}|-|\vec{b}|=2$, we have
$$
\sqrt{(x+2)^{2}+y^{2}}-\sqrt{(x-2)^{2}+y^{2}}=2 .
$$

By the definition of a hyperbola, the equation of the locus of point $P$ is: $x^{2}-\frac{y^{2}}{3}=1(x>0)$.
（2）In the first quadrant, draw $P F \perp x$ axis, the coordinates of point $P$ are $(2,3)$, at this time $\angle P F A=$ $90^{\circ}, \angle P A F=45^{\circ}$, thus $\lambda=2$.

Next, we prove that when point $P$ is in the first quadrant but $P F$ is not perpendicular to the $x$ axis, $\angle P F A=$ $2 \angle P A F$ always holds.
Let $P\left(x_{1}, y_{1}\right)$, then
$$
k_{P A}=\frac{y_{1}}{x_{1}+1}, k_{P F}=\frac{y_{1}}{x_{1}-2}
$$

Thus,
$$
\tan 2 \angle P A F=\frac{2 k_{P A}}{1-\left(k_{P A}\right)^{2}}=\frac{2\left(x_{1}+1\right) y_{1}}{\left(x_{1}+1\right)^{2}-y_{1}^{2}} .
$$

Since point $P$ is on $x^{2}-\frac{y^{2}}{3}=1$, we have
$$
y_{1}^{2}=3\left(x_{1}^{2}-1\right)=3\left(x_{1}+1\right)\left(x_{1}-1\right) .
$$

Substituting this into the above equation and simplifying, we get
$$
\begin{array}{l}
\tan 2 \angle P A F=-\frac{y_{1}}{x_{1}-2}, \\
\tan \angle P F A=-k_{P F}=-\frac{y_{1}}{x_{1}-2},
\end{array}
$$

which means
$$
\tan 2 \angle P A F=\tan \angle P F A,
$$

so
$$
\angle P F A=2 \angle P A F
$$

By symmetry, when point $P$ is in the fourth quadrant, the conclusion also holds.
Therefore, there exists a constant $\lambda=2$ such that $\angle P F A=2 \angle P A F$ always holds.