15. Given point $B(0,1), P, Q$ are any two points on the ellipse $\frac{x^{2}}{4}+y^{2}=1$ different from point $B$, and $B P \perp B Q$.
(1) If the projection of point $B$ on the line segment $P Q$ is $M$, find the equation of the trajectory of $M$;
(2) Find the range of the x-intercept of the perpendicular bisector $l$ of the line segment $P Q$.
15. (1) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then the equation of $P Q$ is $y=k x+m$.
Combining with the ellipse equation, we get
$$
\begin{array}{l}
\left(1+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-4=0 \\
\Rightarrow x_{1}+x_{2}=\frac{-8 k m}{4 k^{2}+1}, x_{1} x_{2}=\frac{4 m^{2}-4}{4 k^{2}+1} .
\end{array}
$$

Since $B P \perp B Q$
$\Rightarrow \frac{y_{1}-1}{x_{1}} \cdot \frac{y_{2}-1}{x_{2}}=-1$
$\Rightarrow x_{1} x_{2}+y_{1} y_{2}-\left(y_{1}+y_{2}\right)+1=0$
$\Rightarrow \frac{\left(k^{2}+1\right)\left(4 m^{2}-4\right)}{4 k^{2}+1}+k(m-1) \frac{-8 k m}{4 k^{2}+1}+$
$(m-1)^{2}=0$
$\Rightarrow 5 m^{2}-2 m-3=0$
$\Rightarrow m=-\frac{3}{5}$ or 1 (discard).
Let $M(x, y)$.
Since $B M \perp P Q$, we have $k=-\frac{x}{y-1}$.
Substituting into the equation of $P Q$ gives
$y=-\frac{x^{2}}{y-1}-\frac{3}{5}$
$\Rightarrow x^{2}+\left(y-\frac{1}{5}\right)^{2}=\left(\frac{4}{5}\right)^{2}(y \neq 1)$.
(2) Since the equation of $P Q$ is $y=k x-\frac{3}{5}$, we have
$$
\frac{x_{1}+x_{2}}{2}=\frac{12 k}{5\left(4 k^{2}+1\right)}, \frac{y_{1}+y_{2}}{2}=\frac{-3}{5\left(4 k^{2}+1\right)} \text {. }
$$

Thus, the equation of the perpendicular bisector of $P Q$ is
$$
y+\frac{3}{5\left(4 k^{2}+1\right)}=-\frac{1}{k}\left[x-\frac{12 k}{5\left(4 k^{2}+1\right)}\right] \text {, }
$$

and its $x$-intercept $b=\frac{9 k}{5\left(4 k^{2}+1\right)}$.
Therefore, $-\frac{9}{20} \leqslant b \leqslant \frac{9}{20}$.