15. Given point $B(0,1)$, and points $P$ and $Q$ are any two points on the ellipse $\frac{x^{2}}{4}+y^{2}=1$ different from point $B$, and $B P \perp B Q$.
(I) If the projection of point $B$ on the line segment $P Q$ is point $M$, find the equation of the trajectory of $M$;
(II) Find the range of the $x$-intercept of the perpendicular bisector $l$ of line segment $P Q$.

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Solution: (I) Let $P\left(x_{1}, y_{1}\right)$, $Q\left(x_{2}, y_{2}\right)$, and the equation of $PQ$ be $y=k x+m$. By combining this with the equation of the ellipse and eliminating $y$, we get: $\left(1+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-4=0$,

Thus, $x_{1}+x_{2}=\frac{-8 k m}{4 k^{2}+1}, x_{1} x_{2}=\frac{4 m^{2}-4}{4 k^{2}+1}$,
From $B P \perp B Q$ we get $\frac{y_{1}-1}{x_{1}} ? \frac{y_{2}-1}{x_{2}}-1$, which means $x_{1} x_{2}+y_{1} y_{2}-\left(y_{1}+y_{2}\right)+1=0$,
Thus, $\frac{\left(k^{2}+1\right)\left(4 m^{2}-4\right)}{4 k^{2}+1}+k(m-1) ? \frac{-8 k m}{4 k^{2}+1} \quad(m-1)^{2}=0$
Simplifying, we get $5 m^{2}-2 m-3=0$, solving which gives $m=1$ (discard) or $m=-\frac{3}{5}$.
Let $M(x, y)$, since $B M^{\wedge} P Q$, we have $k=-\frac{x}{y-1}$,
Substituting into the equation of $PQ$ gives $y=-\frac{x^{2}}{y-1}-\frac{3}{5}$,
Rearranging, we get $x^{2}+\left(y-\frac{1}{5}\right)^{2}=\left(\frac{4}{5}\right)^{2}$, and by the problem statement, the trajectory does not pass through point $B(0,1)$.
Therefore, the equation of the trajectory of the moving point $M$ is: $x^{2}+\left(y-\frac{1}{5}\right)^{2}=\left(\frac{4}{5}\right)^{2}(y \neq 1)$.
(II) The equation of $PQ$ is $y=k x-\frac{3}{5}$, so $\frac{x_{1}+x_{2}}{2}=\frac{12 k}{5\left(4 k^{2}+1\right)}, \frac{y_{1}+y_{2}}{2}=\frac{-3}{5\left(4 k^{2}+1\right)}$

Thus, the equation of the perpendicular bisector of $PQ$ is $y+\frac{3}{5\left(4 k^{2}+1\right)}=-\frac{1}{k}\left(x-\frac{12 k}{5\left(4 k^{2}+1\right)}\right)$,
Its intercept on the $x$-axis is $b=\frac{9 k}{5\left(4 k^{2}+1\right)}$, so, $-\frac{9}{20} \leq b \leq \frac{9}{20}$.
(20 points)