题目 已知点 $B(0,1)$，$P,Q$ 为椭圆 $\dfrac{x^{2}}{4}+y^{2}=1$ 上异于点 $B$ 的任意两点，且 $BP\perp BQ$． 问题1 若点 $B$ 在线段 $PQ$ 上的射影为点 $M$，求点 $M$ 的轨迹方程； 答案1 $x^{2}+\left(y-\dfrac{1}{5}\right)^{2}=\left(\dfrac{4}{5}\right)^{2}(y\ne 1)$ 解析1 设 $P(x_{1},y_{1})$，$Q(x_{2},y_{2})$，$PQ$ 的方程为 $y=kx+m$，与椭圆方程联立消去 $y$ 得\[(1+4k^{2})x^{2}+8kmx+4m^{2}-4=0,\]所以$$x_{1}+x_{2}=\dfrac{-8km}{4k^{2}+1} , x_{1}x_{2}=\dfrac{4m^{2}-4}{4k^{2}+1}.$$由 $BP\perp BQ$ 得$$\dfrac{y_{1}-1}{x_{1}}\cdot \dfrac{y_{2}-1}{x_{2}}=-1,$$即\[x_{1}x_{2}+y_{1}y_{2}-(y_{1}+y_{2})+1=0,\]从而可得\[\dfrac{(k^{2}+1)(4m^{2}-4)}{4k^{2}+1}+k(m-1)\cdot \dfrac{-8km}{4k^{2}+1}+(m-1)^{2}=0,\]化简得$$5m^{2}-2m-3=0,$$解得 $m=1$（舍去）或 $m=-\dfrac{3}{5}$．<br/>设 $M(x,y)$．<br/>因为 $BM\perp PQ$，所以 $k=-\dfrac{x}{y-1}$，代入 $PQ$ 方程得$$y=-\dfrac{x^{2}}{y-1}-\dfrac{3}{5},$$整理得\[x^{2}+\left(y-\dfrac{1}{5}\right)^{2}=\left(\dfrac{4}{5}\right)^{2}.\]由题意知轨迹不经过点 $B(0,1)$．<br/>因此动点 $M$ 的轨迹方程为：$x^{2}+\left(y-\dfrac{1}{5}\right)^{2}=\left(\dfrac{4}{5}\right)^{2}(y\ne 1)$． 备注1 问题2 求线段 $PQ$ 的中垂线 $l$ 在 $x$ 轴上的截距的取值范围． 答案2 $\left[-\dfrac{9}{20},\dfrac{9}{20}\right]$ 解析2 $PQ$ 的方程为 $y=kx-\dfrac{3}{5}$，所以\[\dfrac{x_{1}+x_{2}}{2}=\dfrac{12k}{5(4k^{2}}, \dfrac{y_{1}+y_{2}}{2}=\dfrac{-3}{5(4k^{2}+1)},\]所以 $PQ$ 中垂线方程为\[y+\dfrac{3}{5(4k^{2}+1)}=-\dfrac{1}{k}\left(x-\dfrac{12k}{5(4k^{2}+1)}\right),\]在 $x$ 轴上的截距为$$b=\dfrac{9k}{5(4k^{2}+1)},$$因此 $-\dfrac{9}{20}\leqslant b\leqslant \dfrac{9}{20}$． 备注2