设 $x_0>0$，$x_0 \neq \sqrt{3}$，$ Q\left(x_0, 0\right)$，$P(0,4)$，直线 $P Q$ 与双曲线 $x^2-\dfrac{y^2}{3}=1$ 相交于 $A,B$ 两点．若 $\overrightarrow{P Q}=t \overrightarrow{Q A}=(2-t) \overrightarrow{Q B}$，则 $x_0=$______．
答案 $\dfrac{\sqrt 2}2$．
解析 设 $A(x_1,y_1)$，$B(x_2,y_2)$，则\[\overrightarrow{P Q}=t \overrightarrow{Q A}=(2-t) \overrightarrow{Q B}\iff -4=ty_1=(2-t)y_2,\]因此\[\dfrac{-4}{y_1}+\dfrac{-4}{y_2}=2\iff \dfrac 1{y_1}+\dfrac 1{y_2}=-\dfrac 12,\]设 $PQ:~\dfrac x{x_0}+\dfrac y4=1$，与双曲线方程联立，可得\[x_0^2\left(1-\dfrac y4\right)^2-\dfrac 13y^2-1=0\iff \left(\dfrac{x_0^2}{16}-\dfrac 13\right)y^2-\dfrac 12x_0^2y+x_0^2-1=0,\]从而\[\dfrac{\dfrac 12x_0^2}{x_0^2-1}=-\dfrac 12\iff x_0=\dfrac{\sqrt 2}2.\]