（5分）椭圆$C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a > b > 0)$的左顶点为$A$，点$P$，$Q$均在$C$上，且关于$y$轴对称．若直线$AP$，$AQ$的斜率之积为$\dfrac{1}{4}$，则$C$的离心率为(　　)
A．$\dfrac{\sqrt{3}}{2}$              B．$\dfrac{\sqrt{2}}{2}$              C．$\dfrac{1}{2}$              D．$\dfrac{1}{3}$
分析：设$P(x_{0}$，$y_{0})$，则$Q(-x_{0}$，$y_{0})$，根据斜率公式结合题意可得：$k_{AP}\cdot k_{AQ}=\dfrac{1}{4}$，再结合$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$，整理可得离心率．
解答：解：已知$A(-a,0)$，设$P(x_{0}$，$y_{0})$，则$Q(-x_{0}$，$y_{0})$，
$k_{AP}=\dfrac{{y}_{0}}{{x}_{0}+a}$，
$k_{AQ}=\dfrac{{y}_{0}}{a-{x}_{0}}$，
故$k_{AP}\cdot k_{AQ}=\dfrac{{y}_{0}}{{x}_{0}+a}\cdot \dfrac{{y}_{0}}{a-{x}_{0}}=\dfrac{{y}_{0}^{2}}{a^{2}-{x}_{0}^{2}}=\dfrac{1}{4}$①，
$\because$$\dfrac{{x}_{0}^{2}}{a^{2}}+\dfrac{{y}_{0}^{2}}{b^{2}}=1$，即${y}_{0}^{2}=\dfrac{b^{2}(a^{2}-{x}_{0}^{2})}{a^{2}}$②，
②代入①整理得：$\dfrac{b^{2}}{a^{2}}=\dfrac{1}{4}$，
$e=\dfrac{c}{a}=\sqrt{1-\dfrac{b^{2}}{a^{2}}}=\dfrac{\sqrt{3}}{2}$．
故选：$A$．
解答：本题考查椭圆的简单几何性质，是基础题．