（5分）设椭圆$C_{1}:\dfrac{x^2}{a^2}+y^{2}=1(a > 1)$，$C_{2}:\dfrac{x^2}{4}+y^{2}=1$的离心率分别为$e_{1}$，$e_{2}$．若$e_{2}=\sqrt{3}e_{1}$，则$a=($　　$)$
A．$\dfrac{2\sqrt{3}}{3}$              B．$\sqrt{2}$              C．$\sqrt{3}$              D．$\sqrt{6}$
答案：$A$
分析：利用椭圆$C_{2}:\dfrac{x^2}{4}+y^{2}=1$的方程可求其离心率$e_{2}$，进而可求$e_{1}$，可求$a$．
解：由椭圆$C_{2}:\dfrac{x^2}{4}+y^{2}=1$可得$a_{2}=2$，$b_{2}=1$，$\therefore c_{2}=\sqrt{4-1}=\sqrt{3}$，
$\therefore$椭圆$C_{2}$的离心率为$e_{2}=\dfrac{\sqrt{3}}{2}$，
$\because e_{2}=\sqrt{3}e_{1}$，$\therefore e_{1}=\dfrac{1}{2}$，$\therefore$$\dfrac{{c}_{1}}{{a}_{1}}=\dfrac{1}{2}$，
$\therefore$${a}_{1}^{2}=4{c}_{1}^{2}=4({a}_{1}^{2}-{b}_{1}^{2})=4({a}_{1}^{2}-1)$，
$\therefore a=\dfrac{2\sqrt{3}}{3}$或$a=-\dfrac{2\sqrt{3}}{3}$（舍去）．
故选：$A$．
点评：本题考查椭圆的几何性质，考查运算求解能力，属基础题．