已知椭圆$C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)$的离心率为$\dfrac{\sqrt{3}}{2}$，$A(a,0),\ B(0,b),\ O(0,0)$，$\triangle{OAB}$的面积为$1$．
(1)求椭圆$C$的方程；
(2)设$P$是椭圆$C$上一点，直线$PA$与$y$轴交于点$M$，直线$PB$与$x$轴交于点$N$．求证：$\left|AN\right|\cdot\left|BM\right|$为定值．
解 (1)$\dfrac{x^2}{4}+y^2=1$．
(2)设$P$点坐标为$(2\cos{\theta},\sin\theta)$，可求得$M$点坐标为$\left(0,\dfrac{\sin\theta}{1-\cos\theta}\right)$，$N$点坐标为$\left(\dfrac{2\cos\theta}{1-\sin\theta},0\right)$，故\[\begin{split}\left|AN\right|\cdot\left|BM\right|&=\left|\left(\dfrac{2\cos\theta}{1-\sin\theta}-2\right)\left(\dfrac{\sin\theta}{1-\cos\theta}-1\right)\right|\\&=2\left|\dfrac{\left(\sin\theta+\cos\theta -1\right)^2}{\left(1-\sin\theta\right)\left(1-\cos\theta\right)}\right|\\&=4.\end{split}\]