前言
典例剖析
法1:在直角坐标系下思考计算,
将曲线\(C_1\)转化为普通方程为\(x^2+(y-1)^2=1\),
将曲线\(C_2\)转化为普通方程为\((x-\sqrt{3})^2+y^2=3\),
将二者联立,得到\(\left\{\begin{array}{l}{x^2+(y-1)^2=1}\\{(x-\sqrt{3})^2+y^2=3}\end{array}\right.\)
解得两个交点的坐标为\((0,0)\)和\((\cfrac{\sqrt{3}}{2},\cfrac{3}{2})\);
法2:在极坐标系下思考计算,
将二者联立,得到\(\left\{\begin{array}{l}{\rho=2sin\theta}\\{\rho=2\sqrt{3}cos\theta}\end{array}\right.\quad\) 消去\(\rho\),
得到\(tan\theta=\sqrt{3}\),即\(\theta=\cfrac{\pi}{3}\)。
则