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高考数学必做百题第98题(理科2017版)

  2016-10-11 08:22:55  

098.已知曲线${{C}_{1}}$的参数方程为$\left\{ \begin{align}  & x=4+5\cos t \\ & y=5+5\sin t \\ \end{align} \right.$(t为参数),以坐标原点为极点, $x$轴的正半轴为极轴建立极坐标系,曲线${{C}_{2}}$的极坐标方程为$\rho =2\sin \theta $.

(1)把${{C}_{1}}$的参数方程化为极坐标方程;

(2)求${{C}_{1}}$与${{C}_{2}}$交点的极坐标$\left( \rho \ge 0,\ \ 0\le \theta <2\pi  \right)$。

解:(1)将$\left\{ \begin{align}  & x=4+5\cos t \\ & y=5+5\sin t \\ \end{align} \right.$消去参数,化为普通方程${{(x-4)}^{2}}+{{(y-5)}^{2}}=25$, 

即${{C}_{1}}$:${{C}_{1}}:{{x}^{2}}+{{y}^{2}}-8x-10y+16=0$,

把$\left\{ \begin{align}  & x=\rho \cos \theta  \\ & y=\rho \sin \theta  \\ \end{align} \right.$代入${{C}_{1}}:{{x}^{2}}+{{y}^{2}}-8x-10y+16=0$得 ${{\rho }^{2}}-8\rho \cos \theta -16\rho \sin \theta +16=0$, 

∴${{C}_{1}}$的极坐标方程为

${{\rho }^{2}}-8\rho \cos \theta -16\rho \sin \theta +16=0$; 

(2)${{C}_{2}}$的普通方程为${{x}^{2}}+{{y}^{2}}-2y=0$, 

由$\left\{ \begin{align}  & {{x}^{2}}+{{y}^{2}}-8x-10y+16=0 \\ & {{x}^{2}}+{{y}^{2}}-2y=0 \\ \end{align} \right.$

解得$\left\{ \begin{align}  & x=1 \\ & y=1 \\ \end{align} \right.$或$\left\{ \begin{align}  & x=0 \\ & y=2 \\ \end{align} \right.$,∴${{C}_{1}}$与${{C}_{2}}$的交点的极坐标分别为$(\sqrt{2},\dfrac{\pi }{4}),(2,\dfrac{\pi }{2})$ . 

另解:联立$\left\{ \begin{matrix}   {{\rho }^{2}}\text{-}8\rho \cos \theta -10\rho \sin \theta +16=0  \\   \rho =2\sin \theta \begin{matrix}   {} & {} & {} & \begin{matrix}   {} & {} & \quad   \\\end{matrix}  \\\end{matrix}  \\\end{matrix} \right.$,

代入化简得$\cos \theta \left( \sin \theta -\cos \theta  \right)=0$,

解得$\left\{ \begin{matrix}   \theta =\dfrac{\pi }{2}  \\   \rho =2  \\\end{matrix} \right.$或$\left\{ \begin{matrix}   \theta =\dfrac{\pi }{4}  \\   \rho =\sqrt{2}  \\\end{matrix} \right.$。



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