Question

10．在直角坐標(biāo)系xoy中，曲線C₁的參數(shù)方程為$\left\{\begin{array}{l}x=2\sqrt{2}-\frac{{\sqrt{2}}}{2}t\\ y=\sqrt{2}+\frac{{\sqrt{2}}}{2}t\end{array}$（t為參數(shù)）．在以坐標(biāo)原點為極點，x軸正半軸為極軸的極坐標(biāo)系中，曲線C₂：ρ=4$\sqrt{2}$sinθ．
（Ⅰ）將C₂的方程化為直角坐標(biāo)方程；
（Ⅱ）設(shè)C₁，C₂交于A，B兩點，點P的坐標(biāo)為$（\sqrt{2}，2\sqrt{2}）$，求|PA|+|PB|．

Answer 1

分析 （Ⅰ）利用ρ²=x²+y²，ρcosθ=x，ρsinθ=y，能求出C₂的直角坐標(biāo)方程．
（2）將$\left\{\begin{array}{l}x=2\sqrt{2}-\frac{{\sqrt{2}}}{2}t\\ y=\sqrt{2}+\frac{{\sqrt{2}}}{2}t\end{array}$轉(zhuǎn)化為$\left\{\begin{array}{l}{x=\sqrt{2}-\frac{\sqrt{2}}{2}t}\\{y=2\sqrt{2}+\frac{\sqrt{2}}{2}t}\end{array}\right.$，（t為參數(shù)）．把$\left\{\begin{array}{l}{x=\sqrt{2}-\frac{\sqrt{2}}{2}t}\\{y=2\sqrt{2}+\frac{\sqrt{2}}{2}t}\end{array}\right.$代入${x}^{2}+（y-2\sqrt{2}）^{2}=8$，得t²-2t-6=0，由此能求出|PA|+|PB|．

解答 解：（Ⅰ）∵曲線C₂：ρ=4$\sqrt{2}$sinθ，∴${ρ}^{2}=4\sqrt{2}ρsinθ$，
∴C₂的直角坐標(biāo)方程為：${x}^{2}+{y}^{2}=4\sqrt{2}y$，即${x}^{2}+（y-2\sqrt{2}）^{2}=8$．
（2）將$\left\{\begin{array}{l}x=2\sqrt{2}-\frac{{\sqrt{2}}}{2}t\\ y=\sqrt{2}+\frac{{\sqrt{2}}}{2}t\end{array}$轉(zhuǎn)化為$\left\{\begin{array}{l}{x=\sqrt{2}-\frac{\sqrt{2}}{2}t}\\{y=2\sqrt{2}+\frac{\sqrt{2}}{2}t}\end{array}\right.$，（t為參數(shù)）．
把$\left\{\begin{array}{l}{x=\sqrt{2}-\frac{\sqrt{2}}{2}t}\\{y=2\sqrt{2}+\frac{\sqrt{2}}{2}t}\end{array}\right.$代入${x}^{2}+（y-2\sqrt{2}）^{2}=8$，
得t²-2t-6=0，
則t₁+t₂=2，t₁t₂=-6，
∴|PA|+|PB|=$\sqrt{（{t}_{1}+{t}_{2}）^{2}-4{t}_{1}{t}_{2}}$=$\sqrt{4+24}=2\sqrt{7}$．

點評 本題考查圓的直角坐標(biāo)方程的求法，考查兩線段和的求法，考查極坐標(biāo)方程、參數(shù)方程、直角坐標(biāo)方程等基礎(chǔ)知識，考查推理論證能力、運算求解能力，考查函數(shù)與方程思想、化歸與轉(zhuǎn)化思想，是中檔題．