Question

4．橢圓$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{3}$=1上一動點P，圓E：（x-1）²+y²=1，過圓心E任意作一條直線與圓E交于A，B兩點，圓F：（x+1）²+y²=1，過圓心F任意作一條直線與圓F交于C，D兩點，則$\overrightarrow{PA}•\overrightarrow{PB}$+$\overrightarrow{PC}•\overrightarrow{PD}$最小值（　�。�

• A． 4
• B． 6
• C． 8
• D． 9

Answer 1

分析 如圖所示，由于$\overrightarrow{PA}$=$\overrightarrow{AE}+\overrightarrow{EP}$，$\overrightarrow{PB}$=$\overrightarrow{BE}+\overrightarrow{EP}$，$\overrightarrow{BE}+\overrightarrow{AE}$=$\overrightarrow{0}$，代入可得$\overrightarrow{PA}$$•\overrightarrow{PB}$=${\overrightarrow{EP}}^{2}$-1，同理可得：$\overrightarrow{PC}•\overrightarrow{PD}$=${\overrightarrow{FP}}^{2}$-1．由于$|\overrightarrow{PE}|+|\overrightarrow{PF}|$=4，利用基本不等式的性質(zhì)即可得出．

解答 解：如圖所示，
∵$\overrightarrow{PA}$=$\overrightarrow{AE}+\overrightarrow{EP}$，$\overrightarrow{PB}$=$\overrightarrow{BE}+\overrightarrow{EP}$，$\overrightarrow{BE}+\overrightarrow{AE}$=$\overrightarrow{0}$，
∴$\overrightarrow{PA}$$•\overrightarrow{PB}$=（$\overrightarrow{AE}+\overrightarrow{EP}$）•（$\overrightarrow{BE}+\overrightarrow{EP}$）=$\overrightarrow{AE}•\overrightarrow{BE}$+${\overrightarrow{EP}}^{2}$+$\overrightarrow{EP}（\overrightarrow{BE}+\overrightarrow{AE}）$=${\overrightarrow{EP}}^{2}$-1，
同理可得：$\overrightarrow{PC}•\overrightarrow{PD}$=${\overrightarrow{FP}}^{2}$-1．
∵$|\overrightarrow{PE}|+|\overrightarrow{PF}|$=4，
∴$\overrightarrow{PA}•\overrightarrow{PB}$+$\overrightarrow{PC}•\overrightarrow{PD}$=${\overrightarrow{EP}}^{2}$-1+${\overrightarrow{FP}}^{2}$-1=${\overrightarrow{EP}}^{2}$+${\overrightarrow{FP}}^{2}$-2≥$\frac{（|\overrightarrow{PE}|+|\overrightarrow{PF}|）^{2}}{2}$-2=6．當(dāng)且僅當(dāng)$|\overrightarrow{PE}|$=$|\overrightarrow{PF}|$=2時取等號．
∴$\overrightarrow{PA}•\overrightarrow{PB}$+$\overrightarrow{PC}•\overrightarrow{PD}$最小值是6．
故選：B．

點評 本題考查了橢圓的定義標(biāo)準(zhǔn)方程及其性質(zhì)、向量的三角形法則、基本不等式的性質(zhì)，考查了推理能力與計算能力，屬于中檔題．