(1)易知a=2,b=1,c= ∴F1(-,0),F2(,0) ∴离心率e=,椭圆的准线方程为x=± (2)解法一:设P(x,y),则 •=(--x,-y)•(-x,-y)=x2+y2-3 =x2+1--3 = 因为x∈[-2,2] 故当x=0,即点P为椭圆短轴端点时,•有最小值-2; 当x=±2,即点P为椭圆长轴端点时,,•有最大值1. 解法二: (2)易知a=2,b=1,c= ∴F1(-,0),F2(,0) 设P(x,y),则,•=||•||•cos∠F1PF2 =||||• =[(x+)2+y2+(x-)2+y2-12] =x2+y2-3 (以下同解法一). (3)显然直线x=0不满足题设条件. 可设直l:y=kx-2,A(x1,y1),B(x2,y2) 联立,消去y,整理得:(k2+)x2 +4kx+3=0 ∴x1+x2=-,x1x2= 由△=(4k)2-4(k 2+ )×3=4k2-3>0得:k<或k>① 又∵0°<∠AOB<90° ∴cos∠AOB>0 ∴•=x1x2+y1y2>0 又∵y1y2=(kx1+2)(kx2+2)=k2x1x2+2k(x1+x2)+4 =+= ∵+>0,即k2<4, ∴-2<k<2② 故由①②得-2<k<-,或<k<2. |