设椭圆方程为x^2/a^2+y^2/b^2=1(a>b>0),因离心率为（根号3）/2，故（a^2-b^2)/a^2=3/4,a^2=4b^2.(1)l的方程为y=x+1代入x^2/(4b^2)+y^2/b^2=1化简得5x^2+8x+4-4b^2=0,设P(x1,y1),Q(x2,y2),则x1+x2=-8/5,x1x2=(4-4b^2)/5.由向量PM=（-3/5）向量QM得(-1 展开
设椭圆方程为x^2/a^2+y^2/b^2=1(a>b>0),因离心率为（根号3）/2，故（a^2-b^2)/a^2=3/4,a^2=4b^2.(1)l的方程为y=x+1代入x^2/(4b^2)+y^2/b^2=1化简得5x^2+8x+4-4b^2=0,设P(x1,y1),Q(x2,y2),则x1+x2=-8/5,x1x2=(4-4b^2)/5.由向量PM=（-3/5）向量QM得(-1-x1,-y1)=(-3/5)*(-1-x2,-y2),∴-1-x1=（-3/5）（-1-x2),∴5x1+3x2=-8,解得x1=-8/5,x2=0.∴b^2=1.所求椭圆方程为x^2/4+y^2=1.(1)(2)A(2,0).l的方程为y=(x+1)tanα,(2)代入(1),化简得[1+4(tanα)^2]x^2+8x(tanα)^2+4(tanα)^2-4=0,则x1+x2=-8(tanα)^2/[1+4(tanα)^2],x1x2=[4(tanα)^2-4]/[1+4(tanα)^2].由（2），y1y2=(x1+1)(x2+1)(tanα)^2=(x1x2+x1+x2+1)(tanα)^2,AP·AQ=(x1-2,y1)·(x2-2,y2)=(x1-2)(x2-2)+y1y2=x1x2*[1+(tanα)^2]+[-2+(tanα)^2](x1+x2)+4+(tanα)^2={[4(tanα)^2-4]*[1+(tanα)^2]-8(tanα)^2*[-2+(tanα)^2]}/[1+4(tanα)^2]+4+(tanα)^2=33(tanα)^2/[1+4(tanα)^2]=33{1/4-1/[4+16(tanα)^2]},∴当α=π/2时AP·AQ的最大值为33/4. 收起
