2 直线PA为y=b/(a-c) * x-bc/(a-c) a=2cy=b/c*x-b带入椭圆方程得A交点为x=8/5 * c y=3/5 * bB为（0，-b）设M（x,y）AM=(x-8/5*c,y-3/5*b)BM=(x,y+b)AM*BM=-2(x-8/5*c)*x+(y-3/5*b)*(y+b)=-2x^2-8/5cx+y^2+2/5by-3/5b^2+2=0
