如图,椭圆c1:x^2/4+y^2=1,曲线c2:y=x^2-1与y轴的交点为m,过坐标原点的O的直线l与C2相交于点A

上午上课,刚看到题,好难
(1)
M(0,-1)
设AB：y=kx 代入y=x²-1
得:x²-kx-1=0
设A(x1,y1),B(x2,y2)
∴x1+x2=k,x1x2=-1
∴MA=(x1,y1+1),MB=(x2,y2+1)
∴MA●MB=x1x2+(y1+1)(y2+1)
=-1+(kx1+1)(kx2+1)
=-1+k²x1x2+k(x1+x2)+1
=-k²+k²=0
∵M,A,D共线,M,B,E共线
∴MD●ME=0
（2）
MA:y=(k+1/x1)x-1代入x²/4+y²=1
x²+4[(k+1/x1)x-1]²-4=0
即[1+4(k+1/x1)²]x²-8(k+1/x1)x=0
解得xD=8(k+1/x1)/[1+4(k+1/x1)²]=8(k-x2)/[1+4(k-x2)²]
同理xE=8(k+1/x2)/[1+4(k+1/x2)²]=8(k-x1)/[1+4(k-x1)²]
²(k-x1)(k-x2)=k²-k(x1+x2)+x1x2=-1
xDxE=64(k-x1)(k-x2)/[16(k-x1)²(k-x2)²+4(k-x1)²+4(k-x2)²+1]
=-64/[16+1+8k²-8k(x1+x2)+4(x²1+x²2)]
=-64/[17+4(x1+x2)²-8x1x2]
=-64/[17+4k²+8]
=-64/(25+4k²)∈[-64/25,0)
λ=S1/S2=|x1x2|/|xDxE|∈(0,25/64]
结果还需验证,先提交做记号