1) m=-1 => A(8/5,-3/5) B(0,1) AB=(8根号2)/5
y=kx=> C D(2/根号(1+4k^2),2k/根号(1+4k^2)) (-2/根号(1+4k^2),-2k/根号(1+4k^2))
四边形ACBD面积=AB×(C点到直线l:x=-y+1的距离+D点到直线l:x=my+1的距离)/2
=(8根号2)/5*{[(2+2k)/根号(1+4k^2)+1]+[(2+2k)/根号(1+4k^2)-1]}/根号2/2
=16(1+k)/根号(1+4k^2)/5
求出该表达式的最大值即可.
2)x=my+1代入椭圆方程
=>(1+m^2/4)y^2+(m/2)y-3/4=0
=>y1y2=-3/(4+m^2)
x=my+1
=>y=(x-1)/m
代入椭圆方程
=>(1+m^2)x^2-2x+(1-m^2)=0
=>x1x2=8/(4+m^2) x1x2=4(1-m^2)/(4+m^2)
假设M(a,0)
直线MA与直线MB的斜率之积=y1/(x1-a)*y2/(x2-a)
=y1y2/[(x1-a)(x2-a)]
=-3/(4+m^2)/[(4-4m^2)/(4+m^2)-8a/(4+m^2)+a^2]
=-3/[(a^2-4)m^2+4a^2-8a+4]
要想上式为常数,a^2-4=0 且4a^2-8a+4不等于0 => a=-2,定点(-2,0)