已知方程组x*x/4+y*y=1 y=kx+m 有两组解(x1,y1) (x2,y2)并且有(x1-2)(x2-2)+y

x^2/4+y^2=1 ①
y=kx+m ②
①两边乘以4,得
x^2+4y^2=4 ③
将②代入③,得
x^2+(kx+m)^2=4
整理,得
(k^2+1)x^2+2mkx+m^2-4=0
x1+x2=-b/a=-2mk/(k^2+1)
x1*x2=c/a=(m^2-4)/(k^2+1) ④
(x1-2)(x2-2)+y1y2=0
x1*x2-2(x1+x2)+4+y1y2=0 ⑤
将④代入直线方程中,得
y1*y2=[k(m^2-4)]/(k^2+1)
代入⑤中,得
(m^2-4)/(k^2+1)+2×[(2mk)/(k^2+1)]+4+[k(m^2-4)]/(k^2+1)=0
(m^2-4+4mk+km^2-4k)/(k^2+1)+4=0
[(k+1)m^2+4mk-4k-4]/(k^2+1)+4=0
[(k+1)m^2+4mk-4k-4]/(k^2+1)+(4k^2+4)/(k^2+1)=0
[(k+1)m^2+4k^2+4mk-4k]/(k^2+1)=0
(k+1)m^2+4k^2+4mk-4k=0
k=[(k+1)m^2+4k^2+4mk]/4

haonan