x^2/40+y^2/10=1内一点M(4,-1)

来源：百度知道编辑：UC知道时间：2024/06/19 08:21:35

x^2/40+y^2/10=1内一点M(4,-1)引弦AB,使AB被点M平分,求弦AB的长

AB:y=kx+b......(1)
x^2/40+y^2/10=1......(2)
(1)代入（2），整理得
(1+4k^2)x^2+8kbx+4b^2-40=0
x1+x2=-8kb/(1+4k^2)
已知点M(4,-1)平分引弦AB
(x1+x2)/2=-4kb/(1+4k^2)=4
4k^2+kb+1=0......(3)

y1+y2=2b+k(x1+x2)
(y1+y2)/2=b+k(x1+x2)/2=b-4bk^2/(1+4k^2)=-1
4k^2+b+1=0......(4)
(3)-(4)
(k-1)b=0
b=0不符合已知条件
k=1
y=x+b
-1=4+b
b=-5

y=x-5
x^2/40+y^2/10=1
x1=2,x2=6
y1=-3,y2=1

AB=4√2

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