Question

已知直线l：y=kx+b与椭圆C：3x^2+y^2=3相交于A、B两点，,M是的线段AB的中点,O是坐标原点

已知直线l：y=kx+b与椭圆C：3x^2+y^2=3相交于A、B两点，,M是的线段AB的中点,O是坐标原点

(1)当l与直线x+y=0平行(不重合)时，求直线OM的斜率
(2)如果|OM|=1，证明b^2=((k^2)+3)^2/((k^2)+9),并求线段AD长取最大值时直线l的方程
要详细过程 谢谢！！！

Answer 1

1）
l和直线x+y=0平行
k=-1
l:y=-x+b
3x^2+(b-x)^2=3
4x^2-2bx=3
x1+x2=2b/4=b/2
Mx=(x1+x2)/2=b/4
My=-Mx+b=3b/4
OM 斜率k'=My/Mx=3

2
3x^2+(kx+b)^2=3
(3+k^2)x^2+2bkx=3-b^2
x1+x2=-2bk/((3+k^2)
Mx=(x1+x2)/2=-bk/(3+k^2)
My=kMx+b=-bk^2/(3+k^2)+b
|OM|^2=Mx^2+My^2
|OM|=1
(bk)^2/(3+k^2)^2 +[-bk^2/(3+k^2)+b]^2=1
b^2k^2+b^2k^4 -2b^2k^4(3+k^2)+b^2(3+k^2)^2=(3+k^2)^2
b^2=(k^2+3)^2/(k^2+9)

3
求线段AD? 长

追问

错了 求AB的长度 嘿嘿

追答

|AB|^2=(x1-x2)^2+(y1-y2)^2
|OM|=1
AB在y=1直线上时,|AB|最大
y=1,3x^2+1=3
x1=√6/3,x2=-√6/3
|AB|最大=|x1-x2|=2√6/3

Answer 2

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