马长朋
花朵
共回答了14个问题采纳率:100% 举报
由x+2y-2=0,得
x=2y-2.(1)
x^2/a^2+y^2/b^2=1.(2)
(1)代入(2),得
(a^2+4b^2)y^2-8b^2y+4b^2-a^2*b^2=0
y=[4b^2±ab√(a^2+4b^2-4)]/(a^2+4b^2)
y1-y2=2ab√(a^2+4b^2-4)]/(a^2+4b^2),y1+y2=8b^2/(a^2+4b^2)
x1-x2=2(y1-y2)
AB=√5
(x1-x2)^2+(y1-y2)^2=AB^2
[2(y1-y2)]^2+(y1-y2)^2=AB^2
5(Y1-Y2)^2=5
[2ab√(a^2+4b^2-4)]/(a^2+4b^2)]^2=1.(3)
AB中点坐标为(m,0.5)
(y1+y2)/2=0.5
y1+y2=1
8b^2/(a^2+4b^2)=1
a^2=4b^2,a>b>0
a=2b.(4)
(4)代入(3),得
b=1,a=2
可知此椭圆的方程为:x^2/2+y^2=1,即x^2+2y^2=2
1年前
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