Question

数学学霸进

数学学霸进第21题第二小题，求解法，不要向量法，求其他接法，最好是传统法，谢谢

Answer 1

21、说明：^——表示次方
(2) C：x^2/4+y^2/2=1
x^2+2y^2=4.............(1)
L：y-2=k(x-0)
y=kx+2................(2)
(2)代入(1)：
x^2+2(kx+2)^2=4
(1+2k^2)x^2+8kx+4=0
M(x1,y1)、N(x2,y2)、MN中点A(x0,y0)
x1+x2=-8k/(1+2k^2)
x1x2=4/(1+2k^2)
x0=(x1+x2)/2

=-4k/(1+2k^2)
y0=kx0+2
=-4k^2/(1+2k^2)+2
=2/(1+2k^2)
A(-4k/(1+2k^2),2/(1+2k^2))
(x1-x2)^2=(x1+x2)^2-4x1x2
=64k^2/(1+2k^2)^2-16/(1+2k^2)
=(64k^2-16-32k^2)/(1+2k^2)^2
=(32k^2-16)/(1+2k^2)^2
=16(2k^2-1)/(1+2k^2)^2
|MN|=√[(1+k^2)(x1-x2)^2]
=√[(1+k^2)·16(2k^2-1)/(1+2k^2)^2]
=4/(1+2k^2)√[(1+k^2)(2k^2-1)]
=4/(1+2k^2)√(2k^4+k^2-1)
r=|MN|/2
=2/(1+2k^2)√(2k^4+k^2-1)
圆A方程：(x+4/(1+2k^2))^2+(y-2/(1+2k^2))^2=4(2k^4+k^2-1)/(1+2k^2)^2
圆A过原点O(0,0)
(0+4/(1+2k^2))^2+(0-2/(1+2k^2))^2=4(2k^4+k^2-1)/(1+2k^2)^2
16+4=4(2k^4+k^2-1)
5=2k^4+k^2-1
2k^4+k^2-6=0
(2k^2-3)(k^2+2)=0
2k^2-3=0
k^2=3/2
k=±√6/2
L：y=±√6/2x+2

追问

十分感谢你，不过答案是错的噢