设直线y=kx+b
代入抛物线,得(kx+b)^2=k^2x^2+2kbx+b^2=6x,=> k^2x^2+(2kb-6)x+b^2=0
x1+x2=-(2kb-6)/k^2 ,x1x2=b^2/k^2
|AB|^2=(x1-x2)^2+(y1-y2)^2
=(x1-x2)^2+(kx1-kx2)^2
=(k^2+1)(x1-x2)^2
=(k^2+1)[(x1+x2)^2-4x1x2]
=(k^2+1)[(-(2kb-6)/k^2)^2-4*b^2/k^2]
=12(k^2+1)(3-2kb)/k^4
由|AB|不大于12,得 12(k^2+1)(3-2kb)/k^4≤12^2
(k^2+1)(3-2kb)≤12k^4
这个,四次方程怎么解啊?