在抛物线y2=4x上恒有两点关于直线y=kx+3对称，求k的取值范围
设直线AB:y=-(1/k)x+b关于直线y=kx+3对称,交y2=4x与A(x1,y1),B(x2,y2)
y=-(1/k)x+b与y2=4x联立得到:x^2/k^2-(4+2b/k)x+b^2=0 (1)
(x1+x2)/2=(4k^2+2kb)/2=2k^2+kb
y=-(1/k)x+b与y=kx+3,联立得到,x=(b-3)/(k+1/k)=k(b-3)/(k^2+1)
2k^2+kb=k(b-3)/(k^2+1)2
b=-(2k^3+2k+3)/(k^2)
(1)有解,即 (4+2b/k)^2-4b^2/k^2>0
(4K+2b)^2-4b^2>0
k^2+kb>0
k^2+...全部
设直线AB:y=-(1/k)x+b关于直线y=kx+3对称,交y2=4x与A(x1,y1),B(x2,y2)
y=-(1/k)x+b与y2=4x联立得到:x^2/k^2-(4+2b/k)x+b^2=0 (1)
(x1+x2)/2=(4k^2+2kb)/2=2k^2+kb
y=-(1/k)x+b与y=kx+3,联立得到,x=(b-3)/(k+1/k)=k(b-3)/(k^2+1)
2k^2+kb=k(b-3)/(k^2+1)2
b=-(2k^3+2k+3)/(k^2)
(1)有解,即 (4+2b/k)^2-4b^2/k^2>0
(4K+2b)^2-4b^2>0
k^2+kb>0
k^2+k[-(2k^3+2k+3)/(k^2)]>0
k(k^3+2k+3)0在R上恒成立
所以k(k+1)<0
-1 收起
