Question

设抛物线y^2=x的一弦PQ被直线Y=K(X-1)+1垂直平分.1.求k的取值范围,2.若

Answer 1

设P(y1²,y1)Q(y2²,y2)

PQ被直线y=k（x-1）+1垂直平分

即:(y1-y2)/(y1²-y2²)=-1/k

即:y1+y2=-k

PQ中点E((y1²+y2²)/2,(y1+y2)/2)

将E代入直线y=k（x-1）+1

得到y1²+y2²=-2/k>0,k≠0

即k