Question

过抛物线y^2=4x的焦点F作弦AB,若AF=2BF,则弦AB所在的直线方程

Answer 1

y^2=4x

2p=4,p/2=1

A(x1,y1),B(x2,y2)

AF=2BF y1=-2y2

AF=x1+1

bf=x2+1

x1+1=2(x2+1)

x1=2x2+1

y1^2=4x1=8x2+4

y2^2=4x2

y1^2=4y2^2

8x2+4=16x2

8x2=4

x2=1/2,y2=-√2/2 ,x1=2,y1=√2 k=(y1-y2)/(x1-x2)=(3√2/2)/(3/2)=√2

或

x2=1/2,y2=√2/2,x1=2,y1=-√2 k=(y1-y2)/(x1-x2)=-√2

AB直线：y=√2(x-1) 或y=-√2(x-1)