Question

已知抛物线y^2=4x,AB为过焦点的弦,直线 AB的倾斜角为60度,求AB绝对值的值.

Answer 1

AB的倾斜角为60度,

所以k=tan60=√3

y^2=4x,焦点(1,0)

所以y=√3(x-1)

3(x-1)^2=4x

3x^2-10x+3=0

x1+x2=10/3,x1*x2=1

(x1-x2)^2=(x1+x2)^2-4x1*x2=64/9

(y1-y2)^2=[(√3x1-√3)-(√3x2-√3)]^2=[√3(x1-x2)]^2

=3(x1-x2)^2=64/3

所以AB^2=(x1-x2)^2+(y1-y2)^2

=256/9

|AB|=16/3