Question

已知直线l的参数方程式为x=1+t,y=√3t和曲线E;y^2=2px,直线l与曲线E交于AB两点 若AB=4√3求曲线

Answer 1

直线l：y=√3(x-1)

3(x-1)^2=2px

3x^2-(6+2p)x+3=0

x1+x2=(6+2p)/3

x1*x2=1

x1^2+x2^2=(6+2p)^2/9+2

AB^2=(x2-x1)^2+(y2-y1)^2=48

x2^2-2x1x2+x1^2+[√3(x2-x1)]^2

=x2^2-2+x1^2+3x2^2-6+3x1^2

=4(x2^2+x1^2)-8

=4(6+2p)^2/9-8=48

(6+2p)^2=56*9/4=126

p>0

p=√126/2-3

曲线E;y^2=(√126-6)x