Question

已知双曲线x^2/a^2-y^2/b^2=1(a>0,b>0)的一条渐近线方程是y=√3x且2a^2=c求方程

Answer 1

1.条渐近线方程是y=√3x

a:b=1:√3 b=√3a

a^2+b^2=c^2 4a^2=c^2 c=2a 2a^2=c a=1 b=√3

双曲线方程为x^2-y^2/3=1

2.直线l过坐标原点o 设为 y=kx

x^2-y^2/3=1

m(x1,y1) n(-x1,-y1)

设P(x,y)

kpm=(y-y1)/(x-x1)

kpn=(y+y1)/(x+x1)

kpm*kpn

=(y^2-y1^2)/(x^2-x1^2) y1^2=3x1^2-3 y^2=3x^2-3

=(3x^2-3x1^2)/(x^2-x1^2)

=3