sin(β)=m*cos(α)*cos(β)–m*sin(α)*sin(β);
1/sin(α)=m/tan(α)/tan(β)-m;
sec(α)/m=1/(x*y)-1;
√(1+tan(α)^2)/m=1/(x*y)-1;
√(1+x^2)/m=1/(x*y)-1;
(1+x^2)/m^2=(1/(x*y)-1)^2;
解得:
y1=(m^2*x-√(m^2*x^2+m^2*x^4))/(-x^2+m^2*x^2-x^4),
y2=(m^2*x+√(m^2*x^2+m^2*x^4))/(-x^2+m^2*x^2-x^4),
考虑到x>0,y>0,所以
y2不满足条件.
所以
y=(m^2*x-√(m^2*x^2+m^2*x^4))/(-x^2+m^2*x^2-x^4).