Question

求方程(X^2)/2-(Y^2)/2=1 (X>0)上向量OA*向量OB的最小值

Answer 1

设双曲线的参数方程为

{

x=√2·secθ;

y=√2·tanθ;

则令坐标为:点A(√2·secθ1 ,√2·tanθ1);点B(√2·secθ2 ,√2·tanθ2);

则向量OA·向量OB

=2·secθ1·secθ2 + 2·tanθ1·tanθ2

=2·(1+sinθ1·sinθ2)/(cosθ1·cosθ2)

≥2.