Then substitute it into hyperbolic equation to get k=3× 1=3.
2) There are three such Qs.
First of all, OA=√(3? + 1? )=√ 10
(1) Let OQ=OA=√ 10, then the coordinate of Q point is (√ 10,0).
(--√ 10/0,0) is ignored because q is on the positive semi-axis of the X axis.
② Let OA=AQ. According to the properties of the three lines of an isosceles triangle, it is easy to get the coordinate of point Q as (6,0).
③ Let OQ=AQ, and let the coordinates of Q point be (m, 0), m>0.
So m=√[(3-m)? + 1? ]
The solution is m=5/3.
That is, the coordinate of point Q is (5/3,0).