-
krishna239455
- Master | Next Rank: 500 Posts
- Posts: 114
- Joined: Mon Jun 13, 2011 9:10 am
- Thanked: 1 times
krishna239455 wrote:
Point P and Q lies on the same circle with center at (0, 0).
Thus, (s² + t²) = (-√3)² + 1² = 3 + 1 = 4
Again line segments OP and OQ are perpendicular.
Thus (slope of OP)*(slope of OQ) = -1
Slope of OP = 1/(-√3) = -(1/√3)
=> Slope of OQ = (t - 0)/(s - 0) = t/s = (-1)/(-1/√3) = √3
=> t = √3s
Thus, (s² + (√3s)²) = 4
=> (s² + 3s²) = 4
=> s² = 1
=> s = ±1
As point Q lies in the first quadrant, so s = 1.
The correct answer is B.
