JEE Mains · Maths · STD 11 - 9. straight line
Let \(A,B\) be points on the two half-lines \(x-\sqrt{3}|y|=\alpha\), \(\alpha>0\) at a distance of \(\alpha\) from their point of intersection \(P\). The line segment \(AB\) meets the angle bisector of the given half-lines at the point \(Q\). If \(PQ=\dfrac{9}{2}\) and \(R\) is the radius of the circumcircle of \(\triangle PAB\), then \(\dfrac{\alpha^2}{R}\) is equal to ______
- A 3
- B 6
- C 9
- D 13
Answer & Solution
Correct Answer
(C) 9
Step-by-step Solution
Detailed explanation
The given equations of the half-lines are \(x - \sqrt{3}y = \alpha\) for \(y \ge 0\) and \(x + \sqrt{3}y = \alpha\) for \(y \le 0\). The point of intersection \(P\) is obtained by setting \(y = 0\), which gives \(x = \alpha\). Thus, \(P \equiv (\alpha, 0)\). The angle bisector…
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