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(A) The given lines are $x - \sqrt{3}y = \alpha$ (for $y \ge 0$) and $x + \sqrt{3}y = \alpha$ (for $y < 0$).The point of intersection $P$ is $(\alpha,

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The given lines are $x - \sqrt{3}y = \alpha$ (for $y \ge 0$) and $x + \sqrt{3}y = \alpha$ (for $y The point of intersection $P$ is $(\alpha, 0)$.The angle between the lines is $60^\circ$ because the slopes are $m_1 = 1/\sqrt{3}$ and $m_2 = -1/\sqrt{3}$.Since $PA = PB = \alpha$ and the angle $\angle APB = 60^\circ$, $	riangle PAB$ is an equilateral triangle with side length $a = \alpha$.The height $PQ$ of the equilateral triangle is given by $PQ = \frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{2}\alpha$.Given $PQ = \frac{9}{2}$, we have $\frac{\sqrt{3}}{2}\alpha = \frac{9}{2}$, which implies $\alpha = 3\sqrt{3}$.The circumradius $R$ of an equilateral triangle with side $a$ is $R = \frac{a}{\sqrt{3}}$.Thus, $R = \frac{\alpha}{\sqrt{3}} = \frac{3\sqrt{3}}{\sqrt{3}} = 3$.Finally, $\frac{\alpha^2}{R} = \frac{(3\sqrt{3})^2}{3} = \frac{27}{3} = 9$.

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