(b) $Here,\ h = - \frac{1}{2},\ a = 1,\ b = - 6$ $\therefore\tan = \left| 2\frac{\sqrt{\frac{1}{4} + 6}}{1 - 6} \right| = \frac{2\sqrt{\frac{25}{4}}}{- 5} = | - 1|$ $\therefore\ \theta = \tan^{- 1}(1) = 45{^\circ}$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.