Explanation

(c)
Given, $x = t^{2} + 1$ and $y = t^{2} - t - 6$
$\therefore\ \ \ \frac{dx}{dt} = 2t$ and $\frac{dy}{dx} = 2t - 1$
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t - 1}{2t}$
When, it ,meet $x$-axis, then $y = 0$
$\therefore\ \ \ \ \ \ \ t^{2} - t - 6 = 0$
$\Rightarrow \ \ \ \ \ \ \ \ y = 3,\ - 2$
$\therefore\ \ \ \ \ \left( \frac{dy}{dx} \right)_{at\ t = 3} = \frac{6 - 1}{6} = \frac{5}{6} = m_{1}$ [say]
And $\left( \frac{dy}{dx} \right)_{at\ t = - 2} = \frac{5}{4} = m_{1}$ [say]
$\therefore$ Required angle$= \pm \tan^{- 1}\left\{ \left| \frac{\frac{5}{6} - \frac{5}{4}}{1 + \frac{25}{24}} \right| \right\}$
$= \pm \tan^{- 1}\left\{ \frac{10}{49} \right\}$