Equation of the given curve in parametric form,
$x = t^2 + 1$ and $y = t^2 - t - 6$
Y-coordinate of the point, where the given curve meets X-axis is 0.
When y = 0, then $t^2 - t - 6 = 0$
$ \Rightarrow \; t^2 - 3t + 2t - 6 = 0$
$\Rightarrow \; t(t - 3) + 2(t - 3) = 0$
$ \Rightarrow \; (t - 3)(t + 2) = 0$
$\Rightarrow $ t = 3 or - 2
when t = 3, then x = 10
when t = -2, then x = 5
Hence, the points where the curve meets the X-axis are (10, 0) and (5, 0).
Now, $\frac{ dy}{dx } = \frac{ \frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t-1}{2t} $
Slope of the tangent at point (10, 0)
$ m_{1} = \frac{dy}{dx} \bigg]_{x=10} = \frac{dy}{dx}\bigg]_{t=3} = \frac{5}{6} $
Slope of the tangent at point (5, 0),
$ m_{2}= \frac{dy}{dx}\bigg]_{x=5} = \frac{dy}{dx}\bigg]_{t=-2} =\frac{-5}{-4} = \frac{5}{4} $
If $\theta$ be the angle beween two tangents, then
$\tan\theta= \left|\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right| = \left|\frac{\frac{5}{4} - \frac{5}{6}}{1+ \frac{5}{6} . \frac{5}{4}}\right| $
$ = \left|\frac{\frac{15-10}{12}}{\frac{24+25}{24}}\right| = \left|\frac{\frac{5}{12}}{\frac{49}{24}}\right| = \left|\frac{10}{49}\right| $
$ \Rightarrow \tan\theta = \pm \frac{10}{49} $
$ \therefore \theta = \tan^{-1} \left(\pm \frac{10}{49}\right) =\pm \tan^{-1} \left(\frac{10}{49}\right) $