Explanation

(c)
The point of intersection between the curves $x^{2} = 4(y + 1)$ and $x^{2} = - 4(y + 1)$ is $(0,\ 1)$
The slopes of curve first and curve second at the point $(0,\ - 1)$ are respectively
$m_{1} = \frac{2x}{4} = 0$ and $m_{2} = \frac{- 2x}{4} = 0$
$\therefore\ \tan{\theta = \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} = 0 \Rightarrow \theta = 0^{{^\circ}}}$