Explanation

(d)
Given, curve is $y = x^{2}\ \ \Rightarrow \ \frac{dy}{dx} = 2x$
$\Rightarrow \ \ \ \left( \frac{dy}{dx} \right)_{(1,\ 1)} = 2 = m_{1}$ (say)
And $x = y^{2}\ \Rightarrow \ \ \ 1 = 2y\frac{dy}{dx}$
$\Rightarrow \ \ \frac{dy}{dx} = \frac{1}{2y}\ \Rightarrow \ \ \left( \frac{dy}{dx} \right)_{(1,\ 1)} = \frac{1}{2} = m_{2}$ (say)
$\therefore$ Angle of intersection at the point (1, 1) is given by
$\tan\theta = \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} = \frac{2 - \frac{1}{2}}{1 + 2 \times \frac{1}{2}} = \frac{3}{4}$
$\Rightarrow \ \ \theta = \tan^{- 1}\left( \frac{3}{4} \right)$