Explanation

(d)
Given parabola is $x^{2} = - 2y$
Coordinates of end points of a latusrectum are $A\left( 1,\ - \frac{1}{2} \right)$ and $B\left( - 1,\ - \frac{1}{2} \right)$
Now, $2x = - 2\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx} = - x$
And slope of normal is $- \frac{dx}{dy} = \frac{1}{x}$
The equations of normals at points $A$ and $B$ are
$y + \frac{1}{2} = \frac{1}{1}(x - 1)$
$\Rightarrow 2y - 2x = - 3$ …(i)
And $y + \frac{1}{2} = - \frac{1}{1}(x + 1)$
$\Rightarrow 2y + 2x = - 3$ …(ii)
On solving Eqs. (i) and (ii), we get
$x = 0,\ y = - \frac{3}{2}$