Explanation

(d)
Any point on parabola $y^{2} = 8x$ is $\left( 2t^{2},\ 4t \right).$ The equation of tangent at that point is
$yt = x + 2t^{2}$ …(i)
Given that, $xy = - 1$ …(ii)
On solving Eqs.(i) and (ii), we get
$y\left( yt - 2t^{2} \right) = - 1$
$\Rightarrow ty^{2} - 2t^{2}y + 1 = 0$
$\because\$ It is common tangent. It means they are intersect only at one point and the value of discriminant is equal to zero.
$ie,\ 4t^{4} - 4t = 0$
$\Rightarrow t = 0,1$
$\therefore\ \$The common tangent is $y = x + 2,$ (when $t = 0,$ it is $x = 0$ which can touch $xy = - 1$ at infinity only)