Explanation

(c)
Given $y^{2} = - 8x$
Here, $a = - 2$
We know that if one end of a focal chord is $\left( at^{2},2at\ \right)$, then the other end will be $\left( \frac{a}{t^{2}}, - \frac{2a}{t} \right)$
Here, one end is $( - 1,\ 2\ \sqrt{2)}$
$\therefore\ at^{2} = - 1\ \Rightarrow \ t = \frac{1}{\sqrt{2}}$ $\lbrack\because a = - 2\rbrack$
So, other end$= \left( \frac{- 2}{1/2},\frac{- 2 \times - 2}{1/\sqrt{2}} \right)\ = ( - 4,\ 4\ \sqrt{2})$