Explanation

(b)
Given curve is $y^{2} = 16x$ Let any point be $(h,k)\$But$\ \ 2h = k$, then $k^{2} = 16h\$
$\ \Rightarrow \ 4h^{2} = 16h$
$\Rightarrow \ h = 0,\ h = 4\$
$\Rightarrow \ k = 0,\ k = 8\$
$\therefore$ Points are$(0,\ 0)\ ,\ (4,\ 8)$
Hence, focal distance are respectively
$0 + 4 = 4,\ 4 + 4 = 8\ \ \lbrack\because focal\ distance = h + a\rbrack$