The parabola is $y^2-Kx+8=0$
$y^2=kx-8$
$\qquad= k(x-\large\frac{8}{k})$
$\qquad=4AX$
where $4A=K$
direction of the parabola is $X+A=0$
Comparing it with $x-1=0$ we get,
$1= \large\frac{8}{k}-\frac{K}{4}$
or $k^2+4k-32=0$
$(k+8)(k-4)=0$
$k=-8,4$
Hence C is the correct answer.