Let required ratio be $K:1$
then, $\bigg(\large\frac{3K+4}{K+1},\frac{K-6}{K+1}\bigg)$
is lying on $y^2=4x$
$(K-6)^2=4(3K+4)(K+1)$
=> $11K^2+40\;K-20=0$
=>$K =\large\frac{-40 \pm \sqrt {1600 +4 \times 220}}{22}$
$K= \large\frac{-20 \pm \sqrt {620}}{11}$
$\quad= \large\frac{-20 \pm 2\sqrt {155}}{11}$
Required ratio is $-20 \pm 2 \sqrt {155}:11$
Hence C is the correct answer.