$\begin{array}{1 1}(A)\;\frac{-20 \pm \sqrt {155}}{11} :1\\(B)\;\frac{-2 \pm \sqrt {155}}{11} :1 \\(C)\;-20 \pm 2 \sqrt {155} :11 \\(D)\;-20 \pm \sqrt {155} :11 \end{array}$

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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.

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