# The ratio in which the line segment joining the points $(4,-6)$ and $(3,1)$ is divided by the parabola $y^2=4ax$ is

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

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.