We want to find the equation of line passing through $(3,6)$ and cutting the curve $y=\sqrt{x}$ orthogonally .
I thought that it means we have to find a normal through that point equation of normal is $y=mx -2am -am^3 $.
Putting $(3,6)$ in, we get $(m+2)(m^2-4m+3)$.
Then the equation of the straight line is coming $y+ 2x +4=0$.
EDIT not by OP: How to find the equation of straight line orthogonal to the curve.
Best Answer
At point $Q=(x_0,\sqrt{x_0})$ on the given curve, the normal line has a slope $-2\sqrt{x_0}$. This must be the same as the slope of the line connecting $P=(3,6)$ and $Q$, so we have the equation: $$ -2\sqrt{x_0}={6-\sqrt{x_0}\over 3-x_0}. $$ This leads to a cubic equation, which fortunately has $x_0=4$ as its only real solution. It follows that $Q=(4,2)$ and the equation of line $PQ$ is $y=-4x+18$.
EDIT
Formula $-2\sqrt{x_0}$ for the slope of the normal line at a point $(x_0,y_0)$ on a parabola having equation $y^2=2px$ can be obtained without calculus, imposing the condition that a generic line $y=mx+c$ intersects the parabola at a single point, thus being tangent (see here).
The result is that the slope of the tangent at $(x_0,y_0)$ is $m=p/y_0$, from which it follows that the slope of the normal line is $-1/m=-y_0/p$. In our case $p=1/2$ and $y_0=\sqrt{x_0}$, so that the slope of the normal line at $(x_0,y_0)$ is $-2\sqrt{x_0}$.