How to find a straight line orthogonal to a curve? 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.

You need this result:

If two curves $f(x)$ and $g(x)$ (or a curve and a line), pass each other orthogonally at $x=c$, then $f(c)=g(c)$ and $f'(c)g'(c)=-1$.

Let the equation of the line be $g(x)=ax+b$ and let $f(x)=\sqrt{x}$. Then we get the following system of equations:
$$3a+b=6$$
$$ac+b=\sqrt{c}$$
$$a \cdot \frac{1}{2\sqrt{c}} = -1$$
Subtracting the first from the second, we get $(c-3)a=\sqrt{c}-6$, thus $$a=\frac{\sqrt{c}-6}{c-3}$$
Therefore we get the following equation in $c$:
$$\frac{\sqrt{c}-6}{c-3} \cdot \frac{1}{2\sqrt{c}} = -1$$
$$\sqrt{c}-6 = -2\sqrt{c}(c-3)$$
Let $\sqrt{c}=x$, we get the following equation in $x$: $$x-6 = -2x(x^2-3)$$
$$x-6 = -2x^3+6x$$
$$2x^3-5x-6 =0$$
We notice that $x=2$ is a root, use the rational root theorem.
$$(x-2)(2x^2+4x+3)=0$$
$$x=2 \vee 2x^2+4x+3=0$$
The last equation has $b^2-4ac=16-24=-8$, so there are no real roots.
Thus $c=x^2=4$. Therefore $a=-4$ and $b=18$.Thus $y=-4x+18$.
