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.
 A: 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 $-2\sqrt{x_0}$.
A: Slope of tangent = $ \dfrac{d\sqrt x }{dx} = \dfrac{1}{2 \sqrt x} $. Slope of normal is its negative reciprocal.So equation of normal is:
$$ \frac{y-6}{x-3} = - 2 \sqrt 3. $$
A: 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 $y=-4x+18$.
A: You got the equation and idea right, just substituted the wrong value of $a$.
Here $y=\sqrt{x}$ or $y^2 = x$ (where $y>0$) . It's a standard $y^2= 4ax$ parabola type where $a= 1/4$
General eq. of normal: $y = mx -am^3 -2am$ (for $y^2 = 4ax$ types only) . Point 3,6 must satisfy this, hence, we will get a cubic eq. , by rational roots test, $m = -4$.
Thence we get, $y +4x = 18$.
