Show that the equation of the normal line with the minimum y-coordinate is $ y = \frac{-\sqrt{2}}{2}x + {1\over k}$ 

Question: The curve in the figure is the parabola $y=kx^2$ where $k>0$.
Several normal lines to this parabola are also shown. Consider the points in the first quadrant from which the normal lines are drawn. Notice that as the $x$ coordinate gets smaller, the $y$-coordinate of the intersection of the normal with the other arm of the parabola also decreases until it reaches a minimum, and then it increases. The normal line with the minimum $y$ coordinate is dotted.
$(a)$ Show that the equation of the normal to the parabola at a point $(x_0,y_0)$ is $y = {-x\over 2kx_0} + kx_0^2 + {1\over 2k}$
(b) Show that the equation of the normal line with the minimum y-coordinate is $ y =  \frac{-\sqrt{2}}{2}x + {1\over k} $




What I have done:


$(a)$ Show that the equation of the normal to the parabola at a point $(x_0,y_0)$ is $y = {-x\over 2kx_0} + kx_0^2 + {1\over 2k}$


$$ f(x) = kx^2 $$
$$ f( x_{0}) = kx_{0}^2 $$
$$ f'(x) = 2kx $$
$$ f'(x_{0}) = 2kx_0 $$
$$ Normal = -1/m $$
$$ m= {-1\over 2kx_0} $$
$$ y-y_1 = m (x-x_1) $$
$$ y-kx_0^2 = {-1\over 2kx_0}(x-x_0) $$
$$ y = {-x\over 2kx_0} + kx_0^2 + {1\over 2k} $$


(b) Show that the equation of the normal line with the minimum y-coordinate is $ y =  \frac{-\sqrt{2}}{2}x + {1\over k} $


$$ y =  \frac{-\sqrt{2}}{2}x + {1\over k} $$
$$ {-1\over m} = \frac{-\sqrt{2}}{2} $$
$$ m = \sqrt{2} $$
$$ f'(x) = 2kx $$
$$ 2kx = \sqrt{2} $$
$$ x = {\sqrt{2}\over 2k} $$
$$ f({\sqrt{2}\over 2k}) =  {1\over 2k}$$
$$ y-y_1 = m (x-x_1) $$
$$ y - {1\over 2k} = \frac{-\sqrt{2}}{2} (x-{\sqrt{2}\over 2k}) $$
$$ y =  \frac{-\sqrt{2}}{2}x + {1\over k} $$
However I am stuck trying to prove that this is the minimum y coordinate.
I've attempted 
$$ y = y$$
$$ kx^2 = {-x\over 2kx_0} + kx_0^2 + {1\over 2k}$$
$$ kx^2 + (\frac{1}{2kx_0})x -(kx_0^2 + \frac{1}{2k}) = 0$$
$$ x={-b\pm\sqrt{b^2-4ac} \over 2a}$$  
$$ x={- (\frac{1}{2kx_0})\pm\sqrt{ (\frac{1}{4k^2x_0^2})-4(k)(-kx_0^2 - \frac{1}{2k})} \over 2k}$$  
$$ x={- (\frac{1}{2kx_0})\pm\sqrt{ \frac{1}{4k^2x_0^2}-4(-k^2x_0^2 - \frac{k}{2k})} \over 2k}$$  
$$ x={-\frac{1}{2kx_0}\pm\sqrt{ \frac{1}{4k^2x_0^2}+4k^2x_0^2 + 2} \over 2k}$$  
$$ x={-\frac{1}{2kx_0}\pm\sqrt{ (2kx_0 + \frac{1}{2kx_0})^2} \over 2k}$$  
$$ x={-\frac{1}{2kx_0}\pm { (2kx_0 + \frac{1}{2kx_0})} \over 2k}$$  
$$ x = x_0 $$
$$ x = \frac{-1}{2k^2x_0} - x_0 $$
Now I am lost..

You did almost all the work. From
$$ x={-\frac{1}{2kx_0}\pm { (2kx_0 + \frac{1}{2kx_0})} \over 2k}$$ 
the $x$ you are looking for is the one with the '$-$':
$$x={-\frac{1}{2kx_0}- { (2kx_0 + \frac{1}{2kx_0})} \over 2k}=-\frac{1}{2k^2x_0}-x_0$$
(the other one is simply $x=x_0$).
Now 
$$y=k\left(-\frac{1}{2k^2x_0}-x_0\right)^2$$
so you just have to minimize 
$$\frac{1}{2k^2x_0}+x_0$$
and it's easy to see that the minimum is at $x_0=\frac{1}{\sqrt2k}$.
