Where does the normal to the graph of $y = \sqrt x$ at the point $(a, \sqrt a)$ intersect the $x$-axis? 
Where does the normal to the graph of $y = \sqrt x$ at the point $(a, \sqrt a)$ intersect the $x$-axis?

I know how to find equation with numbers, but got really confused with this one. If anybody could break it down it would be greatly appreciated.
Here is what I've done: Took derivative $= .5x^{-.5}$, then plugged in $x$ value $a$ to get $1/2a^{-1/2}$, so the slope of tangent is $1/2a^{-1/2}$? so the slope of normal is $2a^{-1/2}$? So $y-\sqrt{a} = 2a^{-1/2}(x-a)$?
Thank you in advance

\begin{align*}
  y' &= \frac{1}{2\sqrt{x}} \\
  \text{slope of the normal} &=-2\sqrt{a} \\
  y-\sqrt{a} &= -2\sqrt{a}(x-a) \\
  2\sqrt{a}x+y &= \sqrt{a}(2a+1) \\
  \frac{x}{a+\frac{1}{2}}+\frac{y}{\sqrt{a}(2a+1)} &= 1 \\
  x\text{-intercept} &= a+\frac{1}{2}
\end{align*}
