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

The function $y = \sqrt{x} = x^{\frac{1}{2}}$ has derivative 
$$y' = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$$ 
as you found.  The tangent line to the graph of $y = \sqrt{x}$ at the point $(a, \sqrt{a})$ has slope 
$$y'(a) = \frac{1}{2\sqrt{a}}$$
which is defined for $a > 0$.   
You made a mistake when you calculated the slope of the normal line.  Since the normal line is perpendicular to the tangent line, its slope is the negative reciprocal of the slope of the tangent line.  Thus, the slope of the normal line is $-2\sqrt{a}$. Hence, the equation of the normal line at the point $(a, \sqrt{a})$ is 
$$y - \sqrt{a} = -2\sqrt{a}(x - a)$$
To solve for the $x$-intercept of the normal line, we set $y = 0$ and solve for $x$.
\begin{align*}
0 - \sqrt{a} & = -2\sqrt{a}(x - a)\\
-\sqrt{a} & = -2\sqrt{a}x + 2a\sqrt{a}\\
2\sqrt{a}x & = (2a + 1)\sqrt{a}\\
x & = a + \frac{1}{2}
\end{align*} 
Thus, for $a > 0$, the normal line to the graph of $y = \sqrt{x}$ at the point $(a, \sqrt{a})$ intersects the $x$-axis at the point $(a + \frac{1}{2}, 0)$.
Since 
$$\lim_{x \to 0^{+}} \frac{1}{2\sqrt{x}} = \infty$$ 
the tangent line to the graph of $y = \sqrt{x}$ is vertical, so the normal line is horizontal.  Since the tangent line to the point $(0, \sqrt{0}) = (0, 0)$ is the $y$-axis, the normal line is the $x$-axis.
