Geometric mean proof with two tangents There are two tangent lines on $f(x) = \sqrt{x}$ each with the $x$-value $a$ and $b$ respectively. 
I need to prove that $c$, the $x$ value of the point at which the two lines intersect each other, is equal to $\sqrt{ab}$, the geometric mean of $a$ and $b$. 
I have been trying many different ways of doing this question and I keep getting stuck. 
 A: The tangent line at $(a,\sqrt{a})$ has equation $y=\frac{1}{2\sqrt{a}}x+\frac{\sqrt{a}}{2}$. 
The tangent line at $(b,\sqrt{b})$ has equation $y=\frac{1}{2\sqrt{b}}x+\frac{\sqrt{b}}{2}$.  
Set $\frac{1}{2\sqrt{a}}x+\frac{\sqrt{a}}{2}=\frac{1}{2\sqrt{b}}x+\frac{\sqrt{b}}{2}$ and solve for $x$.
A: The two points of tangency are $(a, \sqrt{a})$ and $(b, \sqrt{b})$. If either of $a$ or $b$ is $0$ then the result holds trivially, so assume $ab \neq 0$.
Because the curve is a parabola, the intersection of the tangents, $(x_0, y_0)$, has $y_0=(\sqrt{a}+\sqrt{b})/2$. That's because projecting perpendicularly onto the directrix, the intersection of two tangents of a parabola will always bisect the segment between the points of tangency (draw a picture). This can be proved using high school geometry and is sometimes known as the Two-Tangent Theorem for the parabola. 
The equation for the tangent at $(a, \sqrt{a})$ is 
$$y- \sqrt{a}=\frac{1}{2\sqrt{a}}(x-a) .
$$
This line passes through $(x_0, y_0)$ so we have
$$
\frac{\sqrt{a}+\sqrt{b}}{2} - \sqrt{a}=\frac{1}{2\sqrt{a}}(x_0-a).
$$
Simplifying,
$$
\frac{\sqrt{b} - \sqrt{a}}{2} =\frac{1}{2\sqrt{a}}(x_0-a).
$$
Clearing denominators gives
$$
\sqrt{ab}-a = x_0 -a,
$$
and so $x_0= \sqrt{ab}$, as desired.
Notice: In fact, if $A$ and $B$ are distinct non-zero points on the curve $y=\sqrt{x}$ and $C$ is the intersection of the tangents at $A$ and $B$, then we have

the $x$-coordinate of C is the geometric mean of the $x$-coordinates of $A$ and $B$

and

the $y$-coordinate of C is the arithmetic mean of the $y$-coordinates of $A$ and $B$.

