Constant length of segment of tangent 
Prove that the segment of the tangent to the curve $y=\frac{a}{2}\ln\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}-\sqrt{a^2-x^2}$ contained between the $y$-axis and the point of tangency has a constant length.

What I have done: First of all I found out slope $m$ of the tangent to the curve at any general point $(h,k)$, which after simplifying comes out to be
$m=\frac{1}{\sqrt{a^2-h^2}}[2h-\frac{a^2}{h}]$
After that I found the equation of the tangent at the general point which comes out to be
$y-k=\left(\frac{1}{\sqrt{a^2-h^2}}(2h-\frac{a^2}{h})\right)(x-h)$
Then I found the $y$-coordinate at $x=0$ which comes out to be $k+\frac{a^2-2h^2}{\sqrt{a^2-h^2}}$
Then I found distance between points $(h,k)$ and $(0,k+\frac{a^2-2h^2}{\sqrt{a^2-h^2}})$ which comes out to be $\sqrt{\frac{3h^4-3a^2h^2+a^4}{a^2-h^2}}$ which shows length of segment depends on $h$ but this was not we had to prove. 
Please tell me where I have made the mistake.
 A: First, note that the given function $$y = f(x;a) = \frac{a}{2} \log \frac{a + \sqrt{a^2-x^2}}{a - \sqrt{a^2-x^2}} - \sqrt{a^2-x^2}$$ satisfies the relationship $$f(ax;a) = a f(x;1)$$ for $a > 0$; thus, $a$ is a scaling factor for positive reals, and it suffices to consider only the special case $a = 1$.  This simplifies the computation considerably.  I leave it as an exercise to show that $$\frac{df(x;1)}{dx} = -\frac{\sqrt{1-x^2}}{x}.$$  Consequently, the equation of the tangent line at $(x,y) = (x_0, f(x_0;1))$ is given by $$y - f(x_0;1) = -\frac{\sqrt{1-x_0^2}}{x_0} (x - x_0).$$  The $y$-intercept of this line is therefore $$b = f(x_0;1) + \sqrt{1-x_0^2}.$$  The squared distance of the segment from the point of tangency to the $y$-intercept is $$D(x_0)^2 = x_0^2 + \left(b - f(x_0;1)\right)^2.$$  But $b - f(x_0;1)$ is simply $\sqrt{1-x_0^2}$, so $D(x_0)^2 = 1$ and the distance is constant.  When we replace the scale factor $a$, we find that the distance is $a$ for $a > 0$; if $a < 0$, then the distance is $-a$, so we can say $D = |a|$ for all $a \ne 0$.
A: Here are some steps:
\begin{align*}
y & =\frac{a}{2}\ln\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}-\sqrt{a^2-x^2}\\
\frac{dy}{dx} & = \frac{a}{2}\left(\frac{a-\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}\right) \, \frac{d}{dx}\left(\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}\right)+\frac{x}{\sqrt{a^2-x^2}}\\
& = \frac{a}{2}\left(\frac{a-\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}\right) \, \left(\frac{(a-\sqrt{a^2-x^2})\color{red}{\frac{-x}{\sqrt{a^2-x^2}}}-(a+\sqrt{a^2-x^2})\color{blue}{\frac{x}{\sqrt{a^2-x^2}}}}{(a-\sqrt{a^2-x^2})^2}\right)+\frac{x}{\sqrt{a^2-x^2}}\\
& =\frac{a}{2}\left(\frac{1}{a+\sqrt{a^2-x^2}}\right)\, \left(\frac{\color{red}{\frac{-2ax}{\sqrt{a^2-x^2}}}}{(a-\sqrt{a^2-x^2})}\right)+\frac{x}{\sqrt{a^2-x^2}}\\
& = \frac{-a^2x}{x^2\sqrt{a^2-x^2}}+\frac{x}{\sqrt{a^2-x^2}}\\
\end{align*}
Can you simplify now?
