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.

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?
