find the value of $a$ and $b$? $$
   D=d(PA)+d(AB)+d(BQ)=\sqrt{a^2+1^2}+\sqrt{(b-a)^2+2^2}+\sqrt{(4-b)^2+1^2}$$
Given  $\frac{\partial D}{\partial a}=0$ and $\frac{\partial D}{\partial b}=0$ 
find the  value of $a$ and $b$
My attempt  : i found the  answer  here but im not getting the  answer
 
My work :  $\frac{\partial D}{\partial a}= \frac{2a}{ \sqrt{ 1^2 + a^ 2}} +\frac{-2b+2a}{ \sqrt{ 2^2 + (b-a)^ 2}} $
Now  we have $\frac{\partial D}{\partial a}=0$
$\frac{2a}{ \sqrt{ 1^2 + a^ 2}} =\frac{2b-2a}{ \sqrt{ 2^2 + (b-a)^ 2}} $
after that  im not able to proceed further


$$\begin{align}
&\frac{\partial D}{\partial a}= \frac{a}{ \sqrt{ 1^2 + a^ 2}} -\frac{b-a}{ \sqrt{ 2^2 + (b-a)^ 2}}=0\\
&\frac{\partial D}{\partial b}=\frac{b-a}{ \sqrt{ 2^2 + (b-a)^ 2}}-\frac{4-b}{ \sqrt{1^2+ (4-b)^2}} =0 
\end{align}
$$
It follows:
$$
\frac{a}{ \sqrt{ 1^2 + a^ 2}}=\frac{4-b}{ \sqrt{1^2+ (4-b)^2}}
\implies  a=4-b.
$$
Can you take it from here?
