If $\frac{1}{x}+\frac{1}{x+a}=\frac{1}{m}+\frac{1}{m+a}$ has roots equal in magnitude but opposite in sign, then

The given equation is $\frac{1}{x}+\frac{1}{x+a}=\frac{1}{m}+\frac{1}{m+a}$ 

$\Rightarrow \frac{x+a+x}{x\left(x+a\right)}=\frac{m+a+m}{m\left(m+a\right)}$

$\Rightarrow m\left(m+a\right)\left(2x+a\right)=\left(2m+a\right)x\left(x+a\right)$

But  coefficient of $x=0$    ($\because$ roots equal in magnitude but opposite in sign )

$2m\left(m+a\right)-\left(2m+a\right)a=0$

$\Rightarrow 2m^2+2ma=2ma+a^2$

          $\therefore 2m^2=a^2$