If  the equation ${(10x-5)}^2+{(10y-4)}^2={\lambda }^2{(3x+4y+1)}^2$  represents a hyperbola, then

 ${(10x-5)}^2+{(10y-4)}^2={\lambda }^2{(3x+4y+1)}^2$
 $\Rightarrow 100{(x-\frac{1}{2})}^2+100{(y-\frac{2}{5})}^2={\lambda }^2{(3x+4y-1)}^2$
Dividing both sides by 25, we get
 $\Rightarrow 4[{(x-\frac{1}{2})}^2+{(y-\frac{2}{5})}^2]=\frac{{\lambda }^2{(3x+4y-1)}^2}{25}$
 $\Rightarrow {(x-\frac{1}{2})}^2+{(y-\frac{2}{5})}^2=\frac{{\lambda }^2}{4}\frac{{(3x+4y-1)}^2}{25}$
So, hence $e^2=\frac{{\lambda }^2}{4}$
For a hyperbola, $e>1$
 $\Rightarrow {\lambda }^2-4>0$
 $\Rightarrow (\lambda -2)(\lambda +2)>0$