If equation ( 10 x - 5 ) 2 + ( 10 y - 4 ) 2 = λ 2 ( 3 x + 4 y - 1 ) 2 represents a hyperbola, then

Given, equation of hyperbola is

${\left(10x-5\right)}^2+{\left(10y-4\right)}^2={\lambda }^2{\left(3x+4y-1\right)}^2$ can be rewritten as

$\frac{\sqrt{{\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{2}{5}\right)}^2}}{\left|\frac{3x+4y-1}{5}\right|}=\left|\frac{\lambda }{2}\right|$

This is of the form of $\frac{PS}{PM}=e$

Where, $P$ is any point on the hyperbola and $S$ is a focus and $M$ is the point of directrix.

Here, $\left|\frac{\lambda }{2}\right|>1\Rightarrow \left|\lambda \right|>2 \left(\because e>1\right)$

$\Rightarrow \lambda <-2$ or $\lambda >2$