Let $\lambda x-2y=\mu$ be a tangent to the hyperbola $a^2{x}^2-y^2=b^2$. Then ${\left(\frac{\lambda }{a}\right)}^2-{\left(\frac{\mu }{b}\right)}^2$ is equal to

Given that $\lambda x-2y=\mu$ is a tangent to the curve $a^2{x}^2-y^2=b^2$ 

So $a^2{x}^2-{\left(\frac{\lambda x-\mu }{2}\right)}^2=b^2$

$\Rightarrow \left(4a^2-{\lambda }^2\right){\mathrm{x}}^2+2\lambda \mu \mathrm{x}-{\mu }^2-4b^2=0$

So for the line to be tangent, the condition is discriminant $=0$

i.e. $4{\lambda }^2{\mu }^2+4\left(4a^2-{\lambda }^2\right)\left({\mu }^2+4b^2\right)=0$

$4{\lambda }^2{b}^2-4a^2{\mu }^2=16a^2{b}^2$

Hence, $\frac{{\lambda }^2}{a^2}-\frac{{\mu }^2}{b^2}=4$