Given: The pole of the line $3x-16y+48=0$ with respect to the hyperbola $9x^2-16y^2=144$ is $\left(\alpha ,\beta \right)$

Equation of polar is given by $T=0$

$\Rightarrow 9\alpha x-16\beta y-144=0$

These two lines are coinciding with each other 

$\Rightarrow \frac{9\alpha }{3}=\frac{-16\beta }{-16}=\frac{-144}{48}$

$\Rightarrow \frac{9\alpha }{3}=\frac{\beta }{1}=\frac{-3}{1}\Rightarrow \alpha =-1, \beta =-3$

$\Rightarrow \alpha -\beta =2$