The points of contact of the line $y=x-1$ with $3x^2-4y^2=12$ is:

The equation of line and hyperbola are $y=x-1 \dots \left(i\right)$ and 

 $3x^2-4y^2=12 ....\left(ii\right)$

From (i) and (ii), we get $3x^2-4(x-1{)}^2=12$

$\Rightarrow 3x^2-4\left(x^2-2x+1\right)=12$ or $x^2-8x+16=0$

$\Rightarrow$    $x=4$

From (i), $y=3$ so points of contact is $(4,3)$

Points of contact are $\left(\pm \frac{a^{2}m}{\sqrt{a^2{m}^2-b^2}},\pm \frac{b^2}{\sqrt{a^2{m}^2-b^2}}\right)$

Here $a^2=4, b^2=3$ and $m=1$. So the required points of contact is $(4,3)$.