If  m is the slope of a common tangent to the curves $\frac{x^2}{16}+\frac{y^2}{9}=1$  and  $x^2+y^2=12$ , then $12m^2$  is equal to :

Given ellipse  $\frac{x^2}{16}+\frac{y^2}{9}=1$
So equation of tangent to the ellipse is,
 $y=mx\pm \sqrt{a^2{m}^2+b^2}$
$y=mx\pm \sqrt{16m^2+9}$                     -------(i)
Now, given circle is  $x^2+y^2=12$
So equation of tangent on equating equations (i) and (ii)
 $\Rightarrow 16m^2+9=12(1+m^2)$
 $16m^2-12m^2=3\Rightarrow 4m^2=3\Rightarrow 12m^2=9$