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 $12{\mathrm{m}}^2$ is equal to

Given,

$\frac{x^2}{16}+\frac{y^2}{9}=1$

we know that equation of tangent to the ellipse is

$\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{ \mathrm{m}}^2+{\mathrm{b}}^2}$

$y=mx\pm \sqrt{16m^2+9} ...\left(1\right)$

$x^2+y^2=12$

Also equation of tangent to the circle is

$y=mx\pm \sqrt{12}\sqrt{1+m^2} ...\left(2\right)$

Now, for common tangent equate eq.$\left(1\right)$ and $\left(2\right)$

$\Rightarrow 16m^2+9=12\left(1+m^2\right)$

$16m^2-12m^2=3$

$4m^2=3$

$12m^2=9$