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

We need to find $$12m^2$$ where $$m$$ is the slope of a common tangent to the ellipse $$\frac{x^2}{16} + \frac{y^2}{9} = 1$$ and the circle $$x^2 + y^2 = 12$$.

For the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a^2 = 16$$ and $$b^2 = 9$$, the equation of a tangent line with slope $$m$$ can be written as

$$y = mx \pm \sqrt{a^2 m^2 + b^2} = mx \pm \sqrt{16m^2 + 9}$$

This follows from the condition that a line $$y = mx + c$$ is tangent to the ellipse when $$c^2 = a^2 m^2 + b^2$$.

For the circle $$x^2 + y^2 = 12$$ (with center at the origin and radius $$2\sqrt{3}$$), the tangent line with slope $$m$$ has the form

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

This follows from the requirement that the perpendicular distance from the center to the line equals the radius, namely $$\frac{|c|}{\sqrt{1+m^2}} = 2\sqrt{3}$$, which leads to $$c^2 = 12(1+m^2)$$.

For a common tangent, the value of $$c^2$$ must satisfy both conditions simultaneously, so we set

$$16m^2 + 9 = 12(1 + m^2)\,.$$

Rewriting this gives

$$16m^2 + 9 = 12 + 12m^2$$

Subtracting terms yields

$$4m^2 = 3\quad\text{and}\quad m^2 = \frac{3}{4}\,.$$

Substituting into $$12m^2$$ gives

$$12m^2 = 12 \times \frac{3}{4} = 9$$

Therefore, the required value is $$9$$.