Statement-1 : An equation of a common tangent to the parabola $${y^2} = 16\sqrt 3 x$$   and the ellipse $$2{x^2} + {y^2} = 4$$    is $$y = 2x + 2\sqrt 3 $$
Statement-2 : If the line $$y = mx + \frac{{4\sqrt 3 }}{m},\,\left( {m \ne 0} \right)$$     is a common tangent to the parabola $${y^2} = 16\sqrt 3 x$$   and the ellipse $$2{x^2} + {y^2} = 4,$$    then $$m$$ satisfies $${m^4} + 2{m^2} = 24$$

Solution :
Given equation of ellipse is $$2{x^2} + {y^2} = 4$$
$$ \Rightarrow \frac{{2{x^2}}}{4} + \frac{{{y^2}}}{4} = 1\,\,\,\,\, \Rightarrow \frac{{{x^2}}}{2} + \frac{{{y^2}}}{4} = 1$$
Equation of tangent to the ellipse $$\frac{{{x^2}}}{2} + \frac{{{y^2}}}{4} = 1$$    is
$$y = mx \pm \sqrt {2{m^2} + 4} .....(1)$$
($$\because $$ equation of tangent to the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$    is $$y=mx+c$$   where $$c = \pm \sqrt {{a^2}{m^2} + {b^2}} $$    )
Now, Equation of tangent to the parabola
$${y^2} = 16\sqrt 3 x$$   is $$y = mx + \frac{{4\sqrt 3 }}{m}.....(2)$$
($$\because $$ equation of tangent to the parabola $${y^2} = 4ax$$   is $$y = mx + \frac{a}{m}$$   )
On comparing (1) and (2), we get
$$\frac{{4\sqrt 3 }}{m} = \pm \sqrt {2{m^2} + 4} $$
Squaring on both the sides, we get
$$\eqalign{ & 16\left( 3 \right) = \left( {2{m^2} + 4} \right){m^2} \cr & \Rightarrow 48 = {m^2}\left( {2{m^2} + 4} \right) \cr & \Rightarrow 2{m^4} + 4{m^2} - 48 = 0 \cr & \Rightarrow {m^4} + 2{m^2} - 24 = 0 \cr & \Rightarrow \left( {{m^2} + 6} \right)\left( {{m^2} - 4} \right) = 0 \cr & \Rightarrow {m^2} = 4\left( {\because {m^2} \ne - 6} \right) \cr & \Rightarrow m = \pm 2 \cr} $$
$$ \Rightarrow $$ Equation of common tangents are $$y = \pm 2x \pm 2\sqrt 3 $$
Thus, statement-1 is true.
Statement-2 is obviously true.