Ellipse

Given equation of ellipse is $2x^2 + y^2 = 4$$\Rightarrow\; \frac{2x^2}{4} +\; \frac{y^2}{4} = 1 \Rightarrow\; \frac{x_2}{2} +\; \frac{y^2}{4} = 1$ Equation of tangent to the ellipse $\; \frac{x^2}{2} +\; \frac{y^2}{4} = 1$ is $y = m x \pm \sqrt{2 m^2 + 4}$.......(1) ( as equation of tangent to the ellipse $\; \frac{x^2}{a^2} +\; \frac{y^2}{b^2} = 1$ is $y = m x + c$ where $c = \pm \sqrt{a^2 m^2 + b^2}$ )Now, Equation of tangent to the parabola$y^2 = 16 \sqrt{3} x$ is $y = m x + \frac{4\sqrt{3}}{m}$......(2) ( as equation of tangent to the parabola$y^2 = 4 a x \; \text{ is } \; y = m x + \frac{a}{m} \; \text{ ) } \;$On comparing (1) and (2), we get$\; \frac{4\sqrt{3}}{m} = \pm \sqrt{2 m^2 + 4}$ Squaring on both the sides, we get $\; 16(3) = (2 m^2 + 4) m^2$$\; \Rightarrow 48 = m^2 (2 m^2 + 4) \Rightarrow 2 m^4 + 4 m^2 - 48 = 0$$\; \Rightarrow m^4 + 2 m^2 - 24 = 0 \Rightarrow (m^2 + 6)(m^2 - 4) = 0$$\; \Rightarrow m^2 = 4 \; \text{ ( as } \; m^2 \neq -6) \Rightarrow m = \pm 2$$\Rightarrow$ Equation of common tangents are $y = \pm 2 x \pm 2 \sqrt{3}$Thus, statement - 1 is true.Statement - 2 is obviously true.