Question

An equation of a common tangent to the parabola $y^{2} = 16\sqrt{3}x$ and the ellipse $2x^{2} + y^{2} = 4$ is $y = 2x + 2\sqrt{3}$ If the line $y = mx + \frac{4\sqrt{3}}{m}, \left(m\ne0\right)$ is a common tangent to the parabola $y^{2} = 16\sqrt{3}x$ and the ellipse $2x^{2} + y^{2} = 4$, then m satisfies $m^{4} + 2m^{2} = 24.$

• A. Statement-1 is false, Statement-2 is true.
• B. Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1.
• C. Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1.
• D. Statement-1 is true, statement-2 is false.

Answer 1

The Correct Option is B

Equation of tangent to the ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{4} = 1$ is $y = mx \pm \sqrt{2m^{2}+4}\quad\quad.....\left(1\right)$ equation of tangent to the parabola $y^{2} = 16 \sqrt{3}x$ is $y = m x + \frac{4\sqrt{3}}{m}\quad\quad.....\left(2\right)$ On comparing $\left(1\right)$ and $\left(2\right)$ $\frac{4\sqrt{3}}{m} = \pm \sqrt{2m^{2}+4}$ $\Rightarrow\quad48 = m^{2} \left(2m^{2} + 4\right)\quad\quad\Rightarrow\quad2m^{4 }+ 4m^{2} - 48 = 0$ $\Rightarrow\quad m^{4 }+ 2m^{2} - 24 = 0\quad\quad\Rightarrow\quad\left(m^{2} + 6\right) \left(m^{2} - 4\right) = 0$ $\Rightarrow\quad m^{2} = 4\quad\quad\Rightarrow\quad m = \pm 2$ $\Rightarrow\quad$ equation of common tangents are $y = \pm 2x \pm 2 \sqrt{3}$ statement -1 is true. statement-2 is obviously true.