Statement 1 : An equation of a common tangent to the parabola $y^{2}=16 \sqrt{3} x$ and the ellipse

Question

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 $y=2 x+2 \sqrt{3}$ \Statement 2 : If the line $y=m x+\frac{4 \sqrt{3}}{m},(m 
eq 0)$ 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 $m^{4}+2 m^{2}=24$.

Professor PC Thomas Classes · Answer

(d) : Statement 1 :\ $y^{2}=16 \sqrt{3} x, y=m x+\frac{4 \sqrt{3}}{m}$\ $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1, x=m_{1} y+\sqrt{4 m_{1}^{2}+2}$\ $\Rightarrow y=\frac{x}{m_{1}}-\sqrt{4+\frac{2}{m_{1}{ }^{2}}}$ Now, $m=\frac{1}{m_{1}}$ and $\left(\frac{4 \sqrt{3}}{m}ight)^{2}=\left(-\sqrt{4+\frac{2}{m_{1}{ }^{2}}}ight)^{2}$\ $\Rightarrow \frac{48}{m^{2}}=4+\frac{2}{m_{1}^{2}}=4+2 m^{2} \Rightarrow \frac{24}{m^{2}}=2+m^{2}$ $$\begin{equation*}\Rightarrow \quad m^{4}+2 m^{2}-24=0 	ag{1} \end{equation*}$$ $\Rightarrow \quad\left(m^{2}+6ight)\left(m^{2}-4ight)=0 \Rightarrow m= \pm 2$\ Statement 2: If $y=m x+\frac{4 \sqrt{3}}{m}$ is a common tangent to $y^{2}=16 \sqrt{3} x$ and ellipse\ $2 x^{2}+y^{2}=4$, then $m$ satisfies\ $m^{4}+2 m^{2}-24=0$\ From (1), statement 2 is a correct explanation for statement 1 .\