Question

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

Answer 1

$y^2=16\sqrt{3}x$ $y=mx+\frac{4\sqrt{3}}{m}$

$\frac{x^2}{2}+\frac{y^2}{4}=1$, $x=m_{1}y+\sqrt{4m_1^2+2}$

$\Rightarrow y=\frac{x}{m_1}-\sqrt{4+\frac{2}{m_1^2}},m=\frac{1}{m_1}$

${\left(\frac{4\sqrt{3}}{m}\right)}^2={(-\sqrt{4+\frac{2}{m_1^2}})}^2$

$=\frac{48}{m^2}+4+\frac{2}{m_1^2}=4=2m^2$

$\Rightarrow m^4+2m^2-24=0$.