JEE Main · 2012 · 29 Apr 2012 · Offline · Parabola

For the parabola $y^2=16\sqrt{3}x$, $4a=16\sqrt{3}$ gives $a=4\sqrt{3}$, so its tangent of slope $m$ is $y=mx+\frac{4\sqrt{3}}{m}$. For this line to touch the ellipse $\frac{x^2}{2}+\frac{y^2}{4}=1$, the tangency condition $c^2=a^2{m}^2+b^2$ with $a^2=2$, $b^2=4$ and $c=\frac{4\sqrt{3}}{m}$ gives $\frac{48}{m^2}=2m^2+4$, that is $m^4+2m^2=24$ — exactly the second statement. Solving, $m^2=4$, so $m=\pm 2$, and $m=2$ gives the common tangent $y=2x+2\sqrt{3}$, confirming the first statement. The second statement supplies the very condition that produces this tangent, so it correctly explains the first.