To find minimum distance between two curves Let $P(x, y, 1)$ and $Q(x, y, z)$ lie on the curves $$\frac{x^2}{9}+\frac{y^2}{4}=4$$ and $$\frac{x+2}{1}=\frac{y-\sqrt{3}}{\sqrt{3}}=\frac{z-1}{2}$$ respectively. Then find the square of the minimum distance between $P$ and $Q$.
My Attempt is:
I tried to find minimum distance between the points $(-2,\sqrt{3})$ and $(6\cos \theta,4\sin \theta)$.

Starting from @Christian Blatter's answer, using $s=2 \tan ^{-1}(x)$ and expanding, we end with
$$2 \sqrt{3}\, x^4+70 \,x^3+72 \sqrt{3} \,x^2-274\, x-26 \sqrt{3}=0$$
Let $x=t-\frac{35}{4 \sqrt{3}}$ to get the depressed quartic
$$t^4-\frac{937 }{8}t^2+\frac{24467}{24 \sqrt{3}} t-\frac{166043}{256}=0$$ which can be exactly solved using radicals.
Following the steps given here, we have
$$\Delta=\frac{386701126204}{27}\quad P=-937\quad Q=\frac{24467}{3 \sqrt{3}}\quad \Delta_0=5935\quad D=-261003$$ So, four real roots with
$$p=-\frac{937}{8}\quad q=\frac{24467}{24 \sqrt{3}}$$
Just finish to get the exact values of $(t_1,t_2,t_3,t_4)$ from which $(x_1,x_2,x_3,x_4)$ and finally $(s_1,s_2,s_3,s_4)$ in terms of messy radicals. 
