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 $(6\cos \theta,4\sin \theta)$.
 A: You can use the method of Lagrange's multipliers. The function formed by the distance between the two points $ (x,y,z)$ and $(x,y,1)$ is examined.
i.e. $\phi = \sqrt{(z-1)^2} $
The constrains are respectively
$$ \frac{x^2} {9} + \frac{y^2} {4} $$
And
$$ \frac{x+2} {1} = \frac{y- \sqrt{3}} {\sqrt{3}} =\frac{z-1} {2} $$
The auxiliary function is formed as 
$$ F(x_1, x_2 , x_3, . . . , x_n, \alpha_1, \alpha_2 . . . , \alpha_k ) = f(x_1, x_2, . . .,x_n) + \sum_{i=0}^k \alpha_i \beta_i ( x_1, x_2, . . . , x_n) $$
Where $\beta_i $ is the function
Now $$\frac{\partial F}{\partial x_1} =0=\frac{\partial F}{\partial x_2} = . . . = \frac{\partial F}{\partial x_n} $$
Which gives the stationary points of F
After these you have to find the extremum points and obtaining the value of $ \alpha_1 , \alpha_2, . . . , \alpha_n $ these are the multipliers
You can further obtain the points for maximum distance
A: You can do it without using Lagrange's method. Consider the parametric representations
$$p(s):=\bigl(6\cos s,4\sin s,1\bigr)\qquad(s\in{\mathbb R}/(2\pi))$$
and
$$q(t):=\bigl(t-2,\sqrt{3}(t+1),2t+1\bigr)\qquad(t\in{\mathbb R})\ .$$
We have to determine $s$ and $t$ such that the vector
$$f(s,t):=p(s)-q(t)$$
is orthogonal to $p'(s)=\bigl(-6\sin s, 4\cos s,0\bigr)$ and to $q'(t)=(1,\sqrt{3},2)=:u$. In this way one obtains the equations 
$$f(s,t)\cdot p'(s)=0,\qquad f(s,t)\cdot u=0\ .$$
Computing $t=h(s)$ from the second equation leads to the single equation
$$g(s):={1\over4}\bigl(-14 \sqrt{3} \cos s - 
   12 \sqrt{3} \cos(2s) - (51 + 86 \cos s) \sin s\bigr)=0\ .$$
The last equation has four solutions $s_i$ (found numerically), and computing the values $$d_i^2:=\bigl|f\bigl(s_i,h(s_i)\bigr)\bigr|^2$$
we obtain exactly the values found by @Cesareo.
Here is my computer output for this problem:

A: 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. 
