To find minimum distance between two curves

Question

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 $(6\cos \theta,4\sin \theta)$.

Answer 1

We can solve this problem proposing a lagrangian. So calling

$$ d^2 = (x_1-x_2)^2+(y_1-y_2)^2+(1-z_2)^2\\ g_1 = \frac{x_1^2}{9}+\frac{y_1^2}{4}-4\\ g_3 = x_2+2-\lambda\\ g_4 = y_2-\sqrt 3-\sqrt 3\lambda\\ g_5 = z_2-1-2\lambda $$

and forming

$$ L(x_1,y_1,x_2,y_2,z_2,\lambda,\mu_1,\mu_2,\mu_3,\mu_4) = d^2+\sum_i\mu_i g_i $$

the stationary condition gives

$$ \nabla L = 0 = \left\{ \begin{array}{c} \frac{2 \mu_1 x_1}{9}+2 (x_1-x_2) \\ \frac{\mu_1 y_1}{2}+2 (y_1-y_2) \\ \mu_2-2 (x_1-x_2) \\ \mu_3-2 (y_1-y_2) \\ \mu_4-2 (1-z_2) \\ \frac{x_1^2}{9}+\frac{y_1^2}{4}-4 \\ -\lambda +x_2+2 \\ -\sqrt{3} \lambda +y_2-\sqrt{3} \\ -2 \lambda +z_2-1 \\ -\mu_2-\sqrt{3} \mu_3-2 \mu_4 \\ \end{array} \right. $$

Solving this system we get

$$ \left( \begin{array}{ccccccccccc} x_1&y_1&x_2&y_2&z_2&\mu_1&\mu_2&\mu_3&\mu_4&\lambda&d^2\\ -5.96291 & -0.444062 & -2.96651 & 0.0580128 & -0.933013 & -4.52256 & -5.99281 & -1.00415 & 3.86603 & -0.966506 & 12.9671 \\ -5.07051 & -2.13853 & -3.22182 & -0.384201 & -1.44364 & -3.28137 & -3.69739 & -3.50865 & 4.88727 & -1.22182 & 12.4667 \\ -1.7813 & 3.81965 & -1.52068 & 2.56225 & 1.95863 & -1.31677 & -0.521237 & 2.51481 & -1.91727 & 0.479317 & 2.56796 \\ 5.72047 & -1.20669 & -1.6712 & 2.30155 & 1.6576 & -11.6293 & 14.7833 & -7.01649 & -1.31521 & 0.328802 & 67.377 \\ \end{array} \right) $$

so the minimum distance is $d = \sqrt{2.56796}$ with

$$ p_1 = ( -1.7813, 3.81965, 1) \in P\\ q_1 = (-1.52068, 2.56225; 1.95863) \in Q $$

Answer 2

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

Answer 3

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:

Answer 4

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.