# Find the minimum and maximum distance between the ellipsoid of ecuation $x^2+y^2+2z^2=6$ and the point $P=(4,2,0)$

I've been asked to find, if it exists, the minimum and maximum distance between the ellipsoid of ecuation $x^2+y^2+2z^2=6$ and the point $P=(4,2,0)$

To start, I've tried to find the points of the ellipsoid where the distance is minimum and maximum using the method of Lagrange multipliers.

First, I considered the sphere centered on the point $P$ as if it were a level surface and I construct:

$f(x,y,z)=(x-4)^2+(y-2)^2+z^2-a^2$

And I did the same with the ellipsoid, getting the following function:

$g(x,y,z)=x^2+y^2+2z^2-6$

Then, I tried to get the points where $\nabla$$f and \nabla$$g$ are paralell:

$\nabla$$f=\lambda$$\nabla$$g \rightarrow (2x-8,2y-4,2z)=\lambda(2x,2y,4z) So, solving the system I got that \lambda=1/2, so in consquence x=8 and y=4 But when I replaced the values of x,y obtained in g=0 to get the coordinate z I get a complex number. I suspect there is something on my reasoning which is incorrect, can someone help me? ## 3 Answers Taking from where you left off. By equating coordinates of both sides, you have: 2z = 4\lambda z\implies z = 0 or \lambda = \dfrac{1}{2}. If \lambda = \dfrac{1}{2} \implies x = 8 which is not possible since x^2 \le 6. So what is left is z = 0 or \lambda = 0. Also \lambda \neq 1 for otherwise 2x-8 = 2x which is not possible. \lambda \neq 0, for if it were 0, then you have: 2x-8 = 0 \implies x = 4, and this is not possible since x^2 \le 6. Thus 2x-8 = 2\lambda x, 2y-4 = 2\lambda y\implies x = \dfrac{4}{1-\lambda}, y = \dfrac{2}{1-\lambda}\implies \dfrac{16}{(1-\lambda)^2}+\dfrac{4}{(1-\lambda)^2}=6\implies (1-\lambda)^2 = \dfrac{20}{6} = \dfrac{10}{3}\implies \lambda = 1\pm \dfrac{\sqrt{30}}{3}. Can you take it from here ?. It looks as if one of the lambdas will yield a min and the other a max. • Following yor suggestions, I've got two points: Q_{1}(x,y,z)=(-12/\sqrt{30}$$,-6/\sqrt{30},0)$ and $Q_{2}(x,y,z)=(12/\sqrt{30}$$,6/\sqrt{30},0). Now, from here to get the minimum and maximum distance I could do it through |Q_{i}P|=\sqrt{(x-4)^2+(y-2)^2+z^2}, right? Jun 17 '17 at 16:12 • Yep.Try it and let me know your answer. Jun 17 '17 at 21:47 • I obtained that |Q_{1}P|=\sqrt{36+10/3\sqrt{30}}, and |Q_{2}P|=\sqrt{36-10/3\sqrt{30}}. What do you think @DeepSea? Jun 19 '17 at 19:16 • I think you got the right answer. Are you good at physics also? or so so? Jun 19 '17 at 23:27 • haha I think I am, but I'm in my first year of the university, so in process of getting way much better! Jun 20 '17 at 13:56 Since by C-S$$2x+y\leq\sqrt{(x^2+y^2)(2^2+1^2)}=\sqrt{5(x^2+y^2)},$$we obtain$$(x-4)^2+(y-2)^2+z^2=x^2+y^2+z^2-4(2x+y)+20\geq \geq x^2+y^2+z^2-4\sqrt{5(x^2+y^2)}+20=6-2z^2+z^2-4\sqrt{5(6-2z^2)}+20==26-z^2-4\sqrt{5(6-2z^2)}\geq26-4\sqrt{30}.$$It's obvious that the equality occurs, which gives the minimal value:$$\sqrt{26-4\sqrt{30}}.$$Since -2x-y\leq\sqrt{5(x^2+y^2)}, by the same way we can get a maximal value:$$\sqrt{26+4\sqrt{30}}.$$Done! hint The ellipse can be parametrized as$$x_e =\sqrt {6}\sin (\phi)\cos (\theta) y_e=\sqrt{6}\sin (\phi)\sin (\theta ) z_e=\sqrt {3}\cos (\phi) $$the square of the distance from a point of the ellipse to the point (4,2,0) is$$D^2=(x_e-4)^2+(y_e-2)^2+z_e^2=3\sin^2 (\phi)+23-8x_e-4y_e $$You can find min and max D, by solving the system,$$\frac {\partial D^2}{\partial \phi}=\frac {\partial D^2}{\partial \theta}=0$\$

• In my course, I didn't learn yet how to parametrize through polar coordinates. In which book can I read something about it? Jun 17 '17 at 16:18
• @NeisySofíaVadori Look for spherical coordinates. Jun 17 '17 at 16:54