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?
Best Answer
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.