The closes point to a curve in space.

Question

I am working on the following problem.

Find the point closest to the origin, of the curve of intersection of the plane $2y+4z =5$ and the cone $z^2 = 4(x^2+y^2)$

I was able to see that the intersection will be a tilted ellipse with coordinates $(x,y,4(x^2+y^2))$.

So I was thinking that I could try putting $F(x,y) = \sqrt{x^2+y^2+16(x^2+y^2)^2}$ and optimize it. But this seems to be too long a calculation and I want to somehow use vectors, such as dot products and gradients because that seems to be easier.

Can anyone help me get started ?

Answer 1

This is a typical exercise for Lagrange multipliers theorem (see http://en.wikipedia.org/wiki/Lagrange_multiplier). You want to refolmulate the problem to the following:

Find the minimum of function $\sqrt{x^2 + y^2 + z^2}$ (that is the distance from origin, you can even use easier function x^2 + y^2 + z^2, because the solution will be the same) on the set which is intersection of the plane $2y+4z=5$ and the cone $z^2=4(x^2+y^2)$.

I guess you know a little about Lagrange multipliers, this shouldn't be hard to compute.

Answer 2

I think it is not necessary to use Lagrange multipliers as it is can be treat a simple geometry problem.

you already know that there is a ellipse and there are two special points $P$ and $Q$.

you can see the $P$ and $Q$ in on the plane $x=0$, if you consider two rings on the cone, one is through $P$, another is through $Q$, all the points on the two rings to original point will have same distances, one is $PO$, another is $QO$,it is trivial that $PO$ is min,$QO$ is max for the ellipse. so you only need to find $P,Q$, recall $x=0$ plane, the problem become a simple one. $P(.5,1)$, so $PO=\sqrt{(0.5)^2+1}=\dfrac{\sqrt{5}}{2}$