Find the point on the cone closest to (1,4,0) Find the point on the cone $z^2=x^2+y^2$ nearest to the point $P(1,4,0)$.
This is a homework problem I've not made much headway on.

Have a look at the scene to come up with an idea:
 (Large Version)
The green surfaces belong to the cone, the red and blue surfaces are spheres around $P$.
E.g you could try to find the intersection between cone and sphere and fiddle with the radius to shrink it to a point.
We can describe the intersection by the system
$$
x^2 + y^2 - z^2 = 0 \\
(x-1)^2 + (y - 4)^2 + z^2 = r^2
$$
This gives
$$
r^2 = (x-1)^2 + (y-4)^2 + x^2 + y^2 = 2x^2 + 2y^2 - 2x - 8y + 17 \iff \\
r^2/2 = x^2 + y^2 - x - 4y + 17/2
= (x - 1/2)^2 + (y-2)^2 + 17/2 - 1/4 - 4 \iff \\
\left(\frac{\sqrt{2r^2-17}}{2}\right)^2 
= \left(x - \frac{1}{2}\right)^2 + \left(y - 2\right)^2
$$
If I made no error, this indicates a circle around $(1/2, 2, 0)$ as projection of the intersection surface onto the $x$-$y$-plane.
Then $z = \pm \sqrt{1/4 + 4} = \pm \sqrt{17}/2$.
That would suggest $Q=(1/2,2,\pm\sqrt{17}/2)$ as closest points, at distance $r = \sqrt{17/2}$.
 (Large Version)
 (Large Version)
