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.

This seems like an exercise in Lagrange Multipliers. You need to minimize the distance function $f(x,y,z) = (x-1)^2 + (y-4)^2 + z^2$ (which is the square distance from $(x,y,z)$ to $(1,4,0)$) subject to the constraint that $g(x,y,z) = x^2 + y^2 - z^2 = 0$. Any point which minimizes such $f$ subject to the constraint will satisfy $$\nabla f = \lambda \nabla g$$ for some $\lambda \in \mathbb R$. This gives us 4 equations: \begin{align*} 2(x-1) &= 2\lambda x \\ 2(y-4) &= 2\lambda y \\ 2z&= -2\lambda z \\ x^2 + y^2 - z^2 &= 0.\end{align*} The third equation gives $z = 0$ or $\lambda = -1$. If $z=0$, then the fourth equation gives $x=0, y= 0$. But then the first two equations couldn't be satisfied, so this is impossible. We conclude that $\lambda = -1$. In this case $x = 1/2$, $y=2$ so by the fourth equation $z = \pm\sqrt{17/4}$. By the symmetry in $z$, both sides of the $\pm$ work, so the points which minimize the distance are $$\left(\tfrac 1 2, 2, \sqrt{17/4}\right) \,\,\,\, \text{ and } \,\,\,\, \left(\tfrac 1 2, 2, -\sqrt{17/4}\right).$$ The distance is then $$d_{\text{min}} = \sqrt{\left(\tfrac 1 2 - 1 \right)^2 + (2-4)^2 + \left( \pm\sqrt{17/4}\right)^2}= \sqrt{\tfrac 1 4 + 4 + \tfrac{17}4} = \sqrt{\tfrac{34}{4}} = \tfrac{\sqrt{34}}{2}.$$
