The shortest distance from surface to a point Many have asked the question about finding the shortest distance from a point to a plane. I have checked those questions and answers and haven't found what I am looking for. Might still have missed some question though. 
I have a problem where I am to calculate the shortest distance from the surface $z = x^2 + 3y^2$ to the point $P = (5, 0, 1)$
I want to use the Langrange multiplier for this. 
This is what I have done: 
I am to minimize $$(x-5)^2 + y^2 + (z - 1)^2$$
s.t.
$$x^2 + 3y^2 - z = 0$$
So the Lagrange function is 
$$L(x, y, z, \lambda) = (x-5)^2 + y^2 (z-1)^2 + \lambda(x^2 + 3y^2 - z)$$
I have taken the gradient of $L$ and am looking for where each partial derivative equals to zero. 
And I am stuck. Is $\lambda = 1/3$? It seems like that, otherwise $y=0$. But if $y\neq 0$ and lambda is $1/3$ then I get that $y=0$ anyway :/ 
 A: I think I can provide a simpler method for this specific problem that does not use Lagrange multipliers. 
Take a cross section of the 3-dimensional solid $z=x^2+3y^2$ at the plane $y=0$. We then have a parabola on the $xz$ plane: $z=x^2, y=0$. Note that point $P$ is also on this plane. To find the smallest distance between the original surface and the point, all we have to do is find the smallest distance between the point and the parabola. We can do this by looking at the normal lines of the parabola, which are of the form $z-m^2=-\frac{1}{2m}(x-m), y=0$ for some point $(m,0,m^2)$. We want this line to contain the point $(5,0,1)$. 
$$1-m^2=-\frac{1}{2m}(5-m)$$
$$2m-2m^3=-5+m$$
$$2m^3-m-5=0$$
$$m=\frac{\sqrt[3]{45-\sqrt{2019}}+\sqrt[3]{45+\sqrt{2019}}}{6^{2/3}}$$
Ergo, the distance we are looking for is the distance between the points $$\left(\frac{\sqrt[3]{45-\sqrt{2019}}+\sqrt[3]{45+\sqrt{2019}}}{6^{2/3}},0,\left(\frac{\sqrt[3]{45-\sqrt{2019}}+\sqrt[3]{45+\sqrt{2019}}}{6^{2/3}}\right)^2\right) \text{and } (5,0,1)$$
I know this may seem way off, but Behrouz Maleki's solution also agrees with this answer (look at case 1 of his answer). 
A: Set
$$f(x,y,z)=(x-5)^2+y^2+(z-1)^2$$
$$g(x,y,z)=z-x^2-3y^2=0$$
By application of Lagrange method, we have 
$$\nabla f=\lambda \nabla g$$
$$(2x-10\,,\,2y\,,\,2z-2)=\lambda(-2x\,,\,-6y\,,\,1)$$
therefore
$$\left\{ \begin{align}
  & y=0\,\,\,\,\quad\Rightarrow \,\,\,z={{x}^{2}} \\ 
 & \lambda =-\frac{1}{3}\,\,\,\Rightarrow \,\,\,x=\frac {15} 2\,\,\,\,,\,\,z=\frac{5}{6} \\ 
\end{align} \right.$$
Case 1
$y=0\quad,\quad z=x^2$
$$f(x,0,x^2)=(x-5)^2+(x^2-1)^2=x^4-x^2-10x+26$$
Now you should minimize $f(x,0,x^2)$
Case 2
$x=\frac {15} 2\,\,\,\,,\,\,z=\frac{5}{6} $
$$g\left(\frac {15} 2,\,y\,,\frac{5}{6}\right)=\frac{5}{6}-\frac{225}{4}-3y^2=0$$
therefore
$$y=\pm \frac{\sqrt{185}}{3}$$
Now you should calculate $f\left( \frac{15}{2},\frac{\sqrt{185}}{3},\frac{5}{6} \right)$ and $f\left( \frac{15}{2},-\frac{\sqrt{185}}{3},\frac{5}{6} \right)$
