Find all the points on the cone $z^2=x^2+4y^2$ that are the closest to the point $(0,0,c)$ 
From all the points on the cone $z^2=x^2+4y^2$ find the closest to the point $(0,0,c)$. State explicitly the minimal distance. $c$ is a constant.

Lagrange multipliers can be used here.
Let the constraint function $g=x^2+4y^2-z^2$.
Let the minimization function $f=(x-0)^2+(y-0)^2+(z-c)^2$ as function of finding the distance in between 2 points ion 3d space. We're also going to keep $f$ in squares so we don't have to get the factor out $z$ from the ellipse equation.
Then we have:
$$
\nabla g=\langle2x, 8y,-2z \rangle\\
\nabla f=\langle 2x, 2y,2(z-c)\rangle
$$
So:
$$
\begin{cases}
2x=k\cdot2x\\
2y=k\cdot 8y\\
2(z-c)=k\cdot(-2z)
\end{cases}
$$
We have that $k=\frac{1}{4}$ then $x=0$ and $z=\frac{4c}{5}$.
If we plug those values into the ellipse equation we get that:
$$
c^2=\frac{125y^2}{4}\Rightarrow c=\pm\frac{\sqrt{125}y}{2}
$$
So the points would be $(0,y,\frac{\sqrt{125}y}{2})$ and $(0,y,-\frac{\sqrt{125}y}{2})$.
First I think something is wrong with my calculations (although I double checked it). Also I'm not sure what does this mean that we have 2 points and how am I supposed to know which one is the minimum.
 A: Minimizing the squared distance,
$$\begin{array}{ll} \text{minimize} & x^2 + y^2 + (z-c)^2\\ \text{subject to} & z^2 = x^2 + 4 y^2\end{array}$$
Let the Lagrangian be
$$\mathcal L (x,y,z,\lambda) := x^2 + y^2 + (z-c)^2 + \lambda (x^2 + 4 y^2 - z^2)$$
Taking the partial derivatives and finding where they vanish, we obtain
$$\begin{array}{rl} (1 + \lambda) \, x &= 0\\ (1 + 4\lambda) \, y &= 0\\ (1 - \lambda) \, z &= c\\ x^2 + 4 y^2 - z^2 &= 0\\\end{array}$$
We have two cases to consider.

$\color{blue}{\boxed{\lambda = -1}}$
In this case, $y = 0$ and $z = \frac c2$. The value of $x$ is given by
$$x = \pm \sqrt{ z^2 - 4 y^2 } = \pm \frac c2$$
Hence, we have the two points
$$(x,y,z) = \left( \pm \frac c2, 0, \frac c2 \right)$$
whose squared distance from $(0,0,c)$ is $\frac{c^2}{2}$.

$\color{blue}{\boxed{\lambda = - \frac 14}}$
In this case, $x = 0$ and $z = \frac{4c}{5}$. The value of $y$ is given by
$$y = \pm \frac 12 \sqrt{ z^2 - x^2 } = \pm \frac{2c}{5}$$
Hence, we have the two points
$$(x,y,z) = \left( 0, \pm \frac{2c}{5}, \frac{4c}{5} \right)$$
whose squared distance from $(0,0,c)$ is $\frac{c^2}{5} < \frac{c^2}{2}$. This is the minimal squared distance.

Example
Let $c = 5$. Hence, the points on the cone closest to $(0,0,5)$ are $(x,y,z) = ( 0, \pm 2, 4 )$. Here is a plot of the cone and the line segment whose endpoints are $(0,0,5)$ and $( 0, 2, 4)$

The length of the line segment is $\sqrt 5$.
A: 
It is infinite elliptic cone May be it can help you to decide.
