How to find the set of all points in a submanifold minimizing the distance to a given point? I have the following set $M$ defined as the set of all points $(x,y,z)$ such that $ x^2+y^2-z^2=-1$. And I'm asked the following three questions :
a) Prove that this is a submanifold of $R^3$
b) Find its tangent space in the point (2,2,3) 
c) Find the set of all point $p_0 \in M$ which minimizes the distance to the point $q= (0,0,4)$  
I would like to know if my procedures for a) and b) are correct, and how exactly I should proceed for c)

a) To prove that this is a submanifold, we just take the derivative of $f(x,y,z) = x^2+y^2-z^2+1$ and set it equal to zero. Then we need to check that the results aren't contained in the set. So $(2x,2y,-2z)=(0,0,0)$ So the only point is $(0,0,0)$, which isn't contained in our set (we get $0=-1$ which is obviously wrong)
b) To find it's tangent space to the point $(2,2,3)$, we input our point into our derivative and find its kernel. So we get $(4,4,-6)=(0,0,0)$. If we solve this, we get $(-1,1,0)$ and $(\frac{3}{2},0,1)$. Now the tangent space to this point is simply the span of these two vectors.
c) Now my problem is here. I couldn't find an example in my book about this step. There were example for a) and b), but nothing for c). Do I somehow need to use $d=\sqrt{(x-x_0)^2+(y-y_0)^2)+(z-z_0)^2)}$ ?  Also, I searched on the internet but didn't find "practical/ computing-type" examples for this question. I found a lot of abstract definitions but no real examples, that's why I'm asking.
Thanks for your help !

Edit : Okay, just in case this helps someone in the future. What is expected in c) is that  we use the distance formula squared (to make life simpler), i.e. $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2$, then our vector from q to p is $(x-0)^2+(y-0)^2+(z-4)^2$, the gradient of which is $(2x, 2y, 2z-8)$. The idea is to use Lagrange multiplier with the set M (i.e. $x^2+y^2-z^2+1$) as a constraint. So we get as in the comments $(2x, 2y, 2z-8)= \lambda (2x, 2y, -2z)$ which yields after solving : $z=2$ and x and y are free variables. We then plug $z=2$ into our set M, which gives the circle $x^2+y^2=3$. Thus all the points that minimize the distance are $(\sqrt{3}cos(\theta), \sqrt{3}sin(\theta), 2)$.  
Maybe this will help someone in the future.
 A: 
b) To find it's tangent space to the point (2,2,3), we input our point into our derivative and find its kernel. So we get (4,4,−6)=(0,0,0). If we solve this, we get (−1,1,0) and (32,0,1). Now the tangent space to this point is simply the span of these two vectors.'

I have three corrections here. 
The first is that "it's" means "it is"; pronouns' possessive forms have no apostrophes. The word you want to use is "its". 
Second, the statement $(4,4,−6)=(0,0,0)$ is false. I think you meant to say that $v$ is in the kernel if and only if $(4,4,−6) \cdot v=(0,0,0)$. 
Third, if we solve that correctly written equation, we don't just get $(−1,1,0)$ and $(32,0,1)$ --- we get all vectors of the form $a(−1,1,0) + b (32,0,1)$ (where $a$ and $b$ are arbitrary real numbers). What kind of result would you have ended up with if you'd made the equally correct claim that "If we solve this, we get $(-1, 1, 0)$ and $(4, -4, 0)$"? 

As for part 3, here's a big hint: if $p \in M$ is as close as possible to $q$, then 


*

*the vector from $p$ to $q$ is perpendicular to the tangent plane at $p$, and 

*the point $p$ satisfies the defining equation for $M$. 


Post-comment addition
Now that you seem to have solve the problem using Lagrange, I'll go ahead and work out what I was suggesting. Let's let $p = (x, y, z)$. Then condition $1$ says that the vector
$$
v = (x, y, z-4)
$$
from $q$ to $p$ must be parallel to the normal vector at $p$, which is proportional to 
$$
n = (x, y, -z)
$$
So we know that $v$ is a multiple of $n$, say
$$
(x, y, z-4) = \lambda(x, y, -z)
$$
...which is exactly the same equation you came up with using Lagrange multipliers. And the algebra is all the same form here on out. 
Why do we get the same equation? Because the gradient [i.e., direction of fastest growth] (at a point $p$) of the "distance from $p$ to a fixed point $q$" function is exactly a multiple of the ray from $q$ to $p$: if you want to increase that distance fastest, you move in the direction of that ray! 
