tangent plane to a surface, does it exist in origo?

Question

I have the following equation and points:

$f(x,y)=\sqrt{x^2+y^2}$ and P=(3, 4, 5)

I calculated the tangent plane to:

$3x+4y-5z=0$

1. Is that correct?
2. And does the tangent plane exist in origo (0, 0, 0)?
3. Is it a critical point in origo?

Thanks a lot in advance:)

Cheers Tom

Answer 1

This figure is a (double) cone with vertex at (0, 0, 0) (in English that is the "origen"). Yes, the point (3, 4, 5) does lie on that line and the tangent plane at that point is given by 3x+ 4y- 5z= 0. yes, that plane contains the origin. In fact the tangent plane at any point on a cone contains the origin. There is, however, no plane that is tangent to the cone [b]at[/b] its vertex. The cone is not "smooth" there.