Finding a normal vector of a curve

Question

Find a normal vector of the curve $ 6x^2+2y^2+z^2 = 225 $ at the given point $ P : (5, 5, 5).$

I don't think my solution is correct since $f \ne 0 $

grad f = $[12x, 4y, 2z]$
grad f(p) = [60 , 20 , 10]

n =$ \frac{ grad f(p) }{|grad f(p)|} = \frac {[60,20,10]}{10\sqrt41}$ = $[\frac{6}{\sqrt41},\frac {2}{\sqrt41},\frac{1}{\sqrt41}] $

is it the right way to attempt it?

Answer 1

In this case, you are taking your f to be $f(x,y,z) = 6x^2 + 2y^2 + z^2$ and $6x^2 + 2y^2 + z^2 = 255$ is a level curve or level surface etc. (Your equation is not a curve) and at a given point gradient is normal to the level surface. Thus $f \neq 0$ should not worry you.