Please, help me with my homework... Prove that the normal to the surface of revolution $z = f(\sqrt{x^2+y^2})$ intersect the axis of rotation.
first, I need to find $z_x$, $z_y$
$z_x$ = $x*\frac{f'(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}$
$z_y$ = $y*\frac{f'(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}$
Next, I need to find the equitation of tangential surface
$z_x(x_0, y_0)*(x-x_0) + z_y(x_0, y_0)*(y-y_0) = 0 $
Next, the normal :
$\frac{x-x_0}{z_x(x_0,y_0)}=\frac{y-y_0}{z_y(x_0,y_0)}$
And what's the next step? How to prove?