What's the normal line when partial derivative with respect to $x = 0$? So I know the equation of a normal line is: $$\frac{x-a}{f_1(a,b)}=\frac{y-b}{f_2(a,b)}=\frac{z-f(a,b)}{-1}$$. But in my textbook there is a litle note that says "with suitable modifications if either $f_1(a,b)=0$ or $f_2(a,b)=0$". Funny that they forget what that modification would look like. Now I have a problem where $f_1=0$, and don't know what to do. I searched for this topic, but frankly the given formula we use is apparantly not so common. Anyhow If someone would know the modification. It would be very helpful.
Thanks in advance.
 A: You have a surface defined by $z=f(x,y)$ and a point $P(a,b, f(a,b))$ in that surface.
The plane tangent to the surface at point P is given by:
$z= f(a,b)+ \frac{\delta f}{\delta x}(a,b)(x-a)+ \frac{\delta f}{\delta y}(a,b)(y-a)$
Let's write it like
$f(a,b)+ \frac{\delta f}{\delta x}(a,b)(x-a)+ \frac{\delta f}{\delta y}(a,b)(y-a) -z =0$
Now compare with a generic plane $Ax+By+Cz+D=0$
We see that $A=\frac{\delta f}{\delta x}(a,b)\;\;$ and $\;\;B=\frac{\delta f}{\delta y}(a,b)\;\;$ and $\;\;C=-1$
We know that $(A,B,C)$ are the components of a vector perpendicular to the plane. So it's the normal to the surface in that point P.
Defining a line in 3D is better done in parametrics. So our normal is:
$x=a+At$
$y=b+Bt$
$z=f(a,b)-t$
If you eliminate "t" in these parametrics you arrive to your original expression
$\frac{x-a}{A}=\frac{y-b}{B}=\frac{z-f(a,b)}{-1}$
Notice that you can avoid division by zero:
$B(x-a)=A(y-b)$
Now, if $A=\frac{\delta f}{\delta x}(a,b)=0$ then the first parametric is just $x=a$, but you can still use the other expression $\frac{y-b}{B}=\frac{z-f(a,b)}{-1}\;\;$ or without denominators $\;\;y-b=B(f(a,b)-z)$
A: We are talking of the normal line, not the "tangent" line. Talking of surfaces we have tangent planes not tangent lines.
The general equation of the normal is
$$n:(x_0,y_0,z_0)+t (f'_{x_0},f'_{y_0},f'_{z_0})$$
