When writing the Lagrange function does it matter if the constraint has minus/plus sign? My university teaches me like this:

However everywhere I look on the internet there is either a 1) minus sign before the lambda 2) or the whole equation is written differently (by placing derivatives of the function and constraint on the opposite sides of the equation).
(Does this have something to do with the tangent lines being parallel and the direction not mattering?)
A bit unrelated, but I figured I'd ask anyways. My university also makes me compute the total differential and then check for the sign to figure out if it's a min or a max. But the weird thing about that is that there is this exception when the computed differential would not be strictly less or strictly more than zero - and then I have to show that dx and dy can't be zero at the same time. Are there any examples of this method anywhere?
Thanks
 A: The essence of Lagrange's method is the following: At a point ${\bf p}$ where the function $f$ has a conditional extremum one necessarily has
$$\nabla f({\bf p})=\lambda\>\nabla g({\bf p})\tag{1}$$
for a certain real number $\lambda$. The geometrical significance of this is that the two gradients should be parallel. Therefore in $(1)$ I could have written $-\lambda$ instead of $\lambda$ as well. Some people, among them the chancellor of your university, encode the condition $(1)$ into an auxiliary and artificial so-called Lagrangian function $$F(x,y,z,\lambda):=f(x,y,z)-\lambda g(x,y,z)$$
(sometimes with $\lambda$ instead of $-\lambda$), and require that the derivatives of this $F$ with respect to all its variables are zero. But note that this function $F$ has no geometric or physical meaning whatsoever.
Towards your second question: When the second differential
$$d^2 f({\bf p})(X,Y):=f_{xx}({\bf p})X^2+2f_{xy}({\bf p})XY+f_{yy}({\bf p})Y^2$$
is not definite additional algebraic analysis is necessary to check whether we have an extremum at ${\bf p}$. Consider the example
$$f(x,y):=x^2+\alpha y^4$$
at ${\bf p}:={\bf 0}$: It is the sign of $\alpha$ that decides the type of singularity we have at ${\bf 0}$. This sign only becomes visible in the fourth   derivatives of $f$ at ${\bf 0}$.
