Why saddle point occurs when discriminant<0 ? is it always the case? Currently I am studying partial derivatives. In the second derivative test   the condition says if discriminant is less than zero there occurs a saddle point, why is this so ? thanks a a lot in advance.
 A: The eigenvectors of the hessian indicate, roughly, the direction for which the function has the highest/lowest second derivative (as seen as a function $f: \mathbb{R} \rightarrow \mathbb{R}$ in that direction). The eigenvalues indicate the value of such second derivative. Since the determinant is the product of eigenvalues, the determinant being negative indicates that there is a direction where it has positive concavity, and a direction where it has negative concavity. Hence, a "saddle" point.
A: When you get critical points of a function $f(x,y)$, you evaluate them in the function $$D(x,y) = f_{xx}(x,y)f_{yy}(x,y) - (f_{xy}(x,y))^2.$$
Notice that the second term is always negative. Now, if the sign of $f_{xx}$ is opposite of $f_{yy}$, the first term is also negative too, being a product of two numbers with opposite signs, and thus $D< 0$. 
Intuitively, we know that the second derivative of a function tells us concavity. The second-order partial $f_{xx}$ tells us the concavity in the $x$-direction, and similarly for $f_{yy}$ in the $y$-direction. 
Imagine that you're standing on the surface at a saddle point. If you were to look in the positive $x$-direction you would be looking up-hill (if we suppose $f_{xx} > 0$ at the saddle point), but if you turn and face the positive $y$-direction, you would be looking down-hill (if we suppose $f_{yy} <0$).
Of course, this is saying that if $f_{xx}$ and $f_{yy}$ are opposite signs, then $D>0$. This isn't quite the conclusion of the second derivative test, but hopefully I appealed to your intuition a little bit. 
