Intuition behind one second partial derivative test case I want the geometrical interpretation of the following:  

If $f_{xx}f_{yy} < f_{xy}^2$ and $f_{xx}$ has the same sign as $f_{yy}$ at a point, then why is that point a saddle point? Because,in the case that they have the same sign,one would expect that point to be a minimum or maximum point,not a saddle point.

Bear in mind,that i have checked the intuitive explanation that is available in wikipedia,but did not understand it.
Also, i am studying physics and not math,so please don't complicate things with symbolisms that i might not understand.
Also, i want an intuitive answer,because i know the maths behind categorizing critical points.
 A: Consider $f:\mathbb R^2 \to \mathbb R$, twice continuously differentiable in a neighborhood of $0$ with $f(0)=0$ and $f'(0)=0$. (We discuss how the value of $f$ changes near the critical point $0$)
Assume the determinant of Hessian $H=f''(0)$ of $f$ at $0$ is negative. Since $f$ is twice continuously differentiable, $H$ is symmetric. Hence, both eigenvalues $\mu_1 \le \mu_2$ of $H$ are real. From $\det H = \mu_1 \mu_2 < 0$ follows that $\mu_1 < 0 < \mu_2$.
Now, let $u$ be the eigenvector to $\mu_1$. By Taylor theorem we have
$$ f(u) \approx \frac12 u^T H u = \frac12 \mu_1 u^T u < 0 $$
for $u$ sufficiently small.
Similarly, for a sufficiently small eigenvector $v$ to $\mu_2$ we have
$$ f(v) \approx \frac12 v^T H v = \frac12 \mu_2 v^T v > 0. $$
Thus, $0$ is a saddle point of $f$.
A: I found a very good answer at Geometric interpretation of $\frac {\partial^2} {\partial x \partial y} f(x,y)$
Its the last answer by @harvard man
also,a link to see before reading it is: http://www.math.harvard.edu/archive/21a_fall_08/exhibits/fxy/index.html posted by @ user26439
(sorry,i cant tag you guys)  
If any of the other users have anything else to say in the lines of the answer in the link,feel free to do so.
