How to linearlize level curves at a saddle point

Question

Let $f(x,y)$ be a real-valued function on a domain $D$ in $\mathbb{R}^2$, and let $(x_s, y_s)$ be a saddle point of $f(x,y)$ in $D$. That is to say, \begin{align} \frac{\partial f}{\partial x}(x_s, y_s) =&0, \\ \frac{\partial f}{\partial y}(x_s, y_s) =&0, \end{align} but that $f$ does not take the maximum or minimum value at $(x_s,y_s)$. We can implicitly define the level curves through the saddle point by the condition, \begin{equation} f(x,y) = f(x_s, y_s). \tag{1} \label{eq: 1} \end{equation} Let us write the explicit forms of these level curves as \begin{align} y =& h_i(x), &i=&1,\dots,N, \tag{2} \label{eq: 2} \end{align} where $N$ is the number of the contour lines passing through the saddle point. How can I calculate the derivative, \begin{align} \frac{d h_i}{dx}(x_s,y_s)&,& i=&1,\dots,N, \tag{3} \label{eq: 3} \end{align} i.e., the slopes of the level curves at the saddle point?

I am confused because it looks like a $0/0$ indefinite form. Please note that by taking derivative of the both sides of eq. (\ref{eq: 1}) with respect to $x$, \begin{equation} \frac{\partial f}{\partial x}(x,y) + \frac{\partial f}{\partial y}(x,y) \frac{dy}{dx} = 0, \tag{4} \label{eq: 4} \end{equation} and if $\partial f/\partial y \not=0$, \begin{equation} \frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}(x,y)}{\frac{\partial f}{\partial y}(x,y)}. \tag{5} \label{eq: 5} \end{equation} However, at the saddle point, \begin{align} \frac{\partial f}{\partial x}(x_s,y_s) =& 0, \tag{6} \label{eq: 6} \\ \frac{\partial f}{\partial y}(x_s,y_s) =& 0, \tag{7} \label{eq: 7} \end{align} and therefore it looks that as $(x,y) \rightarrow (x_s,y_s)$, \begin{equation} \frac{dy}{dx} \rightarrow \frac{0}{0}. \tag{8} \label{eq: 8} \end{equation}

For some examples of $f(x,y)$, I can determine the level curves $\{h_i(x)\}$ explicitly, but still have difficulty to calculate $dh_i/dx$ at $(x_s,y_s)$ as it takes the $0/0$ indefinite form.

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EDITED 2016-07-21 07:40 GMT+2 I replaced all occurrences of contour lines by level curves and added definition of a saddle point.

Answer 1

At the saddle point, you need to expand up to the second order and the Taylor development essentially becomes

$$f-f_0=ax^2+2bxy+cy^2,$$ (a parabolic hyperboloid) which you can factor as the product of two lines.