Consider a PDE of the form $$ a(x,y) \frac{\partial^2 u}{\partial x^2} + 2b(x,y) \frac{\partial^2 u}{\partial x \partial y} + c(x,y) \frac{\partial^2 u}{\partial y} + d(x,y) \frac{\partial u}{\partial x} + e(x,y) \frac{\partial u}{\partial y} + f(x,y) u = g(x,y) \hspace{10mm} (*) $$ Let us express this as $$ A \frac{\partial^2 u}{\partial \xi^2} + 2B \frac{\partial^2 u}{\partial \xi \partial \eta} + C \frac{\partial^2 u}{\partial \eta^2} + F(u_{\xi}, u_{\eta}, u, \xi, \eta,) = 0 $$ where $$ \begin{align*} A =& a \left( \frac{\partial \xi}{\partial x} \right)^2 + 2b \frac{\partial \xi}{\partial x} \frac{\partial \xi}{\partial y} + c \left( \frac{\partial \xi}{\partial y} \right)^2 \\ B =& a \frac{\partial \xi}{\partial x} \frac{\partial \eta}{\partial x} + b \left( \frac{\partial \xi}{\partial x} \frac{\partial \eta}{\partial y} + \frac{\partial \xi}{\partial y} \frac{\partial \eta}{\partial x} \right) + c \frac{\partial \xi}{\partial y} \frac{\partial \eta}{\partial y} \\ C =& a \left( \frac{\partial \eta}{\partial x} \right)^2 + 2b \frac{\partial \eta}{\partial x} \frac{\partial \eta}{\partial y} + c \left( \frac{\partial \eta}{\partial y} \right)^2 \end{align*} $$ Now define the variables $D_\xi$ and $D_\eta$ such that $$ D_{\xi} = \frac{\left( \frac{\partial \xi}{\partial x} \right)}{\left( \frac{\partial \xi}{\partial y} \right)} \hspace{10mm} \text{and} \hspace{10mm} D_{\eta} = \frac{\left( \frac{\partial \eta}{\partial x} \right)}{\left( \frac{\partial \eta}{\partial y} \right)} $$ We may substitute these variables into the expressions for the coefficients $A, B$ and $C$ so that they may be expressed as $$ \begin{align*} A =& \left( a D_{\xi}^2 + 2b D_{\xi} + c \right) \left( \frac{\partial \xi}{\partial y} \right)^2 \\ B =& \left( a D_{\xi} D_{\eta} + b (D_{\xi} + D_{\eta}) + c \right) \frac{\partial \xi}{\partial y} \frac{\partial \eta}{\partial y} \\ C =& \left( a D_{\eta}^2 + 2b D_{\eta} + c \right) \left( \frac{\partial \eta}{\partial y} \right)^2 \end{align*} $$ Thus, to simplify our PDE, we must find some values for $D_\xi$ and $D_\eta$ such that some of the coefficients $A, B$ and $C$ become $0$. We find that for this condition to be met, we must solve the quadratic $$ aD^2 + 2bD + c = 0 $$ for $D$. Since the PDE is parabolic, we know that this quadratic has 1 repeated real root, given by $$ D = -\frac{b}{a} $$ which we may set equal to $D_\epsilon$ to make the $A$ coefficient vanish. Thus, we have defined the change of variables $$ D_{\xi} = \frac{\left( \frac{\partial \xi}{\partial x} \right)}{\left( \frac{\partial \xi}{\partial y} \right)} = - \frac{b}{a} $$ so you would think that the family of characteristics for this PDE would be of the form $$ \frac{dy}{dx} = -\frac{b}{a} $$ However, in our lecture notes it says that the characteristic equation is given by $$ \frac{dy}{dx} = -\frac{b}{a} $$ You can see this in the image below
Why has the sign changed?