Finding vertex of a $f(x,y) = 0$ parabola A parabola whose axis is oblique to the orthogonal coordinate axes is of the form $f(x,y)= 0$, for example
$$f(x,y) = 9x^2 + 24 xy + 16 y^2 + 22x + 46 y + 9=0.$$
Using algebra only it is airly straightforward to find its apex once you  rearrange  the  equation above to
$$f= (3x+4y+5)^2 = 2(4x-3y+8).$$
The intersection of the tangent to the summit and axis gives $\left(-\frac{47}{25},\frac{4}{25}\right)$.
I would like to get the vertex result using calculus only. I suppose the way to do is to minimize the curvature radius (unless I'm answered a better way). 
How do you extend the formula I know for $\{x,f(x)\}$ curves —
$$R=\frac{(1+f'^2)^{3/2}}{f''}$$
to this $f(x,y)=0$ curve? 
Thanks.
 A: What you want for a level set of a $C^2$ function $F$ is the gradient vector $\nabla F,$ let's take that as a row with $F_i = \frac{\partial F}{\partial x_i}$, then the Laplacian $\Delta F,$ the identity matrix $I,$ and the Hessian matrix with entries $F_{ij} = \frac{\partial^2 F}{\partial x_i \partial x_j}.$ Oh, for the identity matrix we use the Kronecker $\delta$ notation, $\delta_{ij}$ is $1$ if $i=j$ but $0$ if $i \neq j.$ For you $1 \leq i,j \leq 2.$ Also the Laplacian is the trace of the Hessian matrix. Go figure.
Next, define a square matrix $B = \Delta F \cdot I - \mbox{Hess} F,$ so
$$ B_{ij} = \Delta F \cdot \delta_{ij} - F_{ij}.  $$ 
We are in $\mathbb R^n,$ for you $n = 2.$  Then the mean curvature of a level set is $$    \frac{1}{(n-1) \, |\nabla F|^3} \; \; \left( \nabla \; F \cdot \; B \; \cdot \; \nabla F^T \right)       $$ 
Finally, for a curve in the plane the mean curvature is the curvature, with the possible disagreement of a $\pm$ sign, about which nothing can be done anyway. If you prefer, instead of $B$ use $-B.$
This is a slightly revised excerpt from my first article, 1991. I do not know of any book where you would be likely to find this. Personal service, is what I'm saying.
