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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.

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yeah, item (17). –  Will Jagy Sep 5 '12 at 4:43

1 Answer 1

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.

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A great thanks for having taken soo much time for this answer. I am unable yey ttdrafting this anThanks for your long answer –  sigismond kmiecik Sep 5 '12 at 20:32
    
A great thanks for having taken soo much time for this answer. A large part of what you wrote is not within my grasp so I cannot make much of it. However I'm amazed in the contrast between the ease of finding any tangents to this conic using partial derivatives on a given point and the sophistication –  sigismond kmiecik Sep 5 '12 at 20:46
    
behind finding its vertex. –  sigismond kmiecik Sep 5 '12 at 20:47

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