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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.

  • 0
    see http://mathworld.wolfram.com/Curvature.html2012-09-04
  • 0
    yeah, item (17).2012-09-05

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